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----------UNIT-1V---------- ORDINARY DIFFERENTIAL EQUATIONS
OF FIRST ORDER AND FIRST DEGREE LINEAR DIFFERENTIAL EQUATIONS OF
SECOND AND HIGHER ORDER
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INDEX
S.no TOPICSTOPICS Page.no
1 Introduction,Exact differential equations
2 Linear and Bernoulli’s equations,Orthogonal trajectories
3 Newton’s law of cooling and growth and decay
4 Introduction, Complementary function, Particular Integrals
5 Particular Integrals, Cauchy’s
6 Variation of Parameters
7 Simple harmonic motion,Electric circuits, length of beams
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INTRODUCTION:
An equation involving a dependent variable and its derivatives with respect to one or more independent variables is called a Differential Equation.
Example 1: y ' '+ 2y = 0
Example 2: y2 – 2y1 +y = 23
Example 3: d2 yd x2 + dydx−¿ y =1
TYPES OF A DIFFERENTIAL EQUATION:
ORDINARY DIFFERENTIAL EQUATION: A differential equation is said to be ordinary, if the derivatives in the equation are ordinary derivatives.
Example: : d2 yd x2 + dydx +¿ y =1
PARTIAL DIFFERENTIAL EQUATION: A differential equation is said to be partial if the derivatives in the equation have reference to two or more independent variables.
Example: ∂4 y∂ x4 +∂ y
∂ x + y = 1
DEFINITIONS:
ORDER OF A DIFFERENTIAL EQUATION: A differential equation is said to be of order n, if thenthderivative is the highest derivative in that equation.
Example: Order of d2 yd x2 + dydx +¿ y =2 is 2 .
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DEGREE OF A DIFFERENTIAL EQUATION: If the given differential equation is a polynomial in y(n), then the highest degree of y(n) is defined as the degree of the differential equation.
Example: Degree of( dydx )4
+ y= 0 is 4 .
Formation of a D.E:
A D.E can be formed by eliminating arbitrary constants from the given D.E by using Differentiation concept. If the given equation contains n arbitrary constants then differentiating it n times successively and eliminating n arbitrary constants we get the corresponding D.E.
SOLUTION OF A DIFFERENTIAL EQUATION:
SOLUTION: Any relation connecting the variables of an equation and not involving their derivatives, which satisfies the given differential equation is called a solution.
GENERAL SOLUTION: A solution of a differential equation in which the number of arbitrary constant is equal to the order of the equation is called a general or complete solution or complete primitive of the equation.
Example: y = Ax + B
PARTICULAR SOLUTION: The solution obtained by giving particular values to the arbitrary constants of the general solution, is called a particular solution of the equation.
Example: y = 3x + 5
PROBLEMS
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(1).Form the differential equation by eliminating arbitrary constants y=a x3+b x2 . Answer: Given equations is y = ax3 +b x2 -------------------(i) Differentiating equation (i) w.r.t ‘ x ‘ We get y1 = 3ax2
+2bx -------------------------------------------(ii) Again differentiating w.r.t ‘ x ‘=>y2 =6ax +2b -------------------------------------------------------(iii)
Now we eliminating arbitrary constants a,b from (i) ,(ii),(iii)
| y −x3 −x2
y 1 −3 x2 −2 xy 2 −6 x −2 |=0
=>x2 y2 -4xy1 + 6y = 0 is the required differential equation .
(2). Form the differential equation by eliminating arbitrary constants y = Ae−3 x+Be2x .
Answer: Given y = Ae−3 x+Be2x --------------(i)
Differentiating equation (i) w.r.t ‘x’
=> y1=−3 Ae−3 x+2 Be2 x-----------------(ii)
=> y2=9 Ae−3 x+4 Be2 x-----------------(iii)
Now eliminating A and B from (i),(ii) and (iii)
=> | e−3 x e2x − y−3e−3 x 2e2x − y1
9e−3 x 4 e2 x − y2|=0
=> (-1)e−3 xe2x| 1 1 y−3 2 y1
9 4 y2|=0
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=> | 1 1 y−3 2 y1
9 4 y2|=0 => y2+ y1−6 y=0 is the required differential equation.
(3) .Form the differential equation by eliminating arbitrary constants xy= Aex+Be−x .
Answer: Given xy = Aex+Be−x --------------(i)
Differentiating equation (i) w.r.t ‘x’
=> x y1+ y=A ex−Be−x-----------------(ii)
=> xy2+ y1+ y1=A ex+B e−x-----------------(iii)
Now eliminating A and B from (i),(ii) and (iii)
=> xy2+2 y1− y=0 by (i)
is the required differential equation.
(4). Form the differential equation by eliminating arbitrary constants
from log( yx ¿=cx.
Answer: Given log( yx ¿=cx.---------------(i)
=> logy – logx = cxNow differentiating w.r.t ‘x’
=> 1ydydx
−1x=c ---------------------------------(ii)
Now ‘c’ value substituting in (i)
we get log( yx ¿= xydydx
−1 is the required differential equation.
(5). Form the differential equation by eliminating arbitrary constants y= c esin−1x .
Answer: Given y= c esin−1x -------(i)
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Where c is the arbitrary constant Now differentiating equation (i) w.r.t ‘x’
We get dydx
=c esin−1 x 1√1−x2
= y√1−x2 by (i)
¿>√1−x2 dydx
− y=0 is the required differential equation.
(6). Form the differential equation by eliminating arbitrary constants A and B from y = ex (Acosx+Bsinx ).Answer: Given y = ex (Acosx+Bsinx ) ----------------(i)Equation (i) differentiating w.r.t ‘x’
dydx
=ex (Acosx+Bsinx )+ex (−Asinx+Bcosx )
dydx
= y+ex (−Asinx+Bcosx ) ------------------(ii)
Again differentiating w.r.t ‘x’
d2 yd x2 =
dydx
+ex (−Acosx−Bsinx )+ex (−Asinx+Bcosx )
¿> d2 yd x2 =dy
dx− y+ dy
dx− y by (i) and (ii)
∴ d2 yd x2 −2 dy
dx+2 y=0 is the required differential equation.
