1 partial di erential equations(p.d.e.) · 2019-08-15 · 1 partial di erential equations(p.d.e.)...

1 Partial Differential Equations(P.D.E.)

Formation of partial differential equations - Lagrange’s Linearequation Solution of standard types of first order partialdifferential equations - Linear partial differential equations ofsecond and higher order with constant coefficients.

1.1 Introduction

In a differential equation if there are two or more independent variablesand the derivatives are partial derivatives then it is called a partialdifferential equation .

Examples:

1.

(∂2z

∂x2

)2

+

(∂2z

∂y2

)3

= xy (z− dependent variable; x and y−

independent variables)

2. x∂z

∂x+ y

∂z

∂y+ t

∂z

∂t= xyt (z− dep. variable; x, y and t− independent

variables)

The order of a partial differential equation if the order of the highestpartial derivative occurring in the equation.

The degree of a partial differential equation is the greatest exponent ofthe highest order.

In the above, e.g.1 is a second order & third degree equation, e.g.2 is afirst order equation & first degree equation.

Note : Throughout this chapter we use the following notations; z will betaken as a dependent variable which depends on two independent variablesx and y so that z = f(x, y).We write

∂z

∂x= p,

∂z

∂y= q,

∂2z

∂x2= r,

∂2z

∂x∂y=

∂2z

∂y∂x= s and

∂2z

∂y2= t.

Thus p2 + q2 = x2 + y2 is a partial differential equation of order 1 andr + xt = s2 − p is a partial differential equation of order 2.

1

2 Unit I - PARTIAL DIFFERENTIAL EQUATIONS (P.D.E.)

1.2 Formation of Partial Differential Equation by Elimination ofArbitrary Constants

Let f(x, y, z, a, b) = 0. (1)be an equation which contains two arbitrary constants ‘a’ and ‘b’.

Partially differentiating (1) with respect to ‘x’ and ‘y’ we get two moreequations.

Eliminating a and b from these three equations, we get φ(x, y, z, p, q) = 0which is a partial differential equation of order 1.

In this case the number of arbitrary constants to be eliminated is equalto the number of independent variables and we obtain a first order partialdifferential equation. If the number of arbitrary constants to be eliminatedis more than the number of independent variables, we get partial differentialequations of second or higher order.

1.2.1 Examples of Formation of P.D.E. by Elimination of Arbitrary Constants

Example 1.1. Form the partial differential equation by eliminating thearbitrary constants from z = ax+ by + a2 + b2.

Solution: Given z = ax+ by + a2 + b2 (1)Differentiating (1) partially w.r.t ‘x’

∂z

∂x= a i.e., p = a (2)

Differentiating (1) partially w.r.t ‘y’

∂z

∂y= b i.e., q = b (3)

From (2) and (3)

a = p and b = q

Substituting in (1), we get z = px+ qy + p2 + q2.

Example 1.2. Eliminating the arbitrary constants a and b from z = a2x+b2y + ab.

Solution: Given z = a2x+ b2y + ab. (1)Differentiating (1) partially w.r.t ‘x’

∂z

∂x= a2 i.e., p = a2 (2)


MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 3

∂z

∂y= b2 i.e., q = b2 (3)

From (2) and (3)

a2 = p and b2 = q

Substituting in (1), we get z = px+ qy +√pq.

Example 1.3. Form the PDE by eliminating the arbitrary constants a andb from z = (x+ a)(y + b).

Solution: Given z = (x+ a)(y + b) (1)Differentiating (1) partially w.r.t ‘x’

∂z

∂x= y + b i.e., p = y + b (2)


∂z

∂y= x+ a i.e., q = x+ a (3)

From (2) and (3)

x+ a = q and y + b = p

Substituting in (1), we get z = pq.

Example 1.4. Form the PDE by eliminating a, b from u = (x2+ a2)(y2+b2).

Solution: Given u = (x2 + a2)(y2 + b2) (1)Differentiating (1) partially w.r.t ‘x’

∂z

∂x= 2x(y2 + b2) i.e., p = 2x(y2 + b2) (2)


∂z

∂y= 2y(x2 + y2) i.e., q = 2y(x2 + y2) (3)

From (2) and (3)

x2 + a2 =q

2yand y2 + b2 =

p

2x

Substituting in (1), we get 4xyu = pq.

Example 1.5. Find the differential equation of all sphere whose centreslie on the z − axis. [UQ]


Solution: Any point on the z− axis is of the form (0, 0, c). The equationof the sphere with centre (0, 0, c) and radius ‘a’ is given by

(x− 0)2 + (y − 0)2 + (z − c)2 = a2 (1)

Differentiating (1) partially w.r.t ‘x’

2x+ 2(z − c)∂z

∂x= 0 i.e., x+ p(z − c) = 0 (2)


2y + 2(z − c)∂z

∂y= 0 i.e., y + q(z − c) = 0 (3)

From (2) and (3)

z − c =−xp

and z − c =−yq

⇒ −xp

=−yq

∴ qx = py.

Example 1.6. Form the PDE from x2 + y2 + (z − c)2 = a2 [UQ]

Solution: Similar to above Example.

Example 1.7. Find the differential equation of all spheres of fixed radiushaving their centres in the xy − plane. [UQ]

Solution: Any point in the xy−plane is of the form (a, b, 0). The equationof the sphere with center (a, b, 0) and radius r is given by (x− a)2 + (y −b)2 + (z − 0)2 = r2. i.e., (x− a)2 + (y − b)2 + z2 = r2. (1)Here a, b are the two arbitrary constants and r is a given constant.Differentiating (1) partially w.r.t ‘x’

2(x− a) + 2z∂z

∂x= 0 i.e., x− a+ pz = 0 (2)


2(y − b) + 2z∂z

∂y= 0

∂z

∂y= 2y(x2 + y2) i.e., y − b+ qz = 0 (3)

From (2) and (3)

x− a = −pz and y − b = −qz

Substituting in (1), we get z2(p2 + q2 + 1) = r2.

Example 1.8. Form the partial differential equation by eliminating thearbitrary constants a and b from (x− a)2 + (y − b)2 + z2 = 1.



Example 1.9. Form the partial differential equation by eliminating thearbitrary constants a and b from log(az − 1) = x+ ay + b.

Solution: Given log(az − 1) = x+ ay + b. (1)Differentiating (1) partially w.r.t ‘x’

1

az − 1.a∂z

∂x= 1 i.e.,

ap

az − 1= 1 (2)


1

az − 1.a∂z

∂y= a i.e.,

q

az − 1= 1 (3)

From (2) ap = az − 1 ⇒ a =1

z − p

From (3) q = az − 1 ⇒ z

z − p− 1

(∵ a =

1

z − p

)(z − p)q = z − z + p

zq = p(1 + q)

Example 1.10. Form the partial differential equation by eliminating a andb from (x− a)2 + (y − b)2 = z2cot2α , where is a constant. [UQ]

Solution: Given (x− a)2 + (y − b)2 = z2cot2α (1)Differentiating (1) partially w.r.t ‘x’

2(x− a) = 2z∂z

∂xcot2α i.e., x− a = pzcot2α (2)


2(y − b) = 2z∂z

∂ycot2α i.e., y − b = qzcot2α (3)

Sub. (2) and (3) in (1), we get

p2z2cot4α + q2z2cot4α = z2cot2α

cot2α(p2 + q2) = 1

p2 + q2 = tan2α

Example 1.11. Form the partial differential equation by eliminating the

arbitrary constants from 2z =x2

a2+y2

b2

Solution: Given 2z =x2

a2+y2

b2(1)



2∂z

∂x=

2x

a2i.e., p =

x

a2(2)


2∂z

∂y=

2y

b2i.e., q =

x

b2(3)

From (2) and (3), we get a2 =x

pand b2 =

y

qSubstituting in (1), we get

2z = p2 + q2


arbitrary constants a and b from z = axey +1

2a2e2y + b.

Solution: Given z = axey +1

2a2e2y + b (1)


∂z

∂x= aey i.e., p = aey (2)


∂z

∂y= axey +

1

2a2e2y.2 i.e., q = axey + a2e2y (3)

Sub. (2) in (3), we get

q = px+ p2


arbitrary constants a and b from z = xy + y√x2 + a2 + b

Solution: Given z = xy + y√x2 + a2 + b (1)


∂z

∂x= y + y

2x

2√x2 + a2

i.e., p = y +xy√x2 + a2

(2)


∂z

∂y= x+

√x2 + a2 i.e., q = x+

√x2 + a2 (3)

From (3), we get√x2 + a2 = q − x

Substituting in (2), we get


p = y +xy

q − x

p =qy

q − x

i.e., pq − px = qy

i.e., px+ qy = pq

Example 1.14. Form the partial differential equation of all spheres whoseradii are the same.

Solution: The equation of all sphere with equal radius can be taken as(x− a)2 + (y − b)2 + (z − c)2 = r2 (1)where a, b, c are arbitrary constants and r is a given constant. Since thenumber of arbitrary constants is more than the number of independentvariables, we will get the p.d.e. of order greater than 1.Differentiating (1) partially w.r.t ‘x’

2(x− a) + 2(z − c)∂z

∂x= 0 i.e., i.e., (x− a) + (z − c)p = 0 (2)


2(y − b) + 2(z − c)∂z

∂y= 0 i.e., (y − b) + (z − c)q = 0 (3)


1 + (z − c)∂2z

∂x2+∂z

∂x.∂z

∂x= 0 i.e., 1 + (z − c)r + p2 = 0 (4)


1 + (z − c)∂2z

∂y2+∂z

∂y.∂z

∂y= 0 i.e., 1 + (z − c)t+ q2 = 0 (5)

From (4) and (5), we get z − c =−p2 − 1

rand z − c =

−q2 − 1

t

i.e., z − c =−p2 − 1

r=

−q2 − 1

tt(p2 + 1) = r(q2 + 1)

Example 1.15. Obtain the partial differential equation by eliminating

a, b, c fromx2

a2+y2

b2+z2

c2= 1

Solution: Since the number of arbitrary constants is more than thenumber of independent variables, we will get the p.d.e of order greaterthan ’1’.

Givenx2

a2+y2

b2+z2

c2= 1 (1)



2x

a2+

2z

c2∂z

∂x= 0 i.e.,

x

a2+z

c2p = 0 (2)


2y

b2+

2z

c2∂z

∂y= 0 i.e.,

y

b2+z

c2q = 0 (3)

Differentiating (2) partially w.r.t. ‘x’1

a2+

1

c2

(z∂2z

∂x2+∂z

∂x.∂z

∂x

)= 0 i.e.,

1

a2+

1

c2(zr2 + p2

)= 0 (4)

Differentiating (3) partially w.r.t ‘y’1

b2+

1

c2

(z∂2z

∂y2+∂z

∂y.∂z

∂y

)= 0 i.e.,

1

b2+

1

c2(zt2 + q2

)= 0 (5)


0 +1

c2

(z∂2z

∂y∂x+∂z

∂x

∂z

∂y

)= 0 i.e.,

1

c2(zs+ pq) = 0 (5)

i.e., zs+ pq = 0

Note : Wemay also get different partial differential equations. The answeris not unique.

1.3 Formation of partial differential equation by elimination ofarbitrary Functions

Note: The elimination of one arbitrary function from a given relationgives a partial differential equation of first order while elimination of twoarbitrary function from a given relation gives a second or higher orderpartial differential equation.

1.3.1 Examples of Formation of P.D.E. by Elimination of Arbitrary Functions

Example 1.16. Eliminate f from z = f(x2 − y2).

Solution: Given z = f(x2 − y2) (1)Differentiating (1) partially w.r.t ‘x’

∂z

∂x= f ′(x2 − y2).2x i.e., p = 2xf ′ (2)


∂z

∂y= f ′(x2 − y2)(−2y) i.e., q = −2yf ′ (3)

From (2) and (3), we get f ′ =p

2xand f ′ =

−q2y

p

2x=

−q2y


i.e., py + qx = 0

Example 1.17. Eliminate f from z = xy + f(x2 + y2).

Solution: Given z = xy + f(x2 + y2). (1)Differentiating (1) partially w.r.t ‘x’

∂z

∂x= y + f ′(x2 + y2).2x i.e., p = y + 2xf ′ (2)


∂z

∂y= x+ f ′(x2 + y2).2y i.e., q = x+ 2yf ′ (3)

From (2) and (3),we get f ′ =p− y

2xand f ′ =

q − x

2y

p− y

2x=q − x

2y

i.e., py − qx = y2 − x2

Example 1.18. Eliminate f from z = x+ y + f(xy).

Solution: Given z = x+ y + f(xy) (1)Differentiating (1) partially w.r.t ‘x’

∂z

∂x= 1 + f ′(xy).y i.e., p = 1 + yf ′ (2)


∂z

∂y= 1 + f ′(xy).x i.e., q = 1 + xf ′ (3)

From (2) and (3), we get f ′ =p− 1

yand f ′ =

q − 1

x

p− 1

y=q − 1

x

i.e., px− qy = x− y

Example 1.19. Obtain PDE by eliminating the arbitrary function f fromxyz = φ(x+ y + z). [UQ]

Solution: Given xyz = φ(x+ y + z) (1)Differentiating (1) partially w.r.t ‘x’

y

(x∂z

∂x+ z

)= φ′(x+ y + z)

(1 +

∂z

∂x

)i.e., y(xp+ z) = (1 + p)φ′ (2)


x

(y∂z

∂y+ z

)= φ′(x+ y + z)

(1 +

∂z

∂y

)i.e., x(yq + z) = (1 + q)φ′ (3)

From (2) and (3), we get


φ′ =y(xp+ z)

1 + pand φ′ =

x(yq + z)

1 + q

y(xp+ z)

1 + p=x(yq + z)

1 + qi.e., y(xp+ z)(1 + q) = x(yq + z)(1 + p).

Example 1.20. Eliminate the arbitrary function f from z = f(yx

).

Solution: Given z = f(yx

)(1)


∂z

∂x= f ′

(yx

).

(−yx2

)i.e., p =

−yx2f ′ (2)


∂z

∂y= f ′

(yx

).1

xi.e., q =

1

xf ′ (3)

From (2) and (3),we get f ′ =−px2

yand f ′ = qx

−px2

y= qx

i.e., px+ qy = 0

Example 1.21. Form a partial differential equation by eliminating f from

z = xf

(x

y

). [UQ]

Solution: Given z = xf

(x

y

)(1)


∂z

∂x= xf ′

(x

y

).1

y+ f

(x

y

)i.e., p =

x

yf ′ + f (2)


∂z

∂y= xf ′

(x

y

).

(−xy2

)i.e., q =

−x2

y2f ′ (3)

From (1), f =z

xFrom (3), f ′ =

−qy2

x2

Substituting these in (2), we get p =x

y

(−qy2

x2

)+z

x⇒ p = −qy

x+z

x

i.e., px+ qy = z

Example 1.22. Form PDE by eliminating arbitrary function f from

z = f(xyz

). [UQ]


Solution: Given z = f(xyz

)(1)


∂z

∂x= f ′

(xyz

)(zy − xy ∂z∂x

z2

)i.e., p =

(zy − xyp

z2

)f ′ (2)


∂z

∂y= f ′

(xyz

)(zx− xy ∂z∂y

z2

)i.e., q =

(zx− xyq

z2

)f ′ (3)


f ′ =pz2

yz − xypand f ′ =

qz2

xz − xyp

pz2

yz − xyp=

qz2

xz − xyppxz − pqxy = qyz − pqxy

i.e., px = qy.

Example 1.23. Form a partial differential equation by eliminating f from

z = x2 + 2f

(1

y+ logx

).