(7). Find the equation of the curve in which the length of the subnormal is proportional to the square of the abscissa.
Answer: Given length of subnormal ∝ x2
i.e y dydx=k x2 where k is aproportionality constant
we apply the variables and separable method
=> ydy = k x2 dx
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integrating on both sides
=> ∫ ydy = ∫ k x2 dx => y2
2=k x
3
3+c
3 y2=2kx3+c is the required solution.
(8). Find the differential equation of all circles whose radius is r.
Answer:Given family of circles is ( x−a )2+( y−b )2= r2 ----------------(i)
Now differentiating equation (i) w.r.t ‘x’
2(x-a) +2(y-b) dydx = 0 ------------------------------------(ii)
1 +(y-b)d2 yd x2 +( dydx )
2
=0
y- b = - [1+( dydx )
2]d2 yd x2
-----------------------------------------(iii)
from (ii) and (iii)
we have x- a −¿ [1+( dydx )
2] dydxd2 yd x2
= 0 => x-a = [1+( dydx )
2] dydxd2 yd x2
------------(iv)
now (x-a) and (y-b) values sub in (i)
[1+( dydx )2]
3
= r2 ( d2 yd x2 )
2
is the required differential equations.
(9). Find the equation of the curve ,in which the length of the subnormal is proportional to the square of the ordinate.
Answer: Given subnormal ∝ y2
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=> y dydx=k y2
=> dydx=ky => 1ydy=k dx Integrating ,
∫ 1ydy=∫ k dx+c => log y = kx +x
Y = ekx . ec => y = c ekx its required solution.
D.E of first order and first degree :
1) Variable Seperable Method:
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PROBLEMS
(1). Solve the differential equation dydx=e2 x−3 y+x2 e−3 y .
Answer: Given dydx=e2x−3 y+x2 e−3 y
dydx=e2x e−3 y+x2 e−3 y
dydx=e−3 y (e2 x+x2 )= e2x+x2
e−3 y
Applying variable and separable method => e−3 y dy=(e2 x+x2 )dx
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Integrating,=> ∫ e−3 ydy=∫ (e2 x+x2 )dx+c∴ (3 e2x+2 x3−2e3 y )=c is the required solution.(2). solve the differential equation ex− y dx+e y−x dy=0.Answer: given differential equation ex− y dx+e y−x dy=0.
=> ex
e y dx=−e y
e x dy => e2x dx=−e2 y dy
Integrating on bothsides=> ∫ e2 xdx=−∫ e2 y dy => e2x+e2 y=c is the required solution.(3).Solve the differential equation (x2+2 y2 )dx−xydy=0.
Answer: Given equation can be written as dydx= x2+2 y2
xy --------(i)
Equation (i) is a homogeneous differential equation
Let y = vx => dydx=v+x dvdx
From (i) v+x dvdx=x2+2v2 x2
v x2 =1+2v2
v
=> x dvdx= x2+2v2 x2
v x2 =1+2v2
v−v=1+v2
v
Applying variable and separable method
=> v
1+v2 dv=1xdx
Integrating on bothsides
∫ 2v1+v2 dv=2∫ 1
xdx => log(1+v2¿=2logx+c
=> log( x2+ y2
x2 )=log x2+logc=log x2c
=> x2+ y2
x2 = x2c∴ x2+ y2−x4 c=0 is the required solution.(4). Solve the differential equation (x2 y-2xy2 ¿dx=9(x¿¿3−3 x2 y)dy .¿
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Answer: Given equation can be written as dydx= x2 y−2 x y2
x3−3 x2 y ---------(i)
Equation (i) is a homogeneous differential equation .
Put y =vx => dydx=v+x dvdx
From (i) v+x dvdx=v x3−2 x3 v2
x3−3x3v= v−2v2
1−3v
¿>x dvdx
= v−2v2
1−3v−v = v2
1−3v
Applying variable and separable method,
¿> 1−3 vv2 dv =dx
x , integrating
We get , ∫ 1−3 vv2 dv =∫ dx
x +c
(5). Solve the differential equation x3 dydx
= y3+ y2 √ y2−x2 .
Answer: Given equation x3d ydx = y3 + y2√ y2−x2
The above equation can be written asd ydx ¿ y3+ y2√ y2−x2
x 3 is the
homogeneous equation. ----------------------------------------(i)
Put y = vx => d ydx = v + x d vdx
From (i) v + x d vdx = v3 x3+v2 x2 √v2x2−x2
x3
=> v + x d vdx =v3+v2√ v2−1
=> x d vdx =v3+v2√ v2−1 –v
=> x d vdx =v (v¿¿2−1)+v2 √v2−1¿
=> x d vdx =v√(v¿¿2−1)[√v2−1+v ]¿
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Applying the variables and separable method dv
v √(v¿¿2−1)[√v2−1+v ]¿ = d xx
Integrating on both sides
=> ∫ √v2−1−v dvv √(v¿¿2−1)[v2−1−v2]
¿ = ∫ d xx
=>∫ v−√v2−1dvv √(v¿¿2−1)
¿ = logx +logc
=>∫ dv√(v¿¿2−1)
¿ - ∫ d vv = logx +logc
=>cosh−1 v –logv =logxc (or) log(√v2−1+v)-logv =logxc (or) =>(√v2−1+v)=vxc Therefore y+ √ y2−x2=cxy.
EXACT DIFFERENTIAL EQUATION:
Let M(x,y)dx + N(x,y)dy = 0 be a first order and first degree differential equation where M and N are real valued functions for some x, y. Then the equation Mdx + Ndy = 0 is said to be an exact differential equation
if ∂M∂ y =∂ N∂ x .
Example: (2y sinx+cosy)dx=(x siny+2cosx+tany)dy
Working rule to solve an exact equation:STEP 1: Check the condition for exactness, if exact proceed to step 2.STEP 2: After checking that the equation is exact, solution can be obtained as ∫M dx+∫(terms not containing x) dy=c-----------------------------------------------------------------------------------------------
PROBLEMS
(1). solve (1+exy )dx+e
xy (1− x
y )dy=0 .
Answer:Given equation (1+exy )dx+e
xy (1− x
y )dy=0 -----------(i)
Equation (i) is the form of Mdx+Ndy = 0
Where M = (1+exy ) and N = e
xy (1− x
y )
Now ∂M∂ y =¿ exy (−x
y2 ) And ∂N∂ x =¿ e
xy (−x
y2 ) Therefore ∂M∂ y =∂ N
∂X
∴ the given equation is exact.
∴ the general solution is given by
∫M dx+∫ ( termsindependent of x∈N )dy=C
Y constant
=> ∫ (1+exy )dx+∫ (0 )dy=C
Y constant
=> x + exy
1y
= c =>
x + y exy = c
is the required solution.
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2.Solve (2y Sin x + Cos y)dx=(x Sin y+ 2 Cos x +Tan y) dy
Sol.In Given Diff.Eqn. M=2y Sin x+ Cos y , N=-x Sin y-2 Cos x- Tan
y
∂m∂ y =2 Sin x –Sin y,
∂N∂ x = - Sin y+ 2 Sin x
There Fore
∂M∂ y
=∂ N∂X
∴ the general solution is given by
∫M dx+∫ ( termsindependent of x∈N )dy=C
∫ (2 y sin x+cos y )dx+∫ (−tan y )dy=C
∴ -2y Cos x + x Cos y+log Sec y = C is the required solution
of given d.e.
INTEGRATING FACTOR:Let Mdx + Ndy = 0 be not an exact differential equation. Then Mdx + Ndy = 0 can be made exact by multiplying it with a suitable function u is called an integrating factor.-----------------------------------------------------------------------------------------------
Example 1: ydx-xdy=0 is not an exact equation. Here 1/x2 is an integrating factor.Example 2: y(x2y2+2)dx+x(2-2x2y2)dy=0 is not an exact equation. Here 1/(3x3y3) is an integrating factorMETHODS TO FIND INTEGRATING FACTORS:METHOD 1: With some experience integrating factors can be found by inspection. That is, we have to use some known differential formulae.Example 1: d(xy)=xdy+ydx Example 2: d[ xy ]=( ydx−xdy )
y2
Example 3: d[log(x2+y2)]=2(xdx+ ydy)(x2+ y 2)
PROBLEMS(1). Solve the differential equation (y-x2¿dx+(x2 cot y−x )dy=0.
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Answer: Given differential equation (y-x2¿dx+(x2 cot y−x )dy=0
The above equation can be written as ydx – xdy = x2dx−x2 cot ydy
Now dividing with x2 ,
we get ydx – xdy
x2 = dx−cot ydy
−( xdy – ydxx2 )= dx−cot ydy
−d ( yx )= dx−cot ydy
Integrating on both sides
( yx )= x+cosec2 y+c
∴ y=−(x2+xcosec2 y+cx) is the required solution.
METHOD 2: If Mdx + Ndy = 0 is a non-exact but homogeneous differential equation and Mx + Ny ≠ 0 then 1/(Mx + Ny) is an integrating factor of Mdx + Ndy = 0.
Example 1: x2ydx-(x3+y3)dy=0 is a non-exact homogeneous equation. Here I.F.=-1/y4
Example 2: y2dx+(x2-xy-y2)dy=0 is a non-exact homogeneous equation. Here I.F.=1/(x2y-y3).
1) Solve x2y dx- (x3+y3)dy=0
Sol. Given x2y dx- (x3+y3)dy=0
Here M=x2y , N=-x3-y3
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∂m∂ y =x2 ,
∂N∂ x =-3x2
∂M∂ y
∂ N∂ X
The equation is not Exact.
To make it Exact we have to multiply by Intigrating Factor.Since M and
N are Homogeneous functions of x and y, the Intigrating Factor is 1
Mx+Ny
1Mx+Ny =
−1y4
Multiply given equation by −1y4
−x2
y3 dx+ x3+ y3
y4 dy=0
Now this is an Exact Differential Eqn.
∴ the general solution is given by
∫ dx+∫dy=C
Log y=x3
3 y3 +C , the required general solution.
2)Solve (x2y-2xy2)dx-(x3-3 x2y)dy=0
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Sol. Given (x2y-2xy2)dx-(x3-3 x2y)dy=0
Here M = x2y – 2xy2, N = -x3+3x2y
∂m∂ y =2xy-4xy ,
∂N∂ x =-3x2+6xy
∂M∂ y ≠ ∂ N∂ X
∴ The equation is not Exact.
Now we make Exact equation ,we have to multiply by Intigrating
Factor.Since M and N are Homogeneous functions of x and y, the
Intigrating Factor is 1
Mx+Ny
1Mx+Ny =
1x2 y2
We get (1y− 2x )dx-(
xy2
− 3y )dy=0 is an Exact D.Eqn.
∴ general Soln.is
∴xy−2 log x+3 log y=C
.
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METHOD 3: If the equation Mdx + Ndy = 0 is of the form y.f(xy) dx + x.g(xy) dy = 0 and Mx – Ny ≠ 0 then 1/(Mx – Ny) is an integrating factor of Mdx + Ndy = 0.
Example 1: y(x2y2+2)dx+x(2-2x2y2)dy=0 is non-exact and in the above form. Here I.F=1/(3x3y3)
Example 2: (xysinxy+cosxy)ydx+(xysinxy-cosxy)xdy=0 is non-exact and in the above form. Here I.F=1/(2xycosxy)
1) Solve ( xy Sin xy + Cos xy)y dx + (xy Sin xy – Cos xy )x dy =0Sol.