Solution: Given z = x2 + 2f

(1

y+ logx

)(1)


∂z

∂x= 2x+ 2f ′

(1

y+ log x

).1

xi.e., p = 2x+

2

xf ′ (2)


∂z

∂y= 2f ′

(1

y+ log x

)(−1

y2

)i.e., q =

−2

y2f ′ (3)


f ′ =x

2(p− 2x) and f ′

−qy2

2

x

2(p− 2x) =

−qy2

2i.e., px− 2x2 = −qy2.i.e., px+ qy2 = 2x2.

Example 1.24. Form PDE by eliminating arbitrary function f and g fromz = f(x+ ay) + g(x− ay).

Solution: Given z = f(x+ ay) + g(x− ay) (1)Differentiating (1) partially w.r.t ‘x’

∂z

∂x= f ′ (x+ ay) + g′ (x− ay) i.e., p = f ′ + g′ (2)



∂z

∂y= af ′(x+ ay)− ag′(x− ay) i.e., q = af ′ − ag′ (3)


∂2z

∂x2= f ′′ + g′′ i.e., r = f ′′ + g′′ (4)


∂2z

∂y2= a2f ′′ + a2g′′ i.e., t = a2 (f ′′ + g′′) (5)

Using (4), we get t = a2r.

Example 1.25. Form partial differential equation by eliminating arbitraryfunction f and g from z = f(2x+ y) + g(3x− y). [UQ]

Solution: Given z = f(2x+ y) + g(3x− y) (1)Differentiating (1) partially w.r.t ‘x’

∂z

∂x= 2f ′ (2x+ y) + 3g′ (3x− y) i.e., p = 2f ′ + 3g′ (2)


∂z

∂y= f ′(2x+ y)− g′(3x− y) i.e., q = f ′ − g′ (3)


∂2z

∂x2= 4f ′′(2x+ y) + 9g′′(3x− y) i.e., r = 4f ′′ + 9g′′ (4)


∂2z

∂y2= f ′′(2x+ y) + g′′(3x− y) i.e., t = f ′′ + g′′ (5)


∂2z

∂x∂y= 2f ′′(2x+ y)− 3g′′(3x− y) i.e., s = 2f ′′ − 3g′′ (6)

Solving (4) and (5), we get f ′′ =1

5(qt− r) and g′′ =

1

5(r − 4t)

Sub. the values of f ′′ and g′′ in (6), we get

s =2

5(9t− r)− 3

5(r − 4t)

5s = 30t− 5r

s = 6t− r


Example 1.26. Form partial differential equation by eliminating arbitraryfunction f and g from z = f(x+ 2y) + g(2x− y).

Solution: Given z = f(x+ 2y) + g(2x− y) (1)Differentiating (1) partially w.r.t ‘x’

p = f ′ + 2g′ (2)


q = 2f ′ − g′ (3)

Differentiating (2) partially w.r.t ‘x’ : r = f ′′ + 4g′′ (4)Differentiating (3) partially w.r.t ‘y’ : t = 4f ′′ + g′′ (5)Differentiating (2) partially w.r.t ‘y’ : s = 2f ′′ − 2g′′ (6)

Solving (4) and (5), we get f ′′ =4t− r

15, g′′ =

4r − t

15Sub. the values of f ′′ and g′′ in (6), we get

s =2

15(4t− r)− 2

15(4r − t)

15s = 10t− 10r

3s+ 2r = 2t

Example 1.27. Eliminate the arbitrary function φ and fromz = xφ(y) + y(x) and form the partial differential equation.

Solution: Given z = xφ(y) + y(x) (1)Differentiating (1) partially w.r.t ‘x’

p = φ(y) + yϕ′(x) (2)


q = xφ′(y) + ϕ(x) (3)


s = φ′(y) + ϕ′(x) (4)

Now, px+ qy = xφ(y) + xyϕ′(x) + xyφ′(y) + yϕ(x)

= xφ(y) + yϕ(x) + xy(φ′(y) + ϕ′(x))

px+ qy = z + xys [using (1) and (4)]

Example 1.28. Form partial differential equation by eliminating arbitraryfunction f and g from z = xf(3x+ 2y) + g(3x+ 2y).


Solution: Given z = xf(3x+ 2y) + g(3x+ 2y) (1)Differentiating (1) partially w.r.t ‘x’

p = 3xf ′ + f + 3g′ (2)


q = 2xf ′ + 2g′ (3)


r = 3(x.3f ′′ + f ′) + 3f ′ + 9g′′r = 9xf ′′ + 9g′′ + 6f ′ (4)

Differentiating (3) partially w.r.t ‘y’ : t = 4xf ′′ + 4g′′ (5)Differentiating (2) partially w.r.t ‘y’ : s = 6xf ′′ + 2f ′ + 6g′′ (6)(4)− 3× (6) ⇒ r − 3s = 9xf ′′ + 9g′′ + 6f ′ − 18xf ′′ − 6f ′ − 18g′′

= −9xf ′′ − 9g′′

r − 3s = −9(xf ′′ + g′′) = −9

(t

4

)[using(5)]

4r − 12s+ 9t = 0

Example 1.29. Form partial differential equation by eliminating arbitrary

function f and g from z = xf(yx

)+ yφ(x). [UQ]

Solution: Given z = xf(yx

)+ yφ(x) (1)


∂z

∂x= xf ′

(yx

)(−yx2

)+ f

(yx

)+ yφ′(x)

∂z

∂x=

−yxf ′(yx

)+ f

(yx

)+ yφ′(x)

i.e., p =−yxf ′ + f + yφ′ (2)


∂z

∂y= xf ′

(yx

).1

x+ φ(x) ⇒ ∂z

∂y= f ′

(yx

)+ φ(x)

i.e., q = f ′ + φ (3)

Now,∂2z

∂x∂y= f ′′

(yx

).

(−yx2

)+ φ′(x) i.e., s =

−yx2f ′′ + φ′ (4)

∂2z

∂y2= f ′′

(yx

).1

xi.e., t =

1

xf ′′ (5)


∴ px+ qy = −yf ′ + xf + xyφ′ + yf ′ + yφ [using (2) and (3)]

px+ qy = xf + yφ+ xyφ′

From (1), px+ qy = z + xyφ′ (6)

Using (5) in (4), s =−yx2

(xt) + φ′ ⇒ φ′ = s+yt

xSub. the value of φ′ in (6), we get

px+ qy = z + xy

(s+

yt

x

)px+ qy = z + xys+ y2t

Example 1.30. Form a p.d.e by eliminating arbitrary function f and gfrom z = f(x+ y)g(x− y).

Solution: Given z = f(x+ y)g(x− y) i.e., z = fg (1)Differentiating (1) partially w.r.t ‘x’

∂z

∂x= fg′ + gf ′ i.e., p = fg′ + gf ′ (2)


∂z

∂y= f (−g′) + gf ′ i.e., q = gf ′ − fg′ (3)

∂2z

∂x2= fg′′ + g′f ′ + gf ′′ + f ′g′ i.e., r = fg′′ + gf ′′ + 2f ′g′ (4)

∂2z

∂y2= −[f(−g′′) + g′f ′] + gf ′′ + f ′(−g′)

t = fg′′ + gf ′′ − 2f ′g′ (5)


p2 − q2 = f 2g′2 + g2f ′2 + 2fgf ′g′ − (g2f ′2 + f 2g′2 − 2fgf ′g′)

= 4fgf ′g′

p2 − q2 = 4zf ′g′ [using(1)] (6)


r − t = 4f ′g′ (7)


p2 − q2 = z(r − t)


1.3.2 Formation of partial differential equation by elimination of arbitrary functionφ from φ(u, v) = 0, where u and v are function of x, y and z.

Let (u, v) = 0 (1)Differentiating (1) partially w.r.t x and y, we get

∂φ

∂u

∂u

∂x+∂φ

∂v

∂v

∂x= 0 (2)

∂φ

∂u

∂u

∂y+∂φ

∂v

∂v

∂y= 0 (3)

To eliminate φ, it is enough to eliminate∂φ

∂uand

∂φ

∂v

From (2) and (3), eliminating∂φ

∂uand

∂φ

∂vFrom (2) and (3), we get ∣∣∣∣∣∣∣

∂u

∂x

∂v

∂x∂u

∂y

∂v

∂y

∣∣∣∣∣∣∣ = 0

Example 1.31. Form the partial differential equation by eliminating thearbitrary function φ from the relation φ(x2 + y2 + z2, xyz) = 0

Solution: Given φ(x2 + y2 + z2, xyz) = 0 (1)Let u = x2 + y2 + z2, v = xyz

Equation (1) becomes φ(u, v) = 0 (2)Eliminating φ from (2), we get∣∣∣∣∣∣∣

∂u

∂x

∂v

∂x∂u

∂y

∂v

∂y

∣∣∣∣∣∣∣ = 0 (3)

Here,∂u

∂x= 2x+ 2zp,

∂u

∂y= 2y + 2zq

∂v

∂x= xyp+ yz,

∂v

∂y= xyq + xz

From (3),

∣∣∣∣ 2x+ 2zp xyp+ yz2y + 2zq xyq + xz

∣∣∣∣ = 0

i.e., (2x+ 2zp)(xyq + xz)− (xyp+ yz)(2y + 2zq) = 0

i.e., pxz2 − y2 − qy(z2 − y2) = z(y2 − x2)

Example 1.32. Form the partial differential equation by eliminating thearbitrary function from f(x+ y + z, xy + z2) = 0. [UQ]


Solution: Given f(x+ y + z, xy + z2) = 0 (1)Let u = x+ y + z, v = xy + z2

Equation (1) becomes φ(u, v) = 0 (2) Eliminating φ from (2), we get∣∣∣∣∣∣∣∂u

∂x

∂v

∂x∂u

∂y

∂v

∂y

∣∣∣∣∣∣∣ = 0 (3)

Here,∂u

∂x= 1 + p,

∂u

∂y= 1 + q

∂v

∂x= y + 2zp,

∂v

∂y= x+ 2zq

From(3), ∣∣∣∣ 1 + p y + 2zp1 + q y + 2zq

∣∣∣∣ = 0

i.e., (1 + p)(x+ 2zq)− (1 + q)(y + 2zp) = 0

i.e., (x− 2z)p− (y − 2z)q = y − x


arbitrary function g from the relation g(yx, x2 + y2 + z2

)= 0

Solution: Given g(yx, x2 + y2 + z2

)= 0 (1)

Let u =y

x, v = x2 + y2 + z2

Equation (1) becomes g(u, v) = 0 (2)Eliminating φ from (2), we get∣∣∣∣∣∣∣

∂u

∂x

∂v

∂x∂u

∂y

∂v

∂y

∣∣∣∣∣∣∣ = 0 (3)

∂u

∂x=

−yx2,∂u

∂y=

1

x∂v

∂x= 2x+ 2zp

∂v

∂y= 2y + 2zq

From (3),

∣∣∣∣∣∣∣−yx2

2x+ 2zp

1

x2y + 2zq

∣∣∣∣∣∣∣ = 0

i.e.,−yx2

(2y + 2zq)− 1

x(2x+ 2zp) = 0

i.e., xzp+ yzq + x2 + y2 = 0


1.3.3 Exercises & Solutions

Find the partial differential equations by eliminating the arbitraryconstants a, b&c as the case may be:

1. z = ax+ by [Ans : z = px+ qy]

2. z = a sin by [Ans: r + t = 0]

3. z = axy + b [Ans: px− qy = 0]

4. z = ax+ a2y2 + b [Ans: q = 2yp2]

5. z = ax4 + by4 [Ans: px+ qy = 4z]

6. ax+ by + cz = 1 [Ans: r = 0 (or) s = 0 (or) t = 0]

7. z = ax+ (1− a)y + b [Ans: p+ q = 1]

8. z = (x− a)2 + (y − b)2 + 1 [Ans: z =(p2

)2+(q2

)2+ 1]

Find the partial differential equations by eliminating the arbitraryconstants from the following:

1. z = f(x2 − y2) [Ans: py + qx = 0]

2. z = f(x+ iy) + g(x− iy) [Ans: r + t = 0]

3. xy + yz + zx = f(z

x+ y)

[Ans: p(x+ y)(x+ 2z)− q(x+ y)(y + 2z) = z(x− y)]

4. z = xf(x+ t) + g(x+ t) [Ans: r = 2s− t]

5. z = x2f(y) + y2g(x) [Ans: xyr = 2(px+ qy − 2z)]

6. z = f(x− y) [Ans: p+ q = 0]

7. z = y2 + 2f(+logy) [Ans: px2 + qy = 2y2]

8. z = x+ yf(x2 − y2) [Ans: z = xq + yp]

9. f(xy + z2, x+ y + z) = 0 [Ans: (2z − x)p− (2z − y)q = x− y]

10. φ = (z2 − xy, xz ) = 0 [Ans: px2 − q(xy − 2z2) = zx]

11. z = f(ax+ by) + g(αx+ βy) [Ans: bβr − (aβ + bα)s+ aαt = 0]


1.4 Lagrange’s Linear Equations

The equation of the form Pp+Qq = R is known as Lagrange’s equation,where P,Q and R are functions of x, y and z.

To solve this equation it is enough to solve the subsidiary equations[(or) Lagrange’s auxiliary equations]

dx

P=dy

Q=dz

Z(1)

If u = a and v = b are two solutions of (1) then the solution of the givenLagrange’s equation is φ(u, v) = 0. Generally, the subsidiary equationscan be solved in two ways1. Method of grouping 2. Method of multipliers

1.4.1 Method of grouping

In the subsidiary equations

dx

P=dy

Q=dz

Z

if the variables can be separated in any pair of equations, then we get asolution of the form u = a and v = b.


1.4.2 Method of Multipliers

1. Single Multiplier (l,m, n):Choose any three multipliers l,m, n may be constants or functions ofx, y, z we have

dx

P=dy

Q=dz

R=ldx+mdy + ndz

lP +mQ+ nRIf it is possible to choose l,m, n such that lP +mQ+ nR = 0 then

ldx+mdy + ndz = 0.

If ldx + mdy + ndz is a perfect differential of some function, sayu(x, y, z), then du = 0∴ u = a is a solution.

* Note : If lP +mQ + nR 6= 0 for any (l,m, n), then go for DoubleMultipliers as follows

2. Double Multipliers (l1,m1, n1), (l2,m2, n2):Choose Double Multipliers (l1,m1, n1), (l2,m2, n2) may be constants orfunctions of x, y, z we have

dx

P=dy

Q=dz

R=l1dx+m1dy + n1dz

l1P +m1Q+ n1R=l2dx+m2dy + n2dz

l2P +m2Q+ n2RIntegrating both sides of the equation,

l1dx+m1dy + n1dz

l1P +m1Q+ n1R=l2dx+m2dy + n2dz

l2P +m2Q+ n2Rwe get u = 0∴ u = a is a solution.

1.4.3 Working Rule to solve the equation Pp + Qq = R

(i) Form the subsidiary equationsdx

P=dy

Q=dz

R

(ii) Solve the subsidiary equations by the method of grouping or themethod of multipliers or both to get two independent solutions u = a

and v = b.

(iii) The general solution is φ(u, v) = 0

1.5 Examples of Lagrange’s Linear Equations

1.5.1 Examples of Lagrange’s Linear Equations by Method of grouping

Example 1.34. Solve px+ qy = z.


Solution: Given px+ qy = z

This equation is of the form Pp+Qq = R, then P = x,Q = y and R = z.The subsidiary equations are

dx

P=dy

Q=dz

R

Takingdx

x=dy

yIntegrating, we get logx = logy + loga

i.e.,x

y= a

Similarly fromdy

y=dz

z, we get

y

z= b

Hence the general solution is φ(x

y,y

z) = 0

Example 1.35. Find the general solution of px+ qy = 0. [UQ]

Solution: Given px+ qy = 0This equation is of the form Pp+Qq = R, then P = x,Q = y and R = 0.The subsidiary equations are

dx

x=dy

y=dz

0

Takingdx

x=dy

yIntegrating, we get logx = logy + loga

i.e.,x

y= a

Takingdy

y=dz

0, we get

dz = 0

Integrating, we get z = b

Hence the general solution is φ(x

y, z) = 0.

Example 1.36. Solvey2z

xp+ xzq = y2. [UQ]


Solution: Giveny2z

xp+ xzq = y2

i.e., y2zp+ x2zq = xy2

The subsidiary equations are

dx

y2z=

dy

x2z=

dz

xy2

Takingdx

y2z=

dy

x2z, we get

dx

y2=dy

x2

i.e., x2dx = y2dy

Integrating, we getx3

3=y3

3+ c1

i.e., x3 − y3 = a

Takingdx

y2z=

dz

xy2, we get

dx

z=dz

xi.e., xdx = zdz


2=z2

2+ c2

i.e., x2 − z2 = b

∴ The general solution is φ(x3 − y3, x2 − z2) = 0.

Example 1.37. Solve yzp+ xzq = xy. [UQ]

Solution: Given yzp+ xzq = xyThe subsidiary equations are

dx

yz=dy

xz=dz

xy

Takingdx

yz=dy

xz, we get

dx

y=dy

x

i.e., xdx = ydy


2=y2

2+ c1

i.e., x2 − y2 = a

Takingdx

yz=dz

xy, we get


xdx = zdz


2=z2

2+ c2

i.e., x2 − z2 = b

∴ The general solution is φ(x2 − y2, x2 − z2) = 0.

Example 1.38. Solve x2p+ y2q = z2.

Solution: Given x2p+ y2q = z2

The subsidiary equations are

dx

x2=dy

y2=dz

z2

Takingdx

x2=dy

y2

Integrating, we get −1

x= −1

y+ c1

i.e.,1

x− 1

y= a

Takingdx

x2=dz

z2Integrating, we get

−1

x= −1

z+ c2

i.e.,1

x− 1

z= b

∴ The general solution is φ(1

x− 1

y,1

x− 1

z) = 0.

Example 1.39. Solve pcotx+ qcoty = cotz.

Solution: Given pcotx+ qcoty = cotzThe subsidiary equations are

dx

cotx=

dy

cot y=

dz

cot z

Takingdx

cotx=

dy

cot y, we have

tanxdx = tan ydyIntegrating, logsecx = logsecy + loga

i.e.,secx

sec y= a

Similarly takingsec y

sec z= b

∴ The general solution is φ

(secx

sec y,sec y

sec z

)= 0.


1.5.2 Examples of Lagrange’s Linear Equations by Method of multipliers

Example 1.40. Solve (y − z)p+ (z − x)q = x− y.

Solution: Given (y − z)p+ (z − x)q = x− y

The subsidiary equations aredx

y − z=

dy

z − x=

dz

x− y(1)

Each fraction of (1) =dx+ dy + dz

y − z + z − x+ x− y=dx+ dy + dz

0

⇒ dx+ dy + dz = 0Integrating, x+ y + z = aUsing x, y, z as multipliers,

each fraction of (1) =xdx+ ydy + zdz

x (y−z)+y (z−x)+z (x−y)=xdx+ ydy + zdz

0

⇒ xdx+ ydy + zdz = 0

Integrating,x2

2+y2

2+z2

2= c

i.e., x2 + y2 + z2 = b

∴ The general solution is f(x+ y + z, x2 + y2 + z2

)= 0.

Example 1.41. Solve (mz − ny)∂z

∂x+ (nx− lz)

∂z

∂y= ly −mx. [UQ]

Solution: Given (mz − ny) p+ (nx− lz) q = ly −mx


mz − ny=

dy

nx− lz=

dz

ly −mx(1)

Using x, y, z as multipliers, each fraction of (1)=

xdx+ ydy + zdz

x (mz − ny) + y (nx− lz) + z (ly −mx)=xdx+ ydy + zdz

0



2+y2

2+z2

2= c1

i.e., x2 + y2 + z2 = aUsing l,m, n as multipliers, each fraction of (1) =

ldx+mdy + ndz

l (mz − ny) +m (nx− lz) + n (ly −mx)=ldx+mdy + ndz

0

ldx+mdy + ndz = 0

Integrating, lx+my + nz = b∴ The general solution is φ(x2 + y2 + z2, lx+my + nz) = 0.

Example 1.42. Solve x(y2 − z2

)p+ y

(z2 − x2

)q = z

(x2 − y2

). [UQ]


Solution: Given x(y2 − z2

)p+ y

(z2 − x2

)q = z

(x2 − y2

)The subsidiary equations are

dx

x (y2 − z2)=

dy

y (z2 − x2)=

dz

z (x2 − y2)

Using x, y, z as multipliers, each fraction of (1)=

xdx+ ydy + zdz

x2 (y2 − z2) + y2 (z2 − x2) + z2 (x2 − y2)=xdx+ ydy + zdz

0



2+y2

2+z2

2= c1

i.e., x2 + y2 + z2 = a

Using1

x,1

y,1

zas multipliers, each fraction of (1) =

lxdx+

1ydy +

1zdz

(y2 − z2) + (z2 − x2) + (x2 − y2)=

lxdx+

1ydy +

1zdz

0l

xdx+

1

ydy +

1

zdz = 0

Integrating, log x+ log y + log z = log b

i.e., xyz = b

∴ The general solution is φ(x2 + y2 + z2, xyz) = 0.

Example 1.43. Solve (y + z)p+ (x+ z)q = x+ y. [UQ]

Solution: Given (y + z)p+ (x+ z)q = x+ y


y + z=

dy

x+ z=

dz

x+ y(1)


2 (x+ y + z)=

dx− dy

− (x− y)=

dy − dz

− (y − z)

d(x− y)

(x− y)=d(y − z)

(y − z)Integrating, we get log(x− y) = log(y − z) + log a

i.e.,x− y

y − z= a

Takingdx+ dy + dz

2 (x+ y + z)=

dx− dy

− (x− y),we have

1

2

d(x+ y + z)

(x+ y + z)= −d(x− y)

(x− y)


Integrating, we get1

2log(x+ y + z) = − log(x− y) + log b√

(x+ y + z)(x− y) = b

The general solution is φ

(x− y

y − z,√

(x+ y + z)(x− y)

)= 0.

Example 1.44. Solve (x2 − yz)p+ (y2 − zx)q = z2 − xy. [UQ]

Solution: The subsidiary equations are

dx

(x2 − yz)=

dy

(y2 − zx)=

dz

z2 − xy(1)

Each fraction of (1) =

dx+ dy + dz

(x2 − yz) + (y2 − zx) + (z2 − xy)=

dx+ dy + dz

x2 + y2 + z2 − xy − yz − zx

Using x, y, z as multipliers, each fraction of (1) =

xdx+ ydy + zdz

x3+y3+z3 − 3xyz=

dx+ dy + dz

x2+y2+z2 − xy − yz − zxxdx+ ydy + zdz

(x+ y + z) (x2 + y2 + z2 − xy − yz − zx)=

dx+ dy + dz

x2+y2+z2 − xy − yz − zxxdx+ ydy + zdz

x+ y + z=dx+ dy + dz

1

xdx+ ydy + zdz = (x+ y + z)d(x+ y + z)

Integrating,x2

2+y2

2+z2

2=

(x+ y + z)2

2+ c

xy + yz + zx = a

Each fraction of (1) =dx− dy

(x2 − yz)− (y2 − zx)=

dy − dz

(y2 − zx)− (z2 − xy)

i.e.,dx− dy

(x2 − y2) + z (x− y)=

dy − dz

(y2 − z2) + x (y − z)

d (x− y)

(x− y) (x+ y + z)=

d (y − z)

(y − z) (x+ y + z)

d (x− y)

(x− y)=d (y − z)

(y − z)Integrating, log(x− y) = log(y − z) + log b

x− y

y − z= b


(xy + yz + zx,

x− y

y − z

)= 0.


Example 1.45. Solve x2(y − z)p+ y2(z − x)q = z2(x− y).


dx

x2 (y − z)=

dy

y2 (z − x)=

dz

z2 (x− y)(1)

Using1

x2,1

y2,1

z2as multipliers


1x2dx+

1y2dy +

1z2dz

(y − z) + (z − x) + (x− y)=

1x2dx+

1y2dy +

1z2dz

0

i.e.,1

x2dx+

1

y2dy +

1

z2dz = 0

Integrating,1

x+

1

y+

1

z= a

Using1

x,1

y,1

zas multipliers


1xdx+

1ydy +

1zdz

x (y − z) + y (z − x) + z (x− y)=

1xdx+

1ydy +

1zdz

0

i.e.,1

xdx+

1

ydy +

1

zdz = 0

Integrating, log x+ log y + log z = log b

xyz = b


(1

x+

1

y+

1

z, xyz

)= 0.

Example 1.46. Solve y2p− xyq = x(z − 2y). [UQ]


p− xyq = x(z − 2y) (1)

Takingdx

y2=

dy

−xy, we have

dx

y=

dy

−xxdx = −ydy

Integrating,x2

2= −y

2

2+ c

i.e., x2 + y2 = a

Takingdy

−xy=

dz

x (z − 2y)we have


dy

−y=

dz

(z − 2y)

i.e., (z − 2y)dy = −ydzi.e., ydz + zdy − 2ydy = 0

i.e., d(yz)− 2ydy = 0

Integrating, yz − y2 = b∴ The general solution is φ

(x2 + y2, yz − y2

)= 0.

Example 1.47. Solve (x2 − y2 − z2)p+ 2xyq − 2xz = 0. [UQ]

Solution: The subsidiary equations aredx

x2 − y2 − z2=

dy

2xy=

dz

2xz(1)

Takingdy

2xy=

dz

2xz, we have

dy

y=dz

z

log y = log z + log a

i.e.,y

z= a

Using x, y, z as multipliersEach fraction of (1) =

xdx+ ydy + zdz

x (x2 − y2 − z2) + 2xy2 + 2xz2=xdx+ ydy + zdz

x3 + xy2 + xz2

=xdx+ ydy + zdz

x (x2 + y2 + z2)

Takingdy

2xy=xdx+ ydy + zdz

x (x2 + y2 + z2)

dy

y=

2 (xdx+ ydy + zdz)

(x2 + y2 + z2)

dy

y=d(x2 + y2 + z2

)(x2 + y2 + z2)

log y = log(x2 + y2 + z2) + logb

i.e.,y

x2 + y2 + z2= b


(y

z,

y

x2 + y2 + z2

)= 0.

Example 1.48. Solve (x− 2z)p+ (2z − y)q = y − x. [UQ]



x− 2z=

dy

2z − y=

dz

y − x(1)


x− 2z + 2z− y + y − x=dx+ dy + dz

0

i.e., dx+ dy + dz = 0

i.e., x+ y + z = a

Each fraction of (1) =ydx+ xdy

xy − 2yz + 2xz− xy=ydx+ xdy

2z(x− y)

Takingydx+ xdy

2z (x− y)=

dz

y − x, we have

ydx+ xdy = −2zdz

i.e., d(xy) = −2zdz

Integrating, xy = −z2 + b

xy + z2 = b

The general solution is φ(x+ y + z, xy + z2

)= 0.

Example 1.49. Solve (z2 − 2yz − y2)p+ (xy + zx)q = xy − zx. [UQ]


dx

z2 − 2yz − y2=

dy

xy + zx=

dz

xy − zx(1)

Using x, y, z as multipliers each fraction of

(1) =xdx+ ydy + zdz

x (z2 − 2yz − y2) + y (xy + zx) + z (xy − zx)=xdx+ ydy + zdz

0we have

xdx+ ydy + zdz = 0


2+y2

2+z2

2= c1

x2 + y2 + z2 = a

Takingdy

xy + zx=

dz

xy − zx, we have

dy

y + z=

dz

y − z

i.e., (y − z)dy = (y + z)dz

ydy − zdy − ydz − zdz = 0

ydy − (ydz + zdy)− zdz = 0

ydy − d(yz)− zdz = 0


Integrating,y2

2− yz − z2

2= c2

i.e., y2 − z2 − 2yz = b

The general solution is φ(x2 + y2 + z2, y2 − z2 − 2yz

)= 0.

Example 1.50. Solve pz − qz = z2 + (x+ y)2.


z=dy

−z=

dz

z2 + (x+ y)2(1)

Takingdx

z=dy

−z, we have dx = −dy

Integrating, x+ y = a

Takingdx

z=

dz

z2 + (x+ y)2, we have

dx

z=

dz

z2 + a2(∵ x+ y = a)

i.e., dx =zdz

z2 + a2

Integrating, x =1

2log(z2 + a2) + c

2x = log(z2 + a2) + 2c

i.e., 2x− log[z2 + (x+ y)2

]= b

The general solution is φ{x+ y, 2x− log

[z2 + (x+ y)2

]}= 0

Example 1.51. Solve p− q = log(x+ y).


1=dy

−1=

dzlog(x+y)

(1)

Takingdx

1=dy

−1, we have dx = −dy

Integrating, x+ y = a

Takingdx

1=

dz

log(x+ y), we have

dx =dz

log a(∵ x+ y = a)

i.e., log adx = dz

Integrating, (loga)x = z + b

i.e., [log(x+ y)]x = z + b

x log(x+ y)− z = bThe general solution is φ(x+ y, xlog(x+ y)− z) = 0.



Solve the following partial differential equations:

1. z(x− y) = x2p− y2q

[Ans : φ

(1

x+

1

y,

z

x+ y

)= 0

]2. (x2 + y2 + yz)p+ (x2 + y2 − xz)q = z(x+ y)[

Ans : φ(x− y − z,x2 + y2

z2) = 0

]3. (3z − 4y)

∂z

∂x+ (4x− 2z)

∂z

∂y= 2y − 3x[Ans : φ(x2 + y2 + z2, 2x+ 3y + 4z) = 0

]4. (y + z)p+ (z + x)q = x+ y[

Ans : φ

(y − z

x− y, (x− y)

√x+ y + z

)= 0

]5. (y2 + z2)p− xyq + xz = 0

[Ans : φ

(yz, x2 + y2 + z2

)= 0]

6. (y − z)p+ (x− y)q = (z − x)[Ans : (x2 + 2yz, x+ y + z) = 0

]7.y − z

yzp+

z − x

xyq =

x− y

xy[Ans : φ(x+ y + z, xyz) = 0]

8. p cotx+ q cot y = cot z

[Ans : φ

(cos z

cos y,cos y

cosx

)= 0

]9. (a− x)p+ (b− y)q = c− z

[Ans : φ

(a− x

b− y,b− y

c− z

)= 0

]1.6 Solution of Partial Differential Equations

A solution or integral of a partial differential equation is a relationbetween the independent and the dependent variables which satisfies thegiven differential equation.Note: There are two distinct types of solution for partial differentialequations, one type of solution containing arbitrary constants and theother type of solution containing arbitrary functions. Both these types fsolution may be given as solutions of the same partial differentialequation.Any solution of a partial differential equation in which the number of

arbitrary constants is equal to the number of independent variables iscalled the complete integral .


Any solution obtained from the complete integral by giving particularvalues to the arbitrary constants is called a particular integral .Any solution obtained from the complete integral by eliminating

arbitrary constants is called a singular integral .Any solution which contains the maximum number of arbitrary

functions is called as a general integral .Let φ(x, y, z, a, b) = 0 be the complete integral of f(x, y, z, p, q) = 0.