M=( xy Sin xy + Cos xy)y, N=(xy Sin xy – Cos xy )x
The I.F= 1
2xyCosxy
∴ ( xySinxy+Cosxy ) y
2 xyCosxydx+( xySinxy−Cosxy )
2 xyCosxydy=0
Is Exact Differential eqn. Solution is
The required sol. Is log (sec xy)+log x-log y=log C
2)Solve y(x2y2+2)dx + x(2-2x2y2)dy=0
Sol.Here M= y3x2+2y ,N=2x-2x3y2
∂M∂ y =3y2x2+2 ,
∂N∂ x =2-6x2y2
∂M∂ y
∂ N∂ X
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1Mx−Ny
= 13 x3 y3 is an integrating Factor.
∴
y3 x2+2 y3 x3 y3 dx+2 x−2 x3 y2
3 x3 y3 =0 is an Exact Diff.Eqn.
∴ the general solution is ∫ 1
3 xdx+ 2
3 y2∫1x3dx−2
3∫1ydy=0
=>log x -2
3x2 y2 -2 log y=C
METHOD 4: If there exists a continuous single variable function f(x) such that ∂M/∂y-∂N/∂x=Nf(x) then e∫f(x)dx is an integrating factor of Mdx + Ndy = 0
Example 1: 2xydy-(x2+y2+1)dx=0 is non-exact and
∂M/∂y - ∂N/∂x=N(-2/x). Here I.F=1/x2
Example 2: (3xy-2ay2)dx+(x2-2axy)=0 is non-exact and ∂M/∂y - ∂N/∂x=N(1/x). Here I.F=x
1) Solve (x2+y2+x)dx +xy dy=0
Sol.Here M=(x2+y2+x),N= xy
∂M∂ y =2y,
∂N∂ x =y
∂M∂ y
∂ N∂ X
The equation is not Exact.
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To make it Exact we have to multiply by Intigrating Factor.Since M and
N are Homogeneous functions of x and y, the Intigrating Factor is
e∫ f (x )dx , where
∂ M∂ y
−∂N∂ x
N= f ( x )
∂M∂ y
−∂N∂ x
N= 2 y− y
xy= 1x= f ( x )
∴e∫ f (x )dx= e
∫ 1xdx
=x ,is Intigrating factor.
∴(x2+y2+x)x dx +x2y dy =0 is Exact Diff.Eqn.
Soln. is ∫( x2+ y2+x ) xdx=C
x44
+ x2 y2
2+ x
3
3=C
2)Solve (3xy-2ay2)dx+(x2-2axy)dy= 0
Sol: given (3xy-2ay2)dx+(x2-2axy)dy= 0
Here M=3xy-2ay2 , N= x2-2axy
∂M∂ y =3x-4ay ,
∂N∂ x =2x-2ay
∂M∂ y
∂ N∂ X
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The equation is not Exact.
To make it Exact we have to multiply by Intigrating Factor.Since M and
N are Homogeneous functions of x and y, the Intigrating Factor is
e∫ f (x )dx , where
∂ M∂ y
−∂N∂ x
N= f ( x )
∂M∂ y
−∂N∂ x
N=3 x−4 ay−2 x+2ay
x2−2axy=1x=f ( x )
∴e∫ f (x )dx= e
∫ 1xdx
=x ,is Intigrating factor.
∴(3xy-2ay2)x dx+(x2-2axy)x dy=0 is exact diff.eqn.
∴ General Sol.is ∫(3x2 y−2axy )dx=C
∴ x3y-ax2y2=C is the required solution.
METHOD 5: If there exists a continuous single variable function f(y) such that
∂N/∂x - ∂M/∂y=Mg(y) then e∫g(y)dy is an integrating factor of Mdx + Ndy = 0
Example 1: (xy3+y)dx+2(x2y2+x+y4)dy=0 is a non-exact equation and ∂N/∂x - ∂M/∂y=M(1/y). Here I.F=y
Example 2: (y4+2y)dx+(xy3+2y4-4x)dy=0 is a non-exact equation and ∂N/∂x - ∂M/∂y=M(-3/y).Here I.F=1/y3
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LINEAR EQUATION:
An equation of the form y′ + Py = Q is called a linear differential equation.
Integrating Factor(I.F.)=e∫pdx
Solution is y(I.F) = ∫Q(I.F)dx+C
Example 1: xdy/dx+y=logx. Here I.F=x and solution is xy=x(logx-1)+C
Example 2: dy/dx+2xy=e-x.Here I.F=ex and solution is yex=x+C
Linear Differential Equation:
Example 1: xdy/dx+y=logx. Here I.F=x and solution is xy=x(logx-1)+C
Example 2: dy/dx+2xy=e-x.Here I.F=ex and solution is yex=x+C
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PROBLEMS
(1). Solve the differential equation (x+1)dydx− y=e3x ( x+1 )2 .
Answer: Given equation is (x+1)dydx− y=e3x ( x+1 )2 .this can be written as
=> dydx−( 1x+1 ) y=e3x (x+1) it’s the form of dydx +Py=Q Here P = −1
X+1 and Q =
e3x (x+1)
I.F = e∫ Pdx=e−∫ 1
x+ 1dx=e− log (1+x)= 11+x
I.F ¿ 11+x
-----------------------------------------------------------------------------------------------
∴ General solution is y ×I.F = ∫Q×I .F dx+c
y × 11+ x = ∫ e3 x ( x+1 ) . 1
1+xdx+c= e3 x
3+c
y = [ e3 x
3+c ] ( x+1 ) .
BERNOULLI’S LINEAR EQUATION:
An equation of the form y′ + Py = Qyn is called a Bernoulli’s linear differential equation. This differential equation can be solved by reducing it to the Leibnitz linear differential equation. For this dividing above equation by yn
Example 1: xdy/dx+y=x2y6.Here I.F=1/x5 and solution is 1/(xy)5=5x3/2+Cx5
Example 2: dy/dx+y/x=y2xsinx. Here I.F=1/x and solution is 1/xy=cosx+C
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ORTHOGONAL TRAJECTORIES: If two families of curves are such that each member of family cuts each member of the other family at right angles, then the members of one family are known as the orthogonal trajectories of the other family.
Example 1: The orthogonal trajectory of the family of parabolas through origin and foci on y-axis is x2/2c+y2/c=1
Example 2: The orthogonal trajectory of rectangular hyperbolas is xy=c2
PROCEDURE TO FIND ORTHOGONAL TRAJECTORIES
Suppose f (x ,y ,c) = 0 is the given family of
-----------------------------------------------------------------------------------------------
curves, where c is the constant.
STEP 1: Form the differential equation by
eliminating the arbitrary constant.
STEP 2: Replace y′ by -1/y′ in the above
equation.
STEP 3: Solve the above differential equation
Orthogonal Trajectories in Cartesian Co-ordinates:
=> −( yx )= x+cosec2 y+c
-----------------------------------------------------------------------------------------------
∴ y=−(x2+xcosec2 y+cx) is the required solution.
PROBLEMS(1). Find the orthogonal trajectories of the family of curves x2+ y2 ¿a2
Answer: given x2+ y2=a2 ------------------(i)Differentiating w.r.t x
we get 2x + 2yd ydx = 0 => d ydx =−xy -------------(ii) it’s the differential
equation of the family of curvesnow we find the orthogonal trajectories
replace d ydx with −d xdy in (ii)
=> d xdy = xy Applying the variables and separable method
=> d xx =dyy Integrating on both sides => ∫ d x
x=∫ dy
y =>log x = log y
+log c => x =cy or y =cx is the required orthogonal trajectories of (i).Orthogonal Trajectories in Polar Co-ordinates:
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Orthogonal Trajectories .
PROBLEMS(1). Obtain the orthogonal trajectories of the family of curve r(1+cosθ)= 2a.
Answer: Given curve r(1+cosθ)= 2a ---------------(i) where a is a parameter
Now differentiating equation (i) w.r.t ‘θ '
drdθ =
−2a(1+cosθ)2 .−sin θ =
2a sin θ(1+cosθ)2 -------------(ii)
Now (ii)/(i) => drdθr
= sin θ
(1+cosθ )2(1+cosθ )
-----------------------------------------------------------------------------------------------
¿> 1r [ drdθ ]= sin θ
1+cosθ-------------------------------(iii)
Equation (iii) represents the orthogonal trajectories of the given family of curve.
Now replacing drdθ with −r2 dθdr in (iii)
We get 1r [−r2 dθ
dr ]= sinθ1+cosθ => −r dθ
dr=
2sin (θ2 )cos( θ2 )2 cos2( θ2 )
= tan( θ2 )
=>drr
=−cot( θ2 )dθ integrating
Log r = -2logsin( θ2 )+log(c2)
=>r= c
2sin 2( θ2 ) = >
r = c1−cosθ
It’s the required orthogonal trajectories.
NEWTON’S LAW OF COOLING:
The rate at which the temperature of a hot body decreases is proportional to the difference between the temperature of the body and the temperature of the surrounding air.
θ′ ∞ (θ – θ0)
-----------------------------------------------------------------------------------------------
PROBLEMS(1) If the air is maintained at 250C and temperature of the body cools from 1400C to 800C in 20 minutes find when the temperature will be 350C.
Answer: Given t = 0 , θ = 1400C
t = 20, θ = 800C and θ = 350C find t
-----------------------------------------------------------------------------------------------
by Newton’s law of cooling
dθdt
=−k (θ−θ0) where θ0 is the temperature of the surrounding
medium(air)
dθdt
=−k (θ−25 ) => dθ(θ−25 )
=−kdt integrating ,
We get log((θ−25 )=−kt+c -------------------(i)
Initially t = 0, θ = 1400C
FROM (i) log((140−25 )=−k (0)+c => c = log(115)∴ log((θ−25 )=−kt+ log(115) => log (115 )−¿log((θ−25 )=kt---------(ii)
When t = 20, θ = 800C
From (ii) log (115 )−¿log((80−25 )=k (20)
20k =log (115 )−¿log((55 ) -------------------(iii)
Now (ii)/(iii) => kt20k
=log (115 )−log ¿¿ = > t20=log (115 )−log ¿¿
When θ = 350C => t20=log (115 )−log ¿¿ = >
t = 66.2(2). If the air is maintained at 300C and the temperature of the body cools from 800C TO 600C in 12 minutes, find the temperature of the body after 24 minutes.Answer: Given θ0=30 and
t = 0 , θ = 800C
t = 12, θ = 600C and t = 24m find θ and
-----------------------------------------------------------------------------------------------
by Newton’s law of cooling
dθdt
=−k (θ−θ0) where θ0 is the temperature of the surrounding
medium(air)
dθdt
=−k (θ−30 ) => dθ(θ−30 )
=−kdt integrating ,
We get log((θ−30 )=−kt+c =>(θ−30 )=ce−kt -------------------(i)
Initially t = 0, θ = 80 0 C :
FROM (i) ((80−30 )=c => c = 50∴ (θ−30 )=50e−kt => θ=30+50 e−kt----------------------(ii)
When t = 12, θ = 60 0 C :
From (ii)60=30+¿ 50e−12 k => 30=50e−12 k
¿> 3050
=e−12 k=¿ 35=e−12 k-------------------(iii)
When t=24 : from (ii) θ=30+50e−24 k => θ=30+50 (e−12 k )2
=> θ=30+50( 35 )
2
= > θ = 480C.