Then the solution obtained by eliminating a and b from the equations

φ(x, y, z, a, b) = 0 (1)

∂φ

∂a= 0 (2)

∂φ

∂b= 0 (3)

is called the singular integral of the partial differential equation.In (1), we put b = g(a) we get

φ(x, y, z, a, g(a)) = 0 (4)

Differentiating (4) w.r.t. ‘a’, we get

∂φ

∂a+∂φ

∂gg′(a) = 0 (5)

The obtained by eliminating ‘a’ from (4) and (5) is called the generalintegral of φ(x, y, z, a, b) = 0

1.6.1 Solution of Partial Differential Equations by Direct Integration

A partial differential equation can be solved by successive integration inall cases where the dependent variable occurs only in the partial derivatives.We illustrate this method in the following Examples.

Example 1.52. Solve∂z

∂x= cos x.

Solution: If z is a function of x only then there is no difference between∂z

∂xand

dz

dx. Therefore to find z we can directly integrate

dz

dx= cos x and

hence the solution is z = sin x + A, where A is arbitrary constant. Buthere z is a function of two variables x and y.∴ Instead of the arbitrary constant A we can take an arbitrary function

of y say f(y).∴ The solution is z = sin x+ f(y).


Example 1.53. Solve∂2z

∂x∂y= sinx.

Solution: Given∂2z

∂x∂y= sinx. Integrating w.r.t ‘x‘

∂z

∂y= −cosx+ f(y),

where f(y) is an arbitrary function of y. Integrating again w.r.t ‘y’ Z =

−ycosx + F (y) + g(x) where F (y) =

∫f(y)dy is an arbitrary function of

y and g(x) is an arbitrary function of x.


∂x∂y= 0.

Solution : Given∂2z

∂x∂y= 0 Integrating w.r.t ‘x‘

∂z

∂y= f(y), where f(y)

is an arbitrary function of y. Integrating again w.r.t ‘y‘ z = F (y) + g(x),where F (y) is an arbitrary function of y and g(x) is an arbitrary functionof x.

Example 1.55. Solve∂2u

∂y∂x= 4xsin(3xy).

Solution: Given∂2u

∂y∂x= 4xsin(3xy) Integrating w.r.t ‘y‘

∂u

∂x=

−4x cos(3xy)

3x+ f(x) =

−4

3cos(3xy) + f(x) Integrating again w.r.t

‘x‘ u =−4

3

sin(3xy)

3y+ F (x) + g(y) ie, u =

−4

9ysin(3xy) + F (x) + g(y),

where F (x) =

∫f(x)dx is an arbitrary function of x and g(y) is an

arbitrary function of y.

Example 1.56. Solve∂z

∂x= 2x+ 3y;

∂z

∂y= 3x+ cosy

Solution:∂z

∂x= 2x+ 3y Integrating w.r.t ‘x‘ z = x2 + 3xy + f(y) (1)

where f(y) is an arbitrary function of ‘y‘ Differentiating (1) partially

w.r.t ‘y‘∂z

∂y= 3x+ f ′(y) By hypothesis,

∂z

∂y= 3x+ cosy

∴ f(y) = siny + c

∴ from (1), z = x2 + 3xy + siny + c , where c is an arbitrary constant.


∂x∂y= sinxsiny for which

∂z

∂y= −2siny, when

x = 0 and z = 0 when y is an odd multiple ofπ

2.


Solution:∂2z

∂x∂y= sinxsiny

Integrating w.r.t ‘x‘∂z

∂y= −cosxsiny + f(y) (1)

Given∂z

∂y= −2siny when x = 0

Hence, −2siny = −siny + f(y)∴ f(y) = −sinyFrom (1),

∂z

∂y= −cosxsiny − siny

Integrating w.r.t ‘y‘z = cosxcosy + cosy + g(x) (2)

Given z = 0 when y = (2n+ 1)π

2

[yis an odd multiple of

π

2

]Hence g(x) = 0From(2), z = (cosx+ 1)cosy.

Example 1.58. Solve∂2u

∂x∂t= e−t cosx, given that u = 0 when t = 0 and

∂u

∂t= 0 when x = 0. Show also that as t→ ∞ , u→ sinx.

Solution: Given∂2u

∂x∂t= e−t cosx

Integrating w.r.t ‘x‘∂u

∂t= e−t sinx+ f(t)

When x = 0,∂u

∂t= 0

∴ f(t) = 0 Hence∂u

∂t= e−t sinx

Integrating w.r.t ‘t‘u = −e−t sinx+ g(x)When t = 0, u = 00 = −sinx+ g(x) i.e., g(x) = sinx∴ u = −e−t sinx+ sinxu = (1− e−t) sin x.Apply t→ ∞, we have e−t → 0⇒ u = sinx.


Solve the following partial differential equation using direct integration


1.∂z

∂x= −x

y

[Ans: z = −x

2

2y+ φ(y)

]2.∂2z

∂x2= sin y

[Ans :z =

x2

2sin y + xf(y) + φ(y)

]3.

∂2z

∂x∂y= x2 + y2

[Ans :z =

x3y

3+y3x

3+ φ(y) + f(x)

]4.∂z

∂x= 4x− 2y;

∂z

∂y= −2x+ 6y

[Ans :z = 2x2 − 2xy + 3y2 + c

]5.∂2z

∂y2− z = 0 when y = 0, z = ex and

∂z

∂y= e−x[

Ans :z = ey coshx+ e−y sinhx]

1.7 Solution of standard types of first order partial differentialequations

A partial differential equation which involves only the first order partial

derivatives p (i.e., =∂z

∂x) and q (i.e., =

∂z

∂x) is called a first order

partial differential equation.The general form of first order partial differential equation is

f(x, y, z, p, q) = 0. We shall see some standard form of such equationsand solve them by special methods.

1.7.1 Type I: f(p, q) = 0 i.e., the equation contain p and q only.

To find Complete IntegralGiven f(p, q) = 0 (1)Let z = ax+ by + c be a trial solution of the equation (1). Then

p =∂z

∂x= a and q =

∂z

∂y= b

From (1), we get

f(a, b) = 0

Hence the complete integral of (1) is

z = ax+ by + c

Solving for b from f(a, b) = 0, we get b = β(a)∴ The complete integral of (1) is

z = ax+ β(a)y + c (2)


since [number of arbitrary constants(a,c)=number of independentvariables(x,y)=2]

To find Singular IntegralTo obtain the singular integral we have to eliminate a and c from the

equations z = ax+ β(a)y + c,∂z

∂a= 0 and

∂z

∂c= 0

i.e., z = ax+ β(a)y + c

x+ β′(a)y = 0

1 = 0

The last equation being absurd and hence there is no singularintegral .

To find General IntegralPut c = χ(a) in complete integral equation (2), we get

z = ax+ β(a)y + χ(a) (3)

Differentiate (3) w.r.t. ‘a’, we get

z = x+ β′(a)y + χ

′(a) (4)

Eliminating ‘a’ from (3) and (4), we get general integral of the givenp.d.e.

Note: For the equation of the type of f(p, q) = 0 there is no singularintegral.

Example 1.59. Solve pq = 1.

Solution: Given pq = 1 (1)To find Complete Integral :Let z = ax+ by + c be a trial solution of the equation (1). Then

p =∂z

∂x= a and q =

∂z

∂y= b

From (1), we get

ab = 1

i.e., =1

a

Hence the complete integral of (1) is

z = ax+1

ay + c (2)


since [number of arbitrary constants(a,c)=number of independentvariables(x,y)=2]To find Singular Integral :To obtain the singular integral we have to eliminate ‘a’ and ‘c’ from the

equations z = ax+ β(a)y + c,∂z

∂a= 0 and

∂z

∂c= 0

i.e., z = ax+1

ay + c

x− 1

a2y = 0

1 = 0

The last equation being absurd and hence there is no singular integral.To find General Integral :Put c = χ(a) in complete integral equation (2), we get

z = ax+1

ay + χ(a) (3)


z = x− 1

a2y + χ

′(a) (4)


Example 1.60. Solve p2 + q2 = 1.

Solution: Given p2 + q2 = 1 (1)Let z = ax+ by + c be a trial solution of (1)

Then p =∂z

∂x= a and q =

∂z

∂y= b

From (1), we get

a2 + b2 = 1

∴ b =√

1− a2

∴ The complete integral is z = ax+√

1− a2y + cThe singular and general integrals are obtained as first example of this

type.

Example 1.61. Solve p+ q = pq.

Solution: Given p+ q = pq (1)Let z = ax+ by + c be a trial solution of (1)


Then p =∂z

∂x= a and q =

∂z

∂y= b

From (1), we get

a+ b = ab

i.e., b =a

a− 1

∴ The complete integral is z = ax+a

a− 1y + c

The singular and general integrals are obtained as in previous procedure.

Example 1.62. Solve pq + p+ q = 0.

Solution: Solution: Similar to above Example

Example 1.63. Find the complete integral of p = eq.

Solution: Given logp = q (1)Let z = ax+ by + c be a trial solution of (1)

Then p =∂z

∂x= a and q =

∂z

∂y= b

From (1), we get

log a = b

∴ The complete integral is z = ax+ (loga)y + c

The singular and general integrals are obtained as first example of thistype.

Example 1.64. Solve√p+

√q = 1.

Solution: Given√p+

√q = 1 (1)

Let z = ax+ by + c be a trial solution of (1)

Then p =∂z

∂x= a and q =

∂z

∂y= b

From (1), we get

√a+

√b = 1

i.e., b =(1−

√a)2

∴ The complete integral is z = ax+(1−

√a)2y + c

The singular and general integrals are obtained as first example of thistype.

Example 1.65. Find the complete integral of p2 + q2 = npq.


Solution: Given p2 + q2 = npq (1)Let z = ax+ by + c be a trial solution of (1)

Then p =∂z

∂x= a and q =

∂z

∂y= b

From (1), we get

a2 + b2 = nab

i.e., b2 − nab + a2 = 0

∴ b =na±

√n2a2 − 4a2

2

b =a

2

[n±

√n2 − 4

]∴ The complete integral is z = ax+

a

2

[n±

√n2 − 4

]y + c

1.7.2 Type II: z = px+ qy + f(p, q) (Clairaut’s Form)

Suppose a partial differential equation of the form

z = px+ qy + f(p, q) (1)

which is said to be of Clairaut’s form.To find complete integral : The complete integral of (1) is bysubstituting p = a and q = b in (1), i.e.,

z = ax+ by + f(a, b) (2)

where a and b are arbitrary constants.To find singular integral : Partially differentiating (2) w.r.t ‘a’ and ‘b’and then equating to zero, we get

x+∂f

∂a= 0 (3)

y +∂f

∂b= 0 (4)

By eliminating a and b from (2), (3) and (4), we get the singular integral.To find General Integral : Put b = ψ(a) in complete integral equation(2), we get

z = ax+ β(a)y + f(a, β(a)) (5)


z = x+ β′(a)y + f

′(a, β(a)) (6)



Example 1.66. Find the complete and singular integrals of z = px+ qy+pq.

Solution: Given z = px+ qy + pq (1)The given partial differential equation is in Clairaut’s form.∴ The complete integral is z = ax+ by + ab (2)

where a and b are arbitrary constants.To find singular integral : Partially differentiating (2) w.r.t ‘a’ and ‘b’then equating to zero, we get

x+ b = 0

y + a = 0

∴ a = −y and b = −x

Substituting the values of ‘a’ and ‘b’ in (2), we get

z = −yx− xy + (−y)(−x) = −yx− xy + yx

∴ z = −xywhich is the singular integral.

To find General Integral : Put b = β(a) in complete integral equation(2), we get

z = ax+ β(a)y + aβ(a) (3)


z = x+ β′(a)y + aβ

′(a) + β(a) (4)


Example 1.67. Solve z = px+ qy + 2√pq.

Solution: Given z = px+ qy + 2√pq (1)

The given partial differential equation is in Clairaut’s form∴ The complete integral is z = ax+ by + 2

√ab (2)

where a and b are arbitrary constants.To find singular integral: Partially differentiating (2) w.r.t a and b thenequating to zero, we get

x+

√b

a= 0 (3)

y +

√a

b= 0 (4)


∴ x = −√b

aand y = −

√a

bFrom the above two equations we get

xy = 1

which is the singular integral.The general integral is obtained as first example of this type.

Example 1.68. Solve z = px+ qy + p2q2.

Solution: The given equation is in Clairaut’s form.∴ The complete integral is z = ax+ by + a2b2 (1)To find singular integral: Partially differentiating (1) w.r.t a and b thenequating to zero, we get

x = −2ab2 (2)

y = −2a2b (3)

xy = 4a3b3 (4)

From (2),x

b= −2ab

From (3),y

a= −2ab

From (1), z = ab[xb+y

a+ ab

]= ab(−2ab− 2ab+ ab)

z = −3a2b2 (5)

Now, z3 = −27a6b6

i.e., z3 = −27(a3b3)2

Using (4), z3 = −27(xy4

)2∴ z3 =

−27

16x2y2

which is the required singular integral.The general integral is obtained as first example of this type.

Example 1.69. Solve z = px+ qy + p2 − q2.

Solution: The given partial differential equation is in Clairaut’s form∴ The complete integral is z = ax+ by + a2 − b2 (1)

To find singular integral: Partially differentiating (2) w.r.t a and b thenequating to zero, we get


x+ 2a = 0

y − 2b = 0

∴ a =−x2

and b =y

2Substituting the values of a and b in (1), we get

z =−x2

2+y2

2+x2

4− y2

4

ie., 4z = y2 − x2 which is the singular integral.The general integral is obtained as first example of this type.

Example 1.70. Solve z = px+ qy + p2 + pq + q2.

Solution: The given equation is in Clairaut’s form.∴ The complete integral is z = ax+ by + a2 + ab+ b2 (1)To find singular integral: Partially differentiating (1) w.r.t a and b thenequating to zero, we get

x+ 2a+ b = 0 (2)

y + a+ 2b = 0 (3)

Solving (2) and (3), we get

a =1

3(y − 2x), b =

1

3(x− 2y)

Substituting the values of a and b in (1), we get

3z = xy − x2 − y2


Example 1.71. Solve z = px+ qy +√

1 + p2 + q2.

Solution: The given equation is in Clairaut’s form.

∴ The complete integral is z = ax+ by +√

1 + a2 + b2 (1)To find singular integral: Partially differentiating (1) w.r.t a and b then


equating to zero, we get

x+a√

1 + a2 + b2= 0 ⇒ x =

−a√1 + a2 + b2

(2)

y +b√

1 + a2 + b2= 0 ⇒ y =

−b√1 + a2 + b2

(3)


x2 + y2 =a2 + b2

1 + a2 + b2

1− (x2 + y2) = 1− a2 + b2

1 + a2 + b2

1− x2 − y2 =1

1 + a2 + b2√1 + a2 + b2 =

1√1− x2 − y2

(4)


a =−x√

1− x2 − y2(5)

and

b =−y√

1− x2 − y2(6)

Substituting (4), (5) and (6) in (1), we get

z =−x2√

1− x2 − y2− y2√

1− x2 − y2+

1√1− x2 − y2

z =√

1− x2 − y2

z2 = 1− x2 − y2

∴ x2 + y2 + z2 = 1


Example 1.72. Find the complete integral ofpqz = p2(xq + p2) + q2(yp+ q2).

Solution: Given pqz = p2(xq + p2) + q2(yp+ q2)

= p2xq + p4 + q2yp+ q4

Dividing by pq we have


z = px+ qy +p4 + q4

pq

This is Clairaut’s form.The complete integral is

z = ax+ by +a4 + b4

ab,

where a and b are arbitrary constants.

1.7.3 Type III: f(z, p, q) = 0 ie., equation not containing x and y explicity

Given f(z, p, q) = 0 (1)Let z = f(x+ ay) be the solution of (1)Put x+ ay = u Then

z = f(u) and∂u

∂x= 1,

∂u

∂y= a

p =∂z

∂x=dz

du

∂u

∂x=dz

du

q =∂z

∂y=dz

du

∂u

∂y= a

dz

du

Substituting the values of p and q in (1), we get

f

(z,dz

du, adz

du

)= 0 (2)

This is an ordinary differential equation of first order

Solving fordz

du, we obtain

dz

du= g(z, a)

i.e.,dz

g(z, a)= du

Integrating

∫dz

g(z, a)= u+ c

φ(z, a) = u+ c

ie., (z, a) = x+ ay + c

which is the complete integral of (1).The singular integral can be obtained by the usual methods.

Example 1.73. Find the complete integral of z = pq.


Solution: Given z = pq (1)Let z = f(x+ ay) be the solution of (1)Put u = x+ ay so that z = f(u)

p =dz

duand q = a

dz

du


z = a

(dz

du

)2

dz

du=

√z

a⇒ dz√

z=

1√adu

Integrating, 2√z =

1√au+ c

2√az = u+

√ac

2√az = u+ b,whereb =

√ac

4az = (u+ b)2

4az = (x+ ay + b)2

which is the complete integral.

Example 1.74. Solve z = p2 + q2.

Solution: Given z = p2 + q2 (1)Let z = f(x+ ay) be the solution of (1)Put u = x+ ay so that z = f(u), then

p =dz

duand q = a

dz

du


z =

(dz

du

)2

+ a2(dz

du

)2

⇒ z =(1 + a2)

(dz

du

)2

dz

du=

√z

1 + a2⇒√1 + a2

dz√z

=du

Integrating,√

1 + a22√z = u+ b

(1 + a2)4z = (u+ b)2

i.e., 4(1 + a2)z = (x+ ay + b)2


Example 1.75. Solve p(1 + q) = qz.


Solution: Given p(1 + q) = qz (1)Let z = f(x+ ay) be the solution of (1)Put u = x+ ay so that z = f(u), then

p =dz

duand q = a

dz

du


dz

du

(1 + a

dz

du

)= a

dz

duz ⇒1 + a

dz

du=az

dz

du=az − 1

a⇒ a

az − 1dz =du

Integrating, log(az − 1) = u+ b

i.e., log(az − 1) = x+ ay + b


Example 1.76. Solve 9(p2z + q2

)= 4.

Solution: Given 9(p2z + q2

)= 4 (1)

Let z = f(x+ ay) be the solution of (1)Put u = x+ ay so that z = f(u), then

p =dz

duand q = a

dz

du


9

[(dz

du

)2

z + a2(dz

du

)2]= 4(

dz

du

)2

=4

9(z + a2)⇒ dz

du=

2

3

1√z + a2

3

2

√z + a2dz = du

Integrating,we get3

2

(z + a2

)3/23/2

= u+ b(z + a2

)3/2= x+ ay + b


Example 1.77. Solve p(1 + q2

)= q(z − a).

Solution: Given p(1 + q2

)= q(z − a) (1)



p =dz

duand q = a

dz

duSubstituting in (1), we get

dz

du

[1 + b2

(dz

du

)2]= b

dz

du(z − a) ⇒

[1 + b2

(dz

du

)2]

=b(z − a)(dz

du

)2

=bz − ba− 1

b2⇒ dz

du=

√bz − ba− 1

bb√

bz − ba− 1dz = du

Integrating, 2√bz − ba− 1 = u+ c

4 (bz − ba− 1)2 = (u+ c)2

i.e., 4 (bz − ba− 1)2 = (x+ by + c)2


Example 1.78. Solve z2(p2 + q2 + 1

)= 1.

Solution: Given z2(p2 + q2 + 1

)= 1 (1)


p =dz

duand q = a

dz

duSubstituting in (1), we get

z2

[(dz

du

)2

+ a2(dz

du

)2

+ 1

]= 1(

dz

du

)2

+ a2(dz

du

)2

+ 1 =1

z2(1 + a2

)(dzdu

)2

=1

z2− 1

(1 + a2

)(dzdu

)2

=1− z2

z2

dz

du=

√1− z2

z√1 + a2

z√1− z2

dz =1√

1 + a2du

Integrating,

∫z√

1− z2dz =

1√1 + a2

∫du (2)

Put 1− z2 = t⇒ −2zdz = dt⇒ zdz = −1

2dt


(2) ⇒ −1

2

∫dt√t=

1√1 + a2

∫du

−1

22√t =

1√1 + a2

u+ b

−√t =

u√1 + a2

+ b

−√t =

x+ ay√1 + a2

+ b

−√

1− z2 =x+ ay√1 + a2

+ b


Example 1.79. Solve z2(p2 + q2 + 1

)= c2.


Example 1.80. Solve q2 = z2p2(1− p2

).

Solution: Given q2 = z2p2(1− p2

)(1)


p =dz

duand q = a

dz

du


a2(dz

du

)2

= z2(dz

du

)2[1−

(dz

du

)2]

a2

z2= 1−

(dz

du

)2

⇒(dz

du

)2

=z2 − a2

z2

dz

du=

√z2 − a2

z⇒ z√

z2 − a2dz =du

Integrating,

∫z√

z2 − a2dz =

∫du

Put z2 − a2 = t⇒ 2zdz = dt⇒ zdz =1

2dt

1

2

∫dt√t=

∫du

1

22√t = u+ b ⇒

√t =u+ b√

z2 − a2 = x+ ay + b



1.7.4 Type IV: f(x, p) = g(y, q) (Separable equations)

A first order partial differential equation is called separable if it can bewritten as f(x, p) = g(y, q).Let f(x, p) = g(y, q) = a , where a is an arbitrary constant.

∴ f(x, p) = a, g(y, q) = a Solving for p and q from these two equations weget

p = f1(x, a) and q = g1(y, a)

Since dz =∂z

∂xdx+

∂z

∂ydy

= pdx+ qdy

dz = f1(x, a)dx+ g1(x, a)dyIntegrating, we get

z =

∫f1(x, a)dx+

∫g1(y, a)dy + b

which gives the complete integral.

Example 1.81. Solve p+ q = x+ y

Solution: Given that p+ q = x+ y (1)i.e., p− x = y − q

Let p− x = y − q = a , where a is an arbitrary constant.Now, p− x = a⇒ p = a+ x

i.e., f1(x, a) = a+ xy − q = a⇒ q = y − z

i.e., f2(y, a) = y − a∴ The complete integral is

z =

∫f1(x, a)dx+

∫f2(y, a)dy + b

z =

∫(a+ x) dx+

∫(y − a)dy + b

z =(a+ x)2

2+

(y − a)2

2+ b

2z = (a+ x)2 + (y − a)2 + 2b

Example 1.82. Solve p− x2 = q + y2

Solution: Let p− x2 = q + y2 = a (1)Now, p− x2 = a⇒ p = a+ x2

i.e., f1(x, a) = a+ x2


q + y2 = a⇒ q = a− y2

i.e., f2(y, a) = a− y2

∴ The complete integral is

z =

∫f1(x, a)dx+

∫f2(y, a)dy + b

z =

∫ (a+ x2

)dx+

∫ (a− y2

)dy + b

z =

(ax+

x3

3

)+

(ay − y3

3

)+ b

Example 1.83. Solve p2 − q2 = x− y

Solution: Given p2 − x = q2 − yLet p2 − x = q2 − y = aNow, p2 − x = a⇒ p =

√a+ x

i.e., f1(x, a) =√a+ x

q2 − y = a⇒ q =√a+ y

i.e., f2(y, a) =√a+ y


z =

∫f1(x, a)dx+

∫f2(y, a)dy + b

z =

∫ √a+ xdx+

∫ √a+ ydy + b

z =2

3(a+ x)3/2 +

2

3(a+ y)3/2 + b

Example 1.84. Solve p2 + q2 = x+ y


Example 1.85. Solve√p+

√q = 2x

Solution: Let√p− 2x = −√

q = a√p− 2x = a⇒ p = a+ 2x2

i.e., f1(x, a) = a+ 2x2

−√q = a⇒ q = a2

i.e., f2(y, a) = a2


z =

∫f1(x, a)dx+

∫f2(y, a)dy + b

z =

∫(a+ 2x)2dx+

∫a2dy + b

z =(a+ 2x)3

3× 2+ a2y + b


Example 1.86. Solve p2y(1 + x2

)= qx2

Solution: Givenp2(1 + x2

)x2

=q

y

Letp2(1 + x2

)x2

=q

y= a

Nowp2(1 + x2

)x2

= a⇒ p =

√ax√

1 + x2

i.e., f1(x, a) =

√ax√

1 + x2q

y= a⇒ q = ay

i.e., f2(y, a) = ay∴ The complete integral is

z =

∫f1(x, a)dx+

∫f2(y, a)dy + b

z =

∫ √ax√

1 + x2dx+

∫aydy + b

Put 1 + x2 = t⇒ 2xdx = dt

z =

√a

2

∫dt√t+ a

∫ydy + b

=

√a

22√t + a

y2

2+ bz =

√a√

1 + x2 + ay2

2+ b

Example 1.87. Solvex

p+y

q= 1

Solution: Givenx

p= 1− y

q

Letx

p= 1− y

q= a

Nowx

p= a ⇒ p =

x

a∴ f1(x, a) =

x

a

1− y

2= a ⇒ q =

y

a+ 1∴ f2(y, a) =

y

a+ 1∴ The complete integral is

z =

∫f1(x, a)dx+

∫f2(y, a)dy + b

z =

∫x

adx+

∫y

a+ 1dy + b

z =x2

2a+

y2

2(a+ 1)+ b


Example 1.88. Solve p+ q = sin x+ sin y

Solution: Given p− sinx = q − sin yLet p− sinx = q − sin y = a

Now p− sinx = a ⇒ p = a+ sinx ∴ f1(x, a) = a+ sinx

q − sin y = a ⇒ q = a+ sin y ∴ f2(y, a) = a+ sin y


z =

∫f1(x, a)dx+

∫f2(y, a)dy + b

z =

∫(a+ sinx) dx+

∫(a + siny) dy + b

z = ax− cosx+ ay − cos y + b


Solve the following partial differential equations

Type I : f(p, q)=0

a) p2 − q2 = 4[Ans: z = ax+

√a2 − 4y + c

]b) p+ sinq = 0 [Ans : z = by − xsinb+ c]

c) p = q [Ans : z = a(x+ y) + c]

Type II : Clairaut’s form:

a) z = px+ qy +p

q− p

[Ans: CI : z = ax+ by +

a

b− a,SI :

y

1− x

]b) (1− x)p+ (2− y)q = 3− z [Ans: CI : z = ax+ by − a− 2b+ 3]

c) z = px+ qy + logpq[Ans: CI : z = ax+ by + logab,SI : z = −2− logxy]

d) z = px+ qy + 3p1/3q1/3[Ans: CI : ax+ by + 3a1/3b1/3,SI : xyz = 1

]Type III : f(z,p, q)=0

a) p2 + pq = z2[Ans: logz =

1√1 + a

(x+ ay) + b

]b) z2 = p2 + q2 + 1

[Ans: cosh−1 z =

1√1 + a2

(x+ ay) + b

]c) p3 + q3 = 8z

[Ans: (1 + a3)z2 =

64

27(x+ ay + b)3

]


d) p+ q = z [Ans: x+ ay = (1 + alogz + b)]

e) p3 = qz[Ans: 4z = a(x+ ay − b)2

]Type IV : f(x,p )=g(y,q )

a) q2 − p = y − x

[Ans: z =

(a+ x)2

2+

2

3(a+ y)3/2 + b

]b) yp = 2yx+ logq

[Ans: z = x2 + ax+

eay

a

]c) p2 + q2 = x+ y

[Ans: z =

2

3

{(x+ a)3/2 + (y − a)3/2

}+ b

]d) q = xyp2

[Ans: z = 2

√ax+

ay2

2+ b

]e) x2p2 = yq2 [Ans: z = alogx+ 2a

√y + b]


1.7.6 Equations Reducible to Standard Forms

1.7.7 Type A : f (xmp, ynq) = 0

An equation of the form f (xmp, ynq) = 0 where m and n are constantscan be transformed into an equation of the first type [i.e., f(p, q) = 0]Put X = x1−m and Y = y1−n where m 6= 1 and n 6= 1.

Then p =∂z

∂x=

∂z

∂X

dX

∂x= P (1−m)x−m where P =

∂z

∂X∴ xmp = (1−m)P

q =∂z

∂y=∂z

∂Y

dY

∂y= Q (1− n) y−n where Q =

∂z

∂Y

∴ ynq = (1− n)Q

Hence the given equation reduces to f [(1−m)P, (1− n)Q] = 0which is of the form F(P, Q) = 0 (Type I)

1.7.8 Type B : f (xmp, ynq, z) = 0

An equation of the form f (xmp, ynq, z) = 0 can also be transformed to thestandard type F(P, Q, z) = 0 (Type II) by the substitution X = x1−m

and Y = y1−n where m 6= 1 and n 6= 1.Note: In the above two types if m = 1 put X = log x and if n = 1 , putY = log y:

∴ p =∂z

∂x=

∂z

∂X

dX

∂x= P

1

xwhere P =

∂z

∂Xi.e., xp = P

q =∂z

∂y=∂z

∂Y

dY

∂x= Q

1

ywhere Q =

∂z

∂Y

i.e., yq = Q∴ The equations reduce to either F (P,Q) = 0 or F (P,Q, z) = 0.

1.7.9 Type C : f(zkp, zkq

)= 0 where k is a constant

An equation of the form f(zkp, zkq

)= 0 can be transformed into an

equation of the first type [i.e., f(p, q) = 0 ].Put Z = zk+1 if k 6= −1

Then∂Z

∂x= (k + 1) zkp

P = (k + 1) zkp where P =∂Z

∂x

i.e., zkp =1

k + 1P


Q = (k + 1) zkq where Q =∂Z

∂y

i.e., zkq =1

k + 1Q

Hence the given equation reduces to f

(1

k + 1P,

1

k + 1Q

)= 0 which is of

the form F(P, Q) = 0Note: If k = −1, then

Put Z = log z

∂Z

∂x=

1

zp

P = z−1p

Similarly,

P = z−1p

Hence the given equation again reduces to the form F (P,Q) = 0.

1.7.10 Type D: f(xmzkp, ynzkq

)= 0

This can be transformed to an equation of the form F (P,Q) = 0 by thesubstitution.

X =

{x1−m if m 6= 1log x if m = 1

Y =

{y1−n if n 6= 1log y if n = 1

Z =

{zk+1 if k 6= −1log z if k = −1

Example 1.89. Solve xp+ yq = 1

Solution: Given xp+ yq = 1 (1)The given differential equation is the form f (xmp, ynq) = 0 with m = 1 ,n = 1.Put X = log x and Y = log y

p =∂z

∂x=

∂z

∂X

∂X

∂x= P

1

x, where P =

∂z

∂X∴ xp = P

Similarly, yq = Q where Q =∂z

∂Y∴ (1) becomes P +Q = 1 (2)

This is of the form F (P,Q) = 0 [Type I]Let z = aX + bY + c be a solution of (2)


∂z

∂X= a⇒ P = a

∂z

∂Y= b⇒ Q = b

Substituting the values of P and Q in (2), we get

a+ b = 1

i.e., b = 1− a

∴ The complete integral of P +Q = 1 is

z = aX + (1− a)Y + b

∴ The required complete integral is

z = a log x+ (1− a) log y + b

Example 1.90. Solve p2x2 + q2y2 = z2

Solution: The given differential equation can be written as(xp)2 + (yq)2 = z2 (1)This is of the form f (xmp, ynq, z) = 0 with m = 1 , n = 1.Put X = log x and Y = log y

p =∂z

∂x=

∂z

∂X

∂X

∂x= P

1

x,whereP =

∂z

∂X

∴ xp = P


∂Y∴ (1) becomes P 2 +Q2 = z2 (2)

This is of the form F (P,Q, z) = 0 [Type II]Let z = φ(X + aY ) be a solution of (2)Let u = X + aY so that z = φ(u)

Then P =dz

duand Q = a

dz

du


∴ (2) becomes

(dz

du

)2

+ a2(dz

du

)2

= z2(dz

du

)2 (1 + a2

)= z2

dz

du=

z√(1 + a2)

dz

z=

du√(1 + a2)

Integrating, log z =u√

(1 + a2)+ b

∴ log z =X + aY√(1 + a2

) + b

log z =log x+ a log y√

(1 + a2)+ b


Example 1.91. Solve p2x2 + q2y2 = z

Solution: This problem is to similar to above problem.