(3). If the air is maintained at 150C and the temperature of the drops from 700C to 400C in 10 minutes. What will be its temperature after 30 minutes.Answer: Given θ0=15 and
t = 0 , θ = 700C
t = 10, θ = 400C and t = 30m find θ and
by Newton’s law of cooling
-----------------------------------------------------------------------------------------------
dθdt
=−k (θ−θ0) where θ0 is the temperature of the surrounding medium(air)
dθd t
=−k (θ−15 ) => dθ
(θ−20 )=−kdt integrating ,
We get log((θ−15 )=−kt+c =>(θ−15 )=ce−kt -------------------(i)
Initially t = 0, θ = 70 0 C :
FROM (i) ((70−15 )=c => c = 55
∴ (θ−15 )=55e−kt => θ=15+55e−kt----------------------(ii)
When t = 10, θ = 40 0 C :
From (ii)40=15+¿ 55e−10 k => 25=55e−10 k
¿> 2555
=e−10 k=¿ 511
=e−10 k-------------------(iii)
When t=30 : from (ii) θ=15+55e−30 k => θ=15+55 (e−10 k )3
=> θ=15+55( 511)
3
= > θ = 20.160C.
(4). If the air is maintained at 200C and the temperature of the body cools from 1000c to 800c in 10minutes,find the temperature of the body after 20minutes and when the temperature will be 400c.Answer: Given θ0=20 and
t = 0 , θ = 1000C
t = 10, θ = 800C and θ = 400C find t and t = 20 find θby Newton’s law of cooling
dθdt
=−k (θ−θ0) where θ0 is the temperature of the surrounding medium(air)
dθdt
=−k (θ−20 ) => dθ
(θ−20 )=−kdt integrating ,
-----------------------------------------------------------------------------------------------
We get log((θ−20 )=−kt+c =>(θ−20 )=ce−kt -------------------(i)
Initially t = 0, θ = 100 0 C :
FROM (i) ((100−20 )=c => c = 80
∴ (θ−20 )=80e−kt => θ=20+80e−kt----------------------(ii)
When t = 10, θ = 80 0 C :
From (ii)80=20+¿ 80e−10 k => 60=80e−10 k
¿> 6080
=e−10 k=¿ 34=e−10 k-------------------(iii)
When t=20 : from (ii) θ=20+80e−20 k => θ=20+80 (e−10 k )2
=> θ=20+80( 34 )
2
= > θ = 650C.
When θ = 40 0 C : from (ii) 40=20+80 e−kt => 20 = 80e−kt
¿> 2080
=e−kt ¿> 14=e−kt => 1
4=(e−k )t
¿> 14=(e−k ) t => 1
4 =[( 34 )
110 ]
t
Taking log on both sides
=> log ( 14 )=log [(3
4 )110]
t
=> t = 10 log (1/4 )
log (3/ 4 )
t = 4.82minutes
LAW OF NATURAL GROWTH:
-----------------------------------------------------------------------------------------------
When a natural substance increases in Magnitude as a result of some action which affects all parts equally, the rate of increase depends on the amount of the substance present.
N′ = k N
LAW OF NATURAL DECAY:
The rate of decrease or decay of any substance is proportion to N the number present at time.
N′ = -k N
PROBLEMS1) A number of N bacteria in culture grew at a rate proportional to N.Value
of N was initially 100 and increased to 332 in 12hour .What would be the
value of N after 1 1
2 hours.
-----------------------------------------------------------------------------------------------
Sol.Given When T=0,N=100
T=1 hr,N=332
At T=1 1
2 ,We have to find N.
dNdt
=KN, for growth
dNN
=Kdt
log N=kt+CN=ekt+CN=Cekt . .. .. .. .(1)
From given When T=0,N=100
(1) => C=100…..(2)
T=1 hr,N=332
(1) =>K=log(332/100)
At T=1 1
2 , N=100e log (332 /100 ). 3/2
N=604.9.
.
2)If 30 % of a Radio active substance disappears in 10 days, how long will
it take for 90 % disappear.
Sol.Given
-----------------------------------------------------------------------------------------------
dydt
αy
dydt
=−ky
Solving y= c e-kt
When t=0 y=N,C=N
Y=N e-kt
LINEAR DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER
-----------------------------------------------------------------------------------------------
INTRODUCTION:
An equation of the form
Dny + k1 Dn-1y +....+ kny = X
Where k1,…..,kn are real constants and X is a continuous function of x is called an ordinary linear equation of order n with constant coefficients.
Its complete solution is
y = C.F + P.I where C.F is a Complementary Function and
P.I is a Particular Integral.
Example: d2y/dx2+3dy/dx+4y = sinx
COMPLEMENTARY FUNCTION:
If roots are real and distinct then
C.F = c1 em1x + …+ ck emkx
Example 1: Roots of an auxiliary equation are 1,2,3 then C.F = c1 ex + c2 e2x + c3 e3x
Example 2: For a differential equation
(D-1)(D+1)y=0, roots are -1 and 1. Hence
-----------------------------------------------------------------------------------------------
C.F = c1 e-x +c2 ex
If roots are real and distinct then
C.F = c1 em1x + …+ ck emkx
Example 1: Roots of an auxiliary equation are 1,2,3 then C.F = c1 ex + c2 e2x + c3 e3x
Example 2: For a differential equation
(D-1)(D+1)y=0, roots are -1 and 1. Hence
C.F = c1 e-x +c2 ex
If roots are real and equal then
C.F = (c1+ c2x +….+ ckxk)emx
Example 1: The roots of a differential equation (D-1)3y=0 are 1,1,1. Hence
C.F.= (c1+c2x+c3x2)ex
Example 2: The roots of a differential equation (D+1)2y=0 are -1,-1.
Hence C.F=(c1+c2x)e-x
If two roots are real and equal and rest are real and different then
C.F=(c1+c2x)em1x+c3em3x+….