Example 1.92. Solve z2 = xypq

Solution: The given equation can be written as z = (xp)(yq) (1)This is of the form f (xmp, ynq, z) = 0 with m = 1 , n = 1.Put X = log x and Y = log y

p =∂z

∂x=

∂z

∂X

∂X

∂x= P

1

x, where P =

∂z

∂X

∴ xp = P


∂Y∴ (1) becomes z2 = PQ (2)


Then P =dz

duand Q = a

dz

du


∴ (2) becomes z2 = a2(dz

du

)2

dz

du=

z√a

dz

z=

√adu

Integrating, log z =√au+ b

i.e., log z =√a (X + aY ) + b

i.e., log z =√a (log x+ a log y) + b


Example 1.93. Solve z2(p2x2 + q2

)= 1

Solution: The given equation can be written as (px)2 + q2 =1

z2(1)

This is of the form f (xmp, q, z) = 0 with m = 1.Put X = log x

p =∂z

∂x=

∂z

∂X

∂X

∂x= P

1

x, where P =

∂z

∂X

∴ xp = P

∴ (1) becomes P 2 + q2 =1

z2(2)

This is of the form F (P, q, z) = 0Let z = φ(X + aY ) be a solution of (2)Let u = X + aY so that z = φ(u)

Then P =dz

duand Q = a

dz

du


∴ (2) becomes

(dz

du

)2

+ a2(dz

du

)2

=1

z2(dz

du

)2 (1 + a2

)=

1

z2

dz

du=

1

z√1 + a2

zdz =1√

1 + a2du

Integrating,z2

2=

1√1 + a2

u+ b

z2

2=

1√1 + a2

(X + ay) + b

z2

2=

1√1 + a2

(log x+ ay) + b√1 + a2z2 = 2 (log x+ ay) + c, where c = 2b


Example 1.94. Solve p2x4 + y2zq = 2z2

Solution: The given equation can be written as (x2p)2 + (y2q)z = 2z2(1)This is of the form f (xmp, ynq, z) = 0 with m = 2, n = 2. [Type II]Put X = x1−m and Y = y1−n

i.e., X =1

xand Y =

1

y

p =∂z

∂x=

∂z

∂X

∂X

∂x=

∂z

∂X

(− 1

x2

)= −P

x2,whereP =

∂z

∂X

∴ x2p = −PSimilarly y2q = −Q, where Q =

∂z

∂Y∴ (1) becomes P 2 −Qz = 2z2 (2)


Then P =dz

duand Q = a

dz

du


∴ (2) becomes

(dz

du

)2

− a

(dz

du

)z = 2z2(

dz

du

)2

− az

(dz

du

)− 2z2 = 0

dz

du=az ±

√a2z2 + 8z2

2

dz

du= z

[a±

√a2 + 8

2

]dz

z=

[a±

√a2 + 8

2

]du

Integrating, log z =

[a±

√a2 + 8

2

]u+ b

log z =

[a±

√a2 + 8

2

](X + aY ) + b

log z =

[a±

√a2 + 8

2

](1

x+a

y

)+ b

which is the required complete integral.

Example 1.95. Solve x4p2 − yzq = z2

Solution: The given equation can be written as (x2p)2 − (yq)z = z2 (1)This is of the form f (xmp, ynq, z) = 0 with m = 2, n = 1.Put X = x1−m and Y = log y

i.e., X =1

xand Y = log y

p =∂z

∂x=

∂z

∂X

∂X

∂x=

∂z

∂X

(− 1

x2

)= P

(− 1

x2

),whereP =

∂z

∂X

∴ x2p = −P

q =∂z

∂y=∂z

∂Y

∂Y

∂y=∂z

∂Y

(1

y

)= Q

(1

y

),whereP =

∂z

∂Yi.e., yq = Q

∴ (1) becomes P 2 −Qz = z2 (2)This is of the form F (P,Q, z) = 0 [Type II]Let z = φ(X + aY ) be a solution of (2)Let u = X + aY so that z = φ(u)

Then P =dz

duand Q = a

dz

du


∴ (2) becomes

(dz

du

)2

− a

(dz

du

)z = z2(

dz

du

)2

− az

(dz

du

)− z2 = 0

dz

du=az ±

√a2z2 + 4z2

2

dz

du= z

(a±

√a2 + 4

2

)dz

z=

(a±

√a2 + 4

2

)du

Integrating, log z =

[a±

√a2 + 4

2

]u+ b

log z =

[a±

√a2 + 4

2

](X + aY ) + b

log z =

[a±

√a2 + 4

2

](1

x+ a log y

)+ b


Example 1.96. Solve z2(p2 + q2) = x2 + y2

Solution: The given equation can be written as (zp)2 + (zq)2 = x2 + y2

(1)Put Z = zk+1, where k = 1 (Type C)i.e., Z = z2

∂Z

∂x= 2z

∂z

∂x

i.e., P = 2zp, where P =∂Z

∂x⇒ zp =

P

2

Similarly, zq =Q

2, where Q =

∂Z

∂y


∴ (1)becomesP 2

4+Q2

4= x2 + y2

P 2 +Q2 = 4(x2 + y2

)P 2 − 4x2 = 4y2 −Q2

Take P 2 − 4x2 = 4y2 −Q2 = 4a2 (say) [TypeIV ]

Now P 2 − 4x2 = 4a2

⇒ P = 2√x2 + a2

∴ f1(x, a) = 2√x2 + a2

Now, 4y2 −Q2 = 4a2

⇒ Q = 2√y2 − a2

∴ f2(y, a) = 2√y2 − a2

∴ Z =

∫f1 (x, a) dx+

∫f2 (y, a) dy + b

Z = 2

∫ √x2 + a2dx+ 2

∫ √y2 − a2dx+ b

Z = 2

[x

2

√x2+a2+

a2

2sinh−1

(xa

)]+2

[y

2

√y2−a2−a

2

2cosh−1

(ya

)]+b

z2 = x√x2 + a2 + a2 sinh−1

(xa

)+ y√y2 − a2 − a2 cosh−1

(ya

)+ b


Example 1.97. Solve p2 + q2 = z2(x2 + y2)

Solution: The given equation can be written as

(pz

)2+(qz

)2= x2 + y2(

z−1p)2

+(z−1q

)2= x2 + y2 (1)

Put Z = log z [∵ k = −1] (TypeC)

∂Z

∂x=

1

z

∂z

∂x

i.e., P = z−1p, where P =∂Z

∂x

Similarly, Q = z−1q, where Q =∂Z

∂y


∴ (1)becomes P 2 +Q2 = x2 + y2

P 2 − x2 = y2 −Q2

Take P 2 − x2 = y2 −Q2 = a2 (say) [TypeIV ]

P 2 − x2 = a2

⇒ P =√x2 + a2

∴ f1(x, a) =√x2 + a2

Now, y2 −Q2 = a2

⇒ Q =√y2 − a2

∴ f2(y, a) =√y2 − a2

∴ Z =

∫f1 (x, a) dx+

∫f2 (y, a) dy + b

Z =

∫ √x2 + a2dx+

∫ √y2 − a2dy + b

Z =

[x

2

√x2+a2+

a2

2sinh−1

(xa

)]+

[y

2

√y2−a2−a

2

2cosh−1

(ya

)]+b

log z =1

2

[x√x2 + a2 + a2 sinh−1

(xa

)+ y√y2 − a2 − a2 cosh−1

(ya

)]+ b




1.x2

p+y2

q= z

[Ans:

3

2z2 =

x3

a+

y3

1− a+ b

]

2. 2x4p2−yzq−3z2=0

[Ans: log z=

a±√a2+24

2

(1

x+a log y

)+y

]

3. xp+ yq = 1

[Ans: z =

1

1 + a(log x+ a log y) + b

]

4.p

x2+

q

y2= z

[Ans: log z =

1

3 (1 + a)

(x3 + ay3

)+ b

]

5. q2y2 = z(z − px)

[Ans: log z =

−1±√1 + 4a2

2a2(log x+ alogy) + b

]


1.8 Homogeneous Linear Partial Differential Equation

A linear partial differential equation in which all the partial derivativesare of the same order is called homogeneous linear partial differentialequation. Otherwise it is called non homogeneous linear partial equation.A homogeneous linear partial differential equation of nth order with

constant coefficients is of the form

a0∂nz

∂xn+ a1

∂nz

∂xn−1∂y+ · · ·+ an

∂nz

∂yn= F (x, y) (1)

where a0, a1, . . . an are constants.This equation also can be written in the form

(a0Dn + a1D

n−1D′ + · · ·+ anD′n)z = F (x, y)

where D =∂

∂x,D′ =

∂

∂y.

i.e., f(D,D′)z = F (x, y) (2)The solution of f(D,D′)z = 0 is called the complementary function of (2).The particular solution of (2) is called the particular integral function of(2) given symbolically by

P.I. =F (x, y)

f(D,D′)

Hence the complete solution z = complementary function + particularintegral.i.e., Complete solution z = C.F. + P.I.

1.8.1 To find the complementary function (C.F)

Put D = m and D′ = 1 in f(D,D′) = 0, then we get an equation, whichis called the auxillary equation of (2).∴ the auxillary equation of (2) is f(m, 1) = 0i.e., a0m

n + a1mn−1 + · · ·+ an = 0

Let the roots of this equation be m1,m2, . . . ,mn.Case (i): If the roots are real (or imaginary) and distinct. ThenC.F. = f1(y +m1x) + f2(y +m2x) + · · ·+ fn(y +mnx)Case (ii):

(a) If any two roots are equal (ie, m1 = m2 = m) and others are distinct,Then C.F. = f1(y+mx)+xf2(y+m2x)+f3(y+m3x)+· · ·+fn(y+mnx).

(b) If any three roots are equal (i.e., m1 = m2 = m3 = m )and others aredistinct. ThenC.F. =f1(y +mx) + xf2(y +mx) + x2f3(y +mx) + f4(y +m4x)+

· · ·+ fn(y +mnx).


1.8.2 To find Particular Integral (P.I)

Type 1 : If F (x, y) = eax+by, then

P.I. =1

f(D,D′)eax+by

=1

f(a, b)eax+by provided f(a, b) 6= 0.

If f(a, b) = 0, then

P.I. = x.1

f ′(D,D′)eax+by

where f ′(D,D′) is the partial derivative of f(D,D′) w.r.t D.

Type 2 : If F (x, y) = sin(ax+ by), then

P.I. =1

f(D,D′)sin(ax+ by)

Replace D2 by −a2, D′2 by −b2 and DD′ by −ab in f(D,D′) providedthe denominator is not equal to zero.If the denominator is zero, then

P.I. = x1

f ′(D,D′)sin(ax+ by)

where f ′(D,D′) is the partial derivative of f(D,D′) w.r.t D. Note :Similar formula holds for F (x, y) = cos(ax+ by).

Type 3 : If F (x, y) = xmyn, then

P.I. =1

f(D,D′)xmyn

= [f(D,D′)]−1xmyn

Expand [f(D,D′)]−1 in ascending powers of D,D’ and then operateon xmyn.

Note : In xmyn if m > n, then try to write f (D,D′) as a function

ofD′

Dand if m < n, then try to write f (D,D′) as a function of

D

D′ .

Note :1

Ddenotes integration w.r.t ‘x’,

1

D′ denotes integration w.r.t

‘y’.

Type 4 : If F (x, y) = eax+byφ(x, y), then


P.I. =1

f(D,D′)eax+byφ(x, y)

= eax+by 1

f(D + a,D′ + b)φ(x, y)

Type 5 : If F (x, y) is any function, resolve f(D,D′) into linear factorssay (D −m1D

′), (D −m2D′), . . . , (D −mnD

′), then

P.I. =1

(D −m1D′)(D −m2D′) . . . (D −mnD′)F (x, y)

Note : (1)1

D −mD′F (x, y) =

∫F (x, c−mx)dx where y = c−mx

(2)1

D +mD′F (x, y) =

∫F (x, c+mx)dx where y = c+mx

1.8.3 Examples of Homogeneous Linear P.D.E.

Example 1.98. Solve(D2 − 5DD′ + 6D′2) z = 0

Solution: The auxillary equation is

m2 − 5m+ 6 = 0 [Replace D by m and D′ by 1]

(m− 2)(m− 3) = 0

i.e., m = 2, 3

C.F. = f1(y + 2x) + f2(y + 3x)

∴ The solution is z = f1(y + 2x) + f2(y + 3x)


∂x3− 4

∂3z

∂x2∂y+ 4

∂3z

∂x∂y2= 0

Solution: Given (D3 − 4D2D′ + 4DD′2)z = 0The auxillary equation is

m3 − 4m2 + 4m = 0

i.e., m(m− 2)2 = 0

i.e., m = 0, 2, 2

C.F. = f1(y + 0x) + f2(y + 2x) + xf3(y + 2x)

∴ The solution is z = f1(y + 0x) + f2(y + 2x) + xf3(y + 2x)

Example 1.100. Solve the equation∂2z

∂x2− 3

∂2z

∂x∂y+ 2

∂2z

∂y2= 0

Solution: Given (D2 − 3DD′ + 2D′2)z = 0The auxillary equation is


m2 + 3m+ 2 = 0

(m− 1)(m− 2) = 0

i.e., m = 1, 2

C.F. = f1(y + x) + f2(y + 2x)

∴ The solution is z = f1(y + x) + f2(y + 2x)

Example 1.101. Solve the equation(2D2 + 5DD′ + 2D′2) z = 0


2m2 + 5m+ 2 = 0

(2m+ 1)(m+ 2) = 0

i.e., m =−1

2,−2

C.F. = f1

(y − 1

2x

)+ f2(y − 2x)

∴ The solution is f1

(y − 1

2x

)+ f2(y − 2x)

Example 1.102. Solve(D2 − 2DD′ +D′2) z = ex+2y


m2 − 2m+ 1 = 0

(m− 1)2 = 0

i.e., m = 1, 1

∴ C.F. is f1 (y − x) + xf2(y − x)

P.I. =1

D2 − 2DD′ +D′2ex+2y

=1

1− 4 + 4ex+2y

= ex+2y

∴ The complete solution z = C.F. + P.I.

i.e., z = f1(y + x) + xf2(y + x) + ex+2y.

Example 1.103. Solve(D2 −D′2) z = ex+2y


m2 − 1 = 0

(m− 1)(m+ 1) = 0

i.e., m = 1,−1


∴ C.F. is f1 (y + x) + f2(y − x)

P.I. =1

D2 −D′2ex+2y

=1

1− 4ex+2y

= −1

3ex+2y


i.e., z = f1(y + x) + f2(y − x)− 1

3ex+2y.

Example 1.104. Solve(D2 − 7DD′2 + 12D′2) z = ex−y


m2 − 7m+ 12 = 0

(m− 3)(m− 4) = 0

i.e., m = 3, 4

∴ C.F. is f1(y + 3x) + f2(y + 4x)

P.I. =1

D2 − 7DD′ + 12D′2ex−y

=1

1 + 7 + 12.ex−y

=1

20ex−y


i.e., z = f1(y + 3x) + f2(y + 4x) +1

20ex−y.

Example 1.105. Solve(6D2 + 6DD′ +D′2) z = (ex + e−2y

)2Solution: The auxillary equation is

6m2 + 6m+ 1 = 0

i.e., m =−6±

√36− 24

12

m =−3±

√3

6

∴ C.F. is f1(y + (−3 +

√3

6)x) + f2(y − (

3 +√3

6)x)


P.I. =1

6D2 + 6DD′ +D′2 (ex + e−2y)2

=1

6D2 + 6DD′ +D′2 (e2x + e−4y + 2ex−2y)

=e2x+0y

6D2 + 6DD′ +D′2 +e0x−4y

6D2 + 6DD′ +D′2 +2ex−2y

6D2 + 6DD′ +D′2

=1

24e2x +

1

16e−4y − ex−2y


Example 1.106. Solve (D +D′)2z = ex−y


m2 + 2m+ 1 = 0

i.e., m = −1,−1

∴ C.F. is f1(y − x) + xf2(y − x)

P.I. =1

D2 + 2DD′ −D′2ex−y

=1

0ex−y

∴ P.I. = x1

2D + 2D′ex−y

= x1

0ex−y

∴ P.I. = x21

2ex−y


i.e., z = f1(y − x) + xf2(y − x) +x2

2ex−y

Example 1.107. Solve(D4 −D′4)2 z = ex+y


m4 − 1 = 0

i.e., (m2 − 1)(m2 + 1) = 0

i.e., m = 1,−1, i,−i

∴ C.F. is f1(y + x) + f2(y − x) + f3(y + ix) + f4(y − ix)


P.I. =1

D4 −D′4ex+y

=1

0.ex+y

∴ P.I. = x1

4D3ex+y

=1

4xex+y


i.e., z=f1(y + x) + f2(y − 4x) + f3(y + ix) + f4(y − ix) +1

4xex+y

Example 1.108. Solve(D3 − 3DD′2 + 2D′3) z = ex+y


m3 − 3m+ 2 = 0

i.e., m = 1, 1,−2

∴ C.F. is f1(y + x) + xf2(y + x) + f3(y − 2x)

P.I. =1

D3 − 3DD′2 + 2D′3ex+y

=1

0.ex+y

∴ P.I. = x1

3D2 − 3D′2ex+y = x

1

0ex+y

P.I. = x21

6Dex+y = x2

1

6ex+y


i.e., z=f1(y + x) + xf2(y + x) + f3(y − 2x) + x21

6ex+y

Example 1.109. Solve the equation(D2 + 3DD′ + 2D′2) z = ex cosh y


m2 + 3m+ 2 = 0

i.e., m = −1,−2

∴ C.F. is f1(y − x) + f2(y − 2x)


P.I. =1

D2 + 3DD′ + 2D′2ex cosh y

=1

D2 + 3DD′ + 2D′2ex

(ey + e−y

2

)=

1

2

[1

D2 + 3DD′ + 2D′2ex+y +

1

D2 + 3DD′ + 2D′2ex−y

]=

1

2

[1

6ex+y +

1

0ex−y

]P.I. =

1

2

[ex+y

6+ x

1

2D + 3D′ex−y

]=

1

2

[ex+y

6+ x

1

−1ex−y

]∴ P.I. =

1

2

[ex+y

6− xex−y

]∴ The complete solution z = C.F. + P.I.

Example 1.110. Solve(D2 + 9D′2) z = cos(2x+ 3y)


m2 + 9 = 0

m2 = −9

i.e., m = ±3i

∴ C.F. is f1(y + 3ix) + f2(y − 3ix)

P.I. =1

D2 + 9D′2 cos(2x+ 3y)

=1

−4− 81cos(2x+ 3y)

[Replacing D2 by − 4, D′2 by − 9

]= − 1

85cos(2x+ 3y)


i.e., z = f1(y + 3ix) + f2(y − 3ix)− 1

85cos(2x+ 3y)

Example 1.111. Solve(D2 − 2DD′ +D′2) z = sin(2x+ 3y)


m2 − 2m+ 1 = 0

m = 1, 1


∴ C.F. is f1(y + x) + xf2(y + x)

P.I. =1

D2 − 2DD′ +D′2 sin(2x+ 3y)

=1

−4+12−9sin(2x+3y)

[Replacing D2 → −4, D′2 → −9, DD′ → −6

]= −sin(2x+ 3y)

∴ The complete solution z = C.F. + P.I.i.e., z = f1(y + x) + xf2(y + x)− sin(2x+ 3y)


∂x2+∂2z

∂y2= cos 2x cos 3y

Solution: Given(D2 +D′2) z = cos 2x cos 3y

The auxillary equation is

m2 + 1 = 0

m2 = −1

m = i,−i

∴ C.F. is f1(y + ix) + f2(y − ix)

P.I. =1

D2 +D′2 cos 2x cos 3y

=1

D2 +D′21

2(cos(2x+ 3y) + cos(2x− 3y))

=1

2

(1

D2 +D′2 cos(2x+ 3y) +1

D2 +D′2 cos(2x− 3y)

)=

1

2

(1

−4− 9cos(2x+ 3y) +

1

−4− 9cos(2x− 3y)

)= − 1

26(cos(2x+ 3y) + cos(2x− 3y))

= − 1

26

(2 cos

(4x

2

)+ cos

(6y

2

))= − 1

13(cos 2xcos3y)

∴ The complete solution z = C.F. + P.I.Aliter :


P.I. =1

f(D2 +D′2)sin ax sin by (or) cos ax cos by

=1

f(−a2,−b2)sin ax sin by (or) cos ax cos by provided f

(−a2,−b2

)6= 0

∴ P.I. =1

D2 +D′2 cos 2x cos 3y

=1

−4− 9cos 2x cos 3y

∴ P.I. = − 1

13cos 2x cos 3y

Example 1.113. Solve(D2 +DD′ − 6D′2) z = cos(2x+ y)


m2 +m− 6 = 0

i.e., (m− 2)(m+ 3) = 0

i.e., m = 2,−3

∴ C.F. is f1(y + 2x) + f2(y − 3x)

P.I. =1

D2 +DD′ − 6D′2 cos(2x+ y)

=1

−4−2+6cos(2x+y)

[Replacing D2 → −4, D′2 → −1, DD′ → −2

]=

1

0cos(2x+ y)

∴ P.I. = x1

2D +D′ cos(2x+ y)

P.I. = x2D −D′

4D2 −D′2 cos(2x+ y)

= x2D −D′

−16 + 1cos(2x+ y)

[Replacing D2 by − 4, D′2 by − 1

]= − x

15[2D cos(2x+ y)−D′ cos(2x+ y)]

= − x

15[−4 sin(2x+ y) + sin(2x+ y)]

=x

5sin(2x+ y)


Example 1.114. Write the P.I. of(D2 +DD′) z = sin(x+ y)


Solution:

P.I. =1

D2 +DD′ sin(x+ y)

=1

−1− 1sin(x+ y)

[Replacing D2 by − 1, DD′ by − 1

]∴ P.I. = −1

2sin(x+ y)

Example 1.115. Solve the equation∂2z

∂x2− ∂2z

∂x∂y= sin x cos 2y

Solution: Given(D2 −DD′) z = sin x cos 2y

The auxillary equation is

m2 −m = 0

i.e., m = 0, 1

∴ C.F. = f1(y + 0x) + f2(y + x)

= f1(y) + f2(y + x)

P.I. =1

D2 −DD′ sinxcos2y

=1

D2 −DD′ .1

2(sin(x+ 2y) + sin(x− 2y))

=1

2

[1

D2 −DD′ sin(x+ 2y) +1

D2 −DD′ sin(x− 2y)

]=

1

2

[1

−1 + 2sin(x+ 2y) +

1

−1− 2sin(x− 2y)

]=

1

2

[sin(x+ 2y)− 1

3sin(x− 2y)

]∴ The complete solution z = C.F. + P.I.

Example 1.116. Solve(D2 + 3DD′ − 4D′2) z = cos(2x+ y)


m2 + 3m− 4 = 0

i.e., (m− 1)(m+ 4) = 0

i.e., m = 1,−4

∴ C.F. = f1(y + x) + f2(y − 4x)


P.I. =1

D2 + 3DD′ − 4D′2 sin y

=1

D2 + 3DD′ − 4D′2 sin(0x+ y)

=1

0 + 0 + 4sin(0x+ y)

[Replacing D2 by 0, D′2 by − 1, DD′ by 0

]P.I. =

1

4sin y


Example 1.117. Solve(D3 +D2D′ −DD′2 −D′3) z = 3 sin(x+ y)


m3 +m2 −m− 1 = 0

i.e., m = 1,−1,−1

∴ C.F. = f1(y + x) + f2(y − x) + xf3(y − x)

P.I. =1

D3 +D2D′ −DD′2 −D′33 sin(x+ y)

=1

−D −D′ +D +D′3 sin(x+ y)[Replacing D2 by − 1, D′2 by − 1, DD′ by − 1

]=

1

03 sin(x+ y)

∴ P.I. = x1

3D2 + 2DD′ −D′23 sin(x+ y)

= x1

−3− 2 + 13 sin(x+ y)[

Replacing D2 by − 1, D′2 by − 1, DD′ by − 1]

= −3

4x sin(x+ y)


Example 1.118. Solve(D3 +D2D′ − 2DD′2) z = sin(x− 2y)


m3 +m2 − 2m = 0

m(m2 +m− 2) = 0

i.e., m = 0, 1,−2


∴ C.F. = f1(y) + f2(y + x) + f3(y − 2x)

P.I. =1

D3 +D2D′ − 2DD′2sin(x− 2y)

=1

−D −D′ + 8Dsin(x− 2y)[

Replacing D2 by − 1, D′2 by − 4]

=1

7D −D′ sin(x− 2y)

=7D +D′

−49 + 4sin(x− 2y)[

Replacing D2 by − 1, D′2 by − 4]

= − 1

45[7D sin(x− 2y) +D′ sin(x− 2y)]

= − 1

45[7 cos(x− 2y)− 2 cos(x− 2y)]

P.I. = −1

9cos(x− 2y)


i.e., z = f1(y) + f2(y + x) + f3(y − 2x)− 1

9cos(x− 2y).

Example 1.119. Solve(D2 − 3DD′ + 2D′2) z = sin x cos y


m2 − 3m+ 2 = 0

(m− 1)(m− 2) = 0

i.e., m = 1, 2

∴ C.F. = f1(y + x) + f2(y + 2x)

P.I. =1

D2 − 3DD′ + 2D′2 sinx cos y

=1

D2 − 3DD′ + 2D′21

2[sin(x+ y) + sin(x− y)]

=1

2

[1

D2 − 3DD′ + 2D′2 . sin(x+ y) +1

D2 − 3DD′ + 2D′2 sin(x− y)

]=

1

2

[1

−1 + 3− 2. sin(x+ y) +

1

−1− 3− 2sin(x− y)

]


=1

2

[1

0. sin(x+ y)− 1

6sin(x− y)

]P.I. =

1

2

[x

1

2D − 3D′ . sin(x+ y)− 1

6sin(x− y)

]=

1

2

[x

2D + 3D′

4D2 − 9D′2 . sin(x+ y)− 1

6sin(x− y)

]=

1

2

[x2D + 3D′

−4 + 9. sin(x+ y)− 1

6sin(x− y)

]=

1

2

[x

5[2D sin(x+ y) + 3D′ sin(x+ y)]− 1

6sin(x− y)

]=

1

2

[x

5[2 cos(x+ y) + 3 cos(x+ y)]− 1

6sin(x− y)

]P.I. =

1

2

[x cos(x+ y)− 1

6sin(x− y)

]


i.e., f1(y + x) + f2(y + 2x) +1

2

[x cos(x+ y)− 1

6sin(x− y)

].

Example 1.120. Solve(D2 + 3DD′ + 2D′2) z = x2y2


m2 + 3m+ 2 = 0

(m+ 1)(m+ 2) = 0

i.e., m = −1,−2

∴ C.F. = f1(y − x) + f2(y − 2x)

P.I. =1

D2 + 3DD′ + 2D′2x2y2

=1

D2[1 + 3D′

D + 2D′2

D2

]x2y2=

1

D2

[1 +

(3D′

D+

2D′2

D2

)]−1

x2y2


=1

D2

[1−

(3D′

D+

2D′2

D2

)+

(3D′

D+

2D′2

D2

)2

− . . .

]x2y2

=1

D2

[1− 3D′

D− 2D′2

D2+

9D′2

D2

]x2y2

=1

D2

[1− 3D′

D+

7D′2

D2

]x2y2

=1

D2

[x2y2 − 3

D(2x2y) +

7

D2(2x2)

]=

1

D2(x2y2)− 1

D3(6x2y) +

1

D4(14x2)

=1

12x4y2 − 1

10x5y +

7

180x6

=1

2

[x

5[2 cos(x+ y) + 3 cos(x+ y)]− 1

6sin(x− y)

]P.I. =

1

2

[x cos(x+ y)− 1

6sin(x− y)

](∵

1

Dnxm =

xm+n

(m+ 1)(m+ 1) . . . (m+ n)

)∴ The complete solution z = C.F. + P.I.

Example 1.121. Solve(D2 − 2DD′ + 2D′2) z = x2y2


m2 − 2m+ 2 = 0

m =2±

√4− 8

2i.e., m = 1± i

∴ C.F. = f1(y + (1 + i)x) + f2(y + (1− i)x)

P.I. =1

D2 − 2DD′ + 2D′2x3y

=1

D2[1− 2D′

D + 2D′2

D2

]x3y=

1

D2

[1−

(2D′

D− 2D′2

D2

)]−1

x3y

=1

D2

[1 +

2D′

D

]x3y =

1

D2

[x3y +

2

D(x3)

]=

1

D2(x3y) +

2

D3(x3) =

x5y

20+x6

60



Example 1.122. Solve(D2 + 6DD′ + 5D′2) z = xy4


m2 + 6m+ 5 = 0

(m+ 1)(m+ 5) = 0

i.e., m = −1,−5

∴ C.F. = f1(y − x) + f2(y − 5x)

P.I. =1

D2 + 6DD′ + 5D′2xy4 =

1

5D′2[

D2

5D′2 +65DD′ + 1

]xy4=

1

5D′2

[1 +

(6D

5D′ +D2

5D′2

)]−1

xy4

=1

5D′2

[1−

(6D

5D′ +D2

5D′2

)+

(6D

5D′ +D2

5D′2

)2

− . . .

]xy4

=1

5D′2

[1− 6D

5D′

]xy4 =

1

5D′2

[xy4 − 6y4

5D′

]=

1

5D′2 (xy4)− 6

25D′3 (y4)

=xy6

150− y7

875


Example 1.123. Find the general integral of the equation∂2z

∂x2+ 3

∂2z

∂x∂y+ 2

∂2z

∂y2= x+ y

Solution: Given(D2 + 3DD′ + 2D′2) z = x+ y The auxillary equation is

m2 + 3m+ 2 = 0

(m+ 1)(m+ 2) = 0

i.e., m = −1,−2

∴ C.F. = f1(y − x) + f2(y − 2x)


P.I. =1

D2 + 3DD′ + 2D′2 (x+ y)

=1

D2(1 + 3D′

D + 2D′2

D2

)(x+ y)

=1

D2

[1 +

(3D′

D+

2D′2

D2

)]−1

(x+ y)

=1

D2

[1−

(3D′

D+

2D′2

D2

)+

(3D′

D+

2D′2

D2

)2

− . . .