-----------------------------------------------------------------------------------------------
Example : The roots of a differential equation (D-2)2(D+1)y=0 are 2,2,-1. Hence C.F.=(c1+c2x)e2x+c3e-X
If roots of Auxiliary equation are complex say p+iq and p-iq then
C.F=epx(c1 cosqx+c2 sinqx)
Example: The roots of a differential equation (D2+1)y=0 are 0+i(1) and 0-i(1). Hence C.F=e0x(c1cosx+c2sinx)
=(c1cosx+c2sinx)
A pair of conjugate complex roots say p+iq and p-iq are repeated twice then
C.F=epx((c1+c2x)cosqx+(c3+c4x)sinqx)Example: The roots of a differential equation
(D2-D+1)2y=0 are ½+i(1.7/2) and ½-i(1.7/2) repeated twice. Hence C.F=e1/2x(c1+c2x)cos(1.7/2)x+ (c3+c4x)sin(1.7/2)x
PROBLEMS(1). solve the differential equation (D3+2D2+D ) y=e2x.
Answer: Given (D3+2D2+D ) y=e2x
Now write the Auxiliary equation is f(m) = 0
-----------------------------------------------------------------------------------------------
=> m3+2m2+m=0 => m(m2+2m+1¿=0
=> m (m+1)2 = 0
m =0,-1 ,-1 Now write the complementary function
yc=c1+(c2+c3 x)e− x.
And Particular Integral y p=1
D3+2D2+De2x =
123+2.22+2
e2x = e2 x
18
∴The general solution is y = yc+ y p
∴y = c1+(c2+c3 x)e−x+¿ e
2 x
18 .
PARTICULAR INTEGRAL:
Method 1: When X = eax put D = a in Particular Integral. If f(a) ≠ 0 then P.I. will be calculated directly. If f(a) = 0 then multiply P.I. by x and differentiate denominator. Again put D = a.Repeat the same process.
Example 1 : y″+5y′+6y=ex. Here P.I=ex/12
Example 2 : 4D2y+4Dy-3y=e2x.Here P.I=e2x/21
PROBLEMS (1). solve the differential equation (D3−1 ) y=(ex+1 )2 .
Answer: Given differential equation (D3−1 ) y=(ex+1 )2
-----------------------------------------------------------------------------------------------
Now write the auxiliary equation f(m) = 0 => m3−1=0
=> (m-1)m2+m+1=0 => m = 1, 1±√32
∴ complementary function = yc=c1 ex+e
x2 [c2cos (√3 x
2 )+c3 sin(√3x2 )].
And P.I = y P=(ex+1 )2
D3−1= e2 x+2ex+1
D 3−1
P.I = y P=e2 x
7+ 2x e2x
3−1
∴ General solution is y = yc+ y p
y¿c1 ex+e
x2 [c2 cos (√3 x
2 )+c3 sin(√3 x2 )]+ e2x
7+ 2 xe2x
3−1.
Method2: When X = Sinax or Cosax or Sin(ax+b) or Cos(ax+b) then put D2= - a2 in Particular Integral.
Example 1: D2y-3Dy+2y=Cos3x. Here P.I=(9Sin3x+7Cos3x)/130
Example 2: (D2+D+1)y=Sin2x. Here
P.I= -1/13(2Cos2x+3Sin2x)
PROBLEMS(1). Solve the differential equation (D2+9 )=cos3 x+sin 2 x.Answer: Given differential equation (D2+9 )=cos3 x+sin 2x
Let f(D) =D2+9
-----------------------------------------------------------------------------------------------
Now write auxiliary equation f(m) = 0 m2+9=0 = > m = ±3i∴ C.F = yc=c1 cos3 x+c2sin 3 x -----------------(i)
And P.I = y p=cos3 x+sin 2x
D2+9= cos3 x
D2+9+ sin 3 xD2+9
¿ xcos3 x2D
+ xsin3 x2D since f(−b2¿=0
y p ¿ xsin3 x6
− xcos3 x6 -------------------(ii)∴ general solution y = yc+ y p
y = c1 cos3 x+c2 sin 3x + xsin3 x6
− xcos3 x6 is the required solution.
(2). solve the differential equation (D2+1¿ y=sinxsin 2x .Answer: Given differential equation (D2+1 ¿ y=sinxsin 2x
Write A.E f(m) = 0 => m2+1=0 => m = ±i The roots are complex.C.F = yc=[ c1 cosx+c2 sinx ]-----------------(i)
And P.I = y P=sinxsin2 xD2+1 =
12. cosx−cos3 x
D2+1 = 12.[ cosxD 2+1
− cos3 xD2+1 ]= P.I 1 +P.I 2
NOW P.I 1 = cosxD2+1 = xcosx2D
= xsinx2
P.I 2= cos3 xD2+1
= cos3 x−9+1
= cos3 x−8
y P= P.I 1 +P.I 2 = 12.[ xsinx2
+ cos3 x8 ]--------------------------(ii)∴ general solution is y = yc+ yP
Y = [c1cosx+c2 sinx ]+ 12.[ xsinx2
+ cos 3x8 ].
(3). solve the differential equation (D4−2D3+2D2−2D+1 ) y=cosx.Answer: Given differential equation (D4−2D3+2D2−2D+1 ) y=cosx.Write auxiliary equation f(m) = 0 => (m4−2m3+2m2−2m+1 ) y=0
-----------------------------------------------------------------------------------------------
=> ¿=> ¿ => (m2+1 ) (m2−2m+1 )=0
=> (m2+1 )(m−1)2=0 ∴ m = ±I,1,1∴ C.F = yc=(c1 cosx+c2 sinx )+(c3+c4 x )ex
And P.I = y p=cosx
D4−2D3+2D2−2D+1= cosx
(D2 )2−2D2 . D+2D2−2D+1
¿cosx
(−1 )2−2 (−12 ) .D+2. (−12 )−2D+1 since D2=−12=−1
¿cosx
1+2.D−2.−2D+1since f (D2 )=f (−b2 )=0
¿ xcosx4D3−6D2+4D−2
= xcosx4D2. D−6D2+4D−2
= xcosx
−4.D+6+4D−2= xcosx
4 ∴ general solution y = yc+ y p
Y = (c1 cosx+c2 sinx )+ (c3+c4 x )ex+ xcosx4 .
(4). Solve the differential equation (D2−4¿ y=2 cos2 x .