](x+ y)

=1

D2

[1− 3D′

D

](x+ y) =

1

D2

[x+ y − 3

D

]=

1

D2(x) +

1

D2(y)− 1

D3(3)

=x3

6+x2y

2− x3

2


Example 1.124. Solve(D2 + 4DD′ − 5D′2) z = x+ y2


m2 + 4m− 5 + 2 = 0

(m+ 5)(m− 1) = 0

i.e., m = 1,−5

∴ C.F. = f1(y + x) + f2(y − 5x)

P.I. =1

D2 + 4DD′ − 5D′2 (x+ y2) =1

−5D′2(−D2

5D′ − 4D5D′ + 1

)(x+ y2)

= − 1

5D′2

[1−

(4D

5D′ +D2

5D′

)]−1

(x+ y2)

= − 1

5D′2

[1 +

(4D

5D′ +D2

5D′

)+

(4D

5D′ +D2

5D′

)2

− . . .

](x+ y2)

= − 1

5D′2

[1 +

4D

5D′

](x+ y2) = − 1

5D′2

[x+ y2 +

4

5D′

]= −1

5

1

D′2 (x) +1

D′2 (y2) +

4

5

1

D′3 (1) = −1

5

(xy2

2+y4

12+

2y3

15

)∴ The complete solution z = C.F. + P.I.

i.e., z = f1(y + x) + f2(y − 5x)− 1

5

(1

2xy2 +

1

12y4 +

2

15y3).


Example 1.125. Solve(D2 − 4DD′ + 4D′2) z = ex−2y cos(2x− y)


m2 − 4m+ 4 = 0

i.e., m = 2, 2

∴ C.F. = f1(y + 2x) + xf2(y + 2x)

P.I. =1

D2 − 4DD′ + 4D′2ex−2y cos(2x− y)

= ex−2y 1

(D + 1)2 − 4(D + 1)(D′ − 2) + 4(D′ − 2)2cos(2x− y)

= ex−2y 1

D2 + 10D − 4DD′ − 20D′ + 4D′2 + 25cos(2x− y)

= ex−2y 1

−4 + 10D − 8− 20D′ − 4 + 25cos(2x− y)

= ex−2y 1

10D − 20D′ + 9cos(2x− y)

= ex−2y ((10D − 20D′)− 9) cos(2x− y)

(10D − 20D′)2 − 81

= ex−2y (10D − 20D′ − 9) cos(2x− y)

100D2 − 400DD′ + 400D′2 − 81

= ex−2y (10D − 20D′ − 9)

−400− 800− 400− 81cos(2x− y)

= −ex−2y

1681(−20 sin(2x− y)− 20 sin(2x− y)− 9 cos(2x− y))

P.I. =1

1681ex−2y [40 sin(2x− y) + 9 cos(2x− y)]


Example 1.126. Solve(D2 −DD′ − 2D′2) z = (y − 1)ex


m2 −m− 2 = 0

(m− 2)(m+ 1) = 0

i.e., m = 2,−1

∴ C.F. = f1(y + 2x) + f2(y − x)


P.I. =(y − 1)ex

D2 −DD′ − 2D′2

= ex1

(D + 1)2 − (D + 1)D′ − 2D′2 (y − 1)

= ex1

D2 + 2D + 1−DD′ −D′ − 2D′2 (y − 1)

= ex[1 +

(D2 + 2D −DD′ −D′ − 2D′2)]−1

(y − 1)

= ex[1−

(D2 + 2D −DD′ −D′ − 2D′2)+ . . .

](y − 1)

= ex [1 +DD′ +D′] (y − 1)

= ex [y − 1 +D(1) + 1]

= exy

∴ The complete solution z = C.F. + P.I.i.e., z = f1(y + 2x) + f2(y − x) + yex.

Example 1.127. Solve(D2 +DD′ − 6D′2) z = y cosx


m2 +m− 6 = 0

(m+ 3)(m− 2) = 0

i.e., m = 2,−3

∴ C.F. = f1(y + 2x) + f2(y − 3x)

P.I. =1

D2 +DD′ − 6D′2y cosx

=1

(D − 2D′)(D + 3D′)y cosx

=1

D − 2D′

∫(c+ 3x) cos xdx where y = c+ 3x

=1

D − 2D′ [(c + 3x) sin x+ 3 cos x]

=1

D − 2D′ [(y − 3x+ 3x) sin x+ 3 cos x] ∵ c = y − 3x

=1

D − 2D′ (y sinx+ 3 cos x)

=

∫[(c1 − 2x) sin x+ 3 cos x]dx where y = c1 − 2x

= [(c1 − 2x)(− cosx)− (−2)(− sinx)] + 3 sinx

= −(c1 − 2x) cos x− 2 sin x+ 3 sin x

= −y cosx+ sinx

P.I. = sin x− y cosx





1.∂2z

∂x2+

∂2z

∂x∂y− 2

∂2z

∂y2= 0 [Ans: z = f1(y + x) + f2(y − 2x)]

2. 2∂2z

∂x2+ 5

∂2z

∂x∂y+ 2

∂2z

∂y2= 0 [Ans: z = f1(y − 2x) + f2(2y − x)]

3.(D3 − 2D2D′) 2e2x + 3x2y[

Ans: z = f1(y) + xf2(y) + f3(y + 2x)1

4e2x +

1

60

(3x5y + x6

)]4. r + s− 6t = cos(2x+ y)[

Ans: z = f1(y − 3x) + f2(y + 2x) +1

5x (sin 2x+ y)

]5. r + 2s+ t = 2(y − x) + sin(x− y)[

Ans: z = f1(y − x) + xf2(y − x) + x2y − x3 +1

2x2 (sinx− y)

]6.(D2 + 2DD +D′2) z = 2 cos y − x sin y

[Ans: z = f1(y − x) + xf2(y − x) + x sin y]

7.(D2 + 3DD′ + 2D′2) z = x+ y[

Ans: z = f1 (y − x) + f2 (y − 2x)− x3

3+x2y

2

]8.(2D2 − 2DD′ +D′2) z = 2e3y + ex+y + y2[Ans: z = f1

(y +

1 + i

2x

)+ f2

(y +

1− i

2x

)+

2

9e3y + ex+y +

x2y2

2

]9.(D2 −DD′) z = cos x cos 2y[

Ans: z = f1(y) + f2(y + x) +1

2cos (x+ 2y)− 1

6cos(x− 2y)

]10.

(D2 − 2DD′) z = e2x + x3y[

Ans: z = f1 (y) + f2 (y + 2x)1

4e2x +

x5y

20+x6

60

]


1.9 Non-Homogeneous Linear P.D.E.

A linear partial differential equation in which all the partial derivativesare not of the same order is called non-homogeneous linear partialdifferential equation.Consider the equation f(D,D′)z = F (x, y) (1)If f(D,D′) is not homogeneous, then (1) is the non-homogeneous linear

equation. As in the of homogenous linear equation , the completesolution is

z = C.F. + P.I.

[Finding P.I. of Non-Homo. P.D.E. is same as in Homo. P.D.E.]

1.9.1 To find the complementary function (C.F.)

Case(1): If (D −m1 − C1) z = 0, then C.F. = eC1xf1 (y +m1x)

Case(2): If (D′ −m1D − C1) z = 0, then C.F. = eC1yf1 (x+m1y)

Case(3): If (D +m1D′ + C1) z = 0 ⇒ [D − (−m1)D

′ − (−C1)] z = 0,

then C.F. = e−C1x (f1(y −m1x))

Case(4): If (D′ +m1D + c1) z = 0 ⇒ [D′ − (−m1)D − (−c1)] z = 0, then

C.F. = e−C1yf1 (x−m1y)

Case(5): If (D −m1D′ − c1) (D −m2D

′ − c2) z = 0, then

C.F. = eC1xf1 (y +m1x) + eC2xf2 (y +m2x)

Case(6): If (D −m1D′ − c1)

2z = 0, then

C.F. = eC1xf1 (y +m1x) + xeC1xf2 (y +m1x)

Case(7): Put D = h and D′ = k in f (D,D′) = 0, then we get an

Aux. eqn.

f(h, k) = 0 (2)

Solve (2) and find the value of h interm of k or k interm of h.

Let the n values of h be denoted by f1(k), f2(k), f3(k), . . . , fn(k).

Then

C.F. =∑

c1ef1(k)x+ky +

∑c2ef2(k)x+ky

+∑

c3ef3(k)x+ky + ...+

∑cnefn(k)x+ky


1.9.2 Examples of Non-Homogeneous Linear P.D.E.

Example 1.128. Solve (D − 2D′ − 1) z = 0

Solution: C.F. = exf1(y + 2x) [∵ C1 = 1,m1 = 2 as case(1)]

Example 1.129. Solve (D′ − 2D − 3) z = 0

Solution: C.F. = e3yf1(x+ 2y) [∵ C1 = 3,m1 = 2 as case(2)]

Example 1.130. Solve (D + 3D′ + 1) z = 0

Solution: Given (D + 3D′ + 1) z = 0 ⇒ [D − (−3)D′ − (−1)]C.F. = e−xf1(y − 3x) [∵ C1 = 1,m1 = 3 as case(3)]

Example 1.131. Solve (D′ + 5D + 6) z = 0

Solution: Given (D′ + 5D + 6) z = 0 ⇒ [D′ − (−5)D − (−6)]C.F. = e−6yf1(x− 5y) [∵ C1 = 6,m1 = 5 as case(4)]

Example 1.132. Solve (D − 4D′ − 1) (D′ + 5D − 3) z = 0

Solution: Given (D − 4D′ − 1) [D′ − (−5)D − 3] z = 0C.F. = exf1(y + 4x) + e3yf2(x− 5y)

Example 1.133. Solve (D − 2D′ + 2)3z = 0

Solution: Given (D − 2D′ + 2)3z = 0 ⇒ [D − 2D′ − (−2)]

3z = 0

C.F. = e−2xf1(y + 2x) + xe−2xf2(y + 2x) + x2e−2xf3(y + 2x)

Example 1.134. Solve(D2 −DD′ +D′ − 1

)z = 0

Solution:Consider

(D2 −DD′ +D′ − 1

)=(D2 − 1

)−DD′ +D′

= (D − 1)(D + 1)−D′(D − 1)

= (D − 1)(D −D′ + 1)

= [D − 0D′ − 1][D −D′ − (−1)]

C.F. = exf1(y) + e−xf2(y + x)

(OR)

Put D = h and D′ = k in

f (D,D′) = 0 ⇒(D2 +DD′ +D′ − 1

)z = 0

then we get an Aux. eqn.f(h, k) = 0 ⇒

(h2 + hk + k − 1

)= 0 (1)

Solve (1) and find the value of h interm of k or k interm of h.h2 + hk + k − 1 = 0


Now, h =k ±

√k2 − 4(k − 1)

2=k ±

√(k − 2)2

2=k ± (k − 2)

2i.e., h = k − 1, 1

C.F. =∑

c1e(k−1)x+ky +

∑c2e1x+ky

=∑

c1e−xek(x+y) +

∑c2ex+ky

= e−xf1(x+ y) + exf2(y)

Example 1.135. Solve(D2 −D′2 − 3D + 3D′) z = 0

Solution: Put D = h and D′ = k in

f (D,D′) = 0 ⇒(D2 −D′2 − 3D + 3D′) z = 0

then we get an Aux. eqn.f(h, k) = 0 ⇒

(h2 − k2 − 3h+ 3k

)= 0 (1)

Solve (1) and find the value of h interm of k(h2 − k2 − 3(h− k)

)= 0

(h− k)(h+ k)− 3(h− k) = 0

(h− k)(h+ k − 3) = 0

h = k, 3− k

C.F. =∑

c1ekx+ky +

∑c2e(3−k)x+ky =

∑c1ek(x+y) +

∑c2e3x+k(y−x)

= f1(x+ y) + e3xf2(y − x)

Example 1.136. Solve(2D2 −DD′ −D′2 + 6D + 3D′) z = 0

Solution: Put D = h and D′ = k

A.E.(2h2 − hk − k2 + 6h+ 3k

)= 0

i.e.,[2h2 + h(6− k) + (3k − k2)

]= 0

h =(k − 6)±

√(6− k)2 − 8(3k − k2)

4

=(k−6)±

√k2−12k+36−24k+8k2

4=

(k − 6)±√9k2 − 36k + 36

4

=(k − 6)± (3k − 6)

4=

(k−6)+(3k−6)

4,(k−6)−(3k−6)

4

= k − 3,−k2


C.F. =∑

c1e(k−3)x+ky +

∑c2e

−k

2x+ ky

=∑

c1e−3xek(x+y) +

∑c2e

k

(y−x

2

)

= e−3xf1(x+ y) + f2

(y − x

2

)1.10 Assignment III[Partial Differential Equations]

1. (i) Form a partial differential equation by eliminating a and b from theexpression (x− a)2 + (y − b)2 + z2 = c2.(ii) Find the partial differential equation of all planes which are at aconstant distance a1 from the origin.

2. (i) Form the partial differential equation by eliminating arbitraryfunction from z = xf(2x+ y) + g(2x+ y).(ii) Form the PDE by eliminating the arbitrary function φ fromφ(x2 + y2 + z2, ax+ by + cz

)= 0.

3. Solve the following:(i) p(1 + q)z = qz

(ii) x2p+ y2q = 0 (iii) xp+ yq = 0

4. Solve the following:(i) z = px+ qy + log pq (ii) z = px+ qy + pq

(iii) z = px+ qy +√

1 + p2 + q2 (iv) z = px+ qy + p2q2

5. Solve the following:(i) p(1 + q) = qz (ii) x4p2 − yzq = z2

(iii) x4p2 + y2qz = 2z2 (iv) z2(p2x2 + q2

)= 1

6. Solve the following:(i) p2 + q2 = x2 + y2 (ii) z2

(p2 + q2

)= x2 + y2

(iii) 4z2q2 = y + 2zp− x (iv) p2 + q2 = z2(x2 + y2

)7. Solve the following:

(i) x2(y − z)p = y2(z − x)q = z2(x− y)(ii)

(x2 − yz

)p+

(y2 − zx

)q = z2 − xy

(iii) (y − xz)p+ (yz − x)q = (x+ y)(x− y)


(iv)(x2 − y2 − z2

)p+ 2xyq = 2zx

(v) x(z2 − y2

)p+ y

(x2 − z2

)q = z

(y2 − x2

)(vi) x

(y2 + z

)p+ y

(x2 + z

)q = z

(x2 − y2

)(vii) (3z − 4y)p+ (4x− 2z)q = (2y − 3x)(viii) (x− 2z)p+ (2z − y)q = y − x(ix) x(y − z)p+ y(z − x)q = z(x− y)

8. Solve the following:(i)(D3 +D2D′ −DD′2 −D′3) z = e2x+y + cos(x+ y)

(ii)∂3z

∂x3− 2

∂3z

∂x2∂y= ex+2y + 4 sin (x+ y)

(iii)(D3 − 7DD′2 − 6D′3) z = cos (x+ 2y) + 4

(iv)(D2 + 2DD′ +D′2) z = sinh(x+ y) + ex+2y

(v)(D2 + 2DD′ +D′2) z = x2y + ex−y

(vi)(D2 −DD′ − 2D′2) z = 2x+ 3y + e3x+4y

(vii)(D2 −D′2) z = ex−y sin(x+ 2y)

(viii)(D2 − 5DD′ + 6D′2) z = y sinx

(ix) r + s− 6t = y cosx

9. Solve the following:(i)(D2 −D′2 − 3D + 3D′) z = xy + 7

(ii)(D2 + 2DD′ − 2D − 2D′2) z = sin(x+ 2y)

(iii)(2D2 −DD′ −D′2 + 6D + 3D′) z = xey

(iv)(D2 + 2DD′ +D′2 − 2D − 2D′) z = e3x+y + 4

(v)(D2 +D′2 + 2DD′ + 2D + 2D′ + 1

)z = e2x+y

1 partial di erential equations(p.d.e.) · 2019-08-15 · 1 partial di erential equations(p.d.e.)...

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