Answer: Given equation is (D2−4¿ y=2 cos2 x
HERE f(D) = D2−4¿
Now write the auxiliary equation f(m) = 0 => m2−4=0∴ the roots m = 2,-2 . here the roots are real and different.∴ C.F = yc=c1 e2x+c2e
−2x -----------------------(i)
P.I = y p=2 cos2 xD2−4
=1+cos2 xD2−4
= 1D2−4
+ cos2 xD2−4
=P I 1+P I 2
-----------------------------------------------------------------------------------------------
P I 1=1
D2−4=−1
4
P I 2=cos2 xD2−4
=cos 2x−8 ( SINCE D2=−4¿
∴ y p=PI 1+P I 2 = −14 + cos2 x
−8
∴ General solution is y = yc+¿ y p = c1 e2 x+c2 e
−2x+−14 + cos2x
−8
∴ y =c1 e2 x+c2 e
−2x−14−¿ cos2x
8 .
Method3: When X = xk or in the form of polynomial then convert f(D) into the form of binomial expansion from which we can obtain Particular Integral.
Example 1: (D2+D+1)y=x3.Here P.I=x3-3x2+6
Example 2: (D2+D)y=x2+2x+4. Here P.I=x3/3+4x
PROBLEMS(1). solve the differential equation (D2+D+1¿ y=x3.Answer: Given equation is (D2+D+1¿ y=x3
Let f(D) = (D2+D+1 ¿
Now write auxiliary equation f(m) = 0 => m2+m+1=0
=> m = −1± i√32 = −1
2± i √3
2 are complex and conjugate.
C.F = yc=e− x
2 [c1 cos √3 x2
+c2sin √3 x2 ]------------(i)
And P.I = y P= x3
D2+D+1 = [1+(D2+D ) ]−1
x3
-----------------------------------------------------------------------------------------------
= [1−D+D3 ] x3
y P = x3−3 x2+6∴ general solution y = yC+ y P
y=e− x
2 [c1 cos √3 x2
+c2 sin √3 x2 ]+x3−3 x2+6.
(2).solve the differential equation (D3−3D−2 )Y=x2.
Answer: Given differential equation (D3−3D−2 ) y=x2.Write auxiliary equation f(m) = 0 => (m3−3m−2 ) y=0
=> (m+1 ) (m2−m−2 )=0
=> (m+1 ) (m+1 )(m−2)=0 ∴ m = -1,-1,2∴ C.F = yc=(c1+c2 x )e−x+c3e2x
And P.I = y p=x2
D3−3D−2=
1−2 [1−D3+3D
2 ]−1
x2
¿ 1−2 [1+ D3+3D
2+(D3+3D
2 )2] x2
¿ 1−2 [x2+3 x+ 9
2 ]∴ general solution y = yc+ y p
Y = (c1+c2 x )e− x+c3 e2x+ 1
−2 [ x2+3x+ 92 ] .
Method4:When X = eaxv then put D = D+a and take out eax to the left of f(D). Now using previous methods we can obtain Particular Integral.
-----------------------------------------------------------------------------------------------
Example 1: (D4-1)y=ex Cosx. then P.I=-exCosx/5
Example 2: (D2-3D+2)y=xe3x+Sin2x. then
P.I=e3x/2(x- 3/2)+1/20(3Cos2x-Sin2x
Method5:When X = x.v then
P.I = [{x – f "(D)/f(D)}/f(D)]v
Example 1: (D2+2D+1)y=x Cosx. Here P.I=x/2Sinx+1/2(Cosx-Sinx)
Example 2: (D2+3D+2)y=x ex Sinx. Here P.I=ex[x/10(Sinx-Cosx)-1/25Sinx+Cosx/10]
PROBLEMS(1). solve the differential equation (D4+2D2+1 ) y=x2 cos2 x .
Answer: Given equation is (D4+2D2+1 ) y=x2 cos2 x .
Write the auxiliary is f(m) = 0
=> m4+2m2+1=0 => (m2+1 )2 = 0 => m = ± i,± i
Here the roots are complex and conjugate and repeted twice.∴ C.F =yc=(c1+c2 x )cos x+(c3+c4 x ) sin x---------(i)
And P.I = y p=x2cos2x .
D4+2D2+1 = x2
(D2+1 )2 [ 1+cos2x2 ]
P.I = y p=¿ 12 [ x2
(D2+1 )2+ x2
(D2+1 )2cos2 x ]
-----------------------------------------------------------------------------------------------
¿ 12
(1+D−2 )−2x2+ 1
2R . P[ x2 ei2 x
(D2+1 )2 ] ¿
12
(1−2D2+3D4−…) x2+ 12R . Pei2x [ 1
[ (D+2i )2+1 ]2x2]
= 12
(x2−4 )+ 12R .Pe i2x [ 1
[D2+4Di−3 ]2x2]
= 12
(x2−4 )+ 12R .Pe i2x .−1
3 [ 1
[1−D2+4 Di3 ]
2 x2]
= 12
(x2−4 )−16 R .P ei2 x [1−
D 2+4Di3 ]
−2
x2
= 12
(x2−4 )−16R .Pe i2x [1+2 D2+4Di
3+3(D2+4 Di
3 )2
+..] x2
= 12
(x2−4 )−16R .P e i2x [x2+ 4+16 xi
3− 32
3 ] = 1
2(x2−4 )−1
6R .P ¿
y p = 12
(x2−4 )−16 [( x2−26
9 )cos2x−16 x3
sin 2 x].∴ the general solution is y ¿ yc+ y p
y= (c1+c2 x ) cos x+(c3+c4 x ) sin x+ 12
(x2−4 )− 16 [(x2−26
9 )cos2 x−16 x3
sin 2 x ].(2). solve the differential equation (D2+4 ) y=x sinx.
Answer: Given differential equation is (D2+4 ) y=x sinx
Write the auxiliary equation f(m) = 0 => m2+4 = 0
∴ m = ±2i are complex roots and conjugare.-----------------------------------------------------------------------------------------------
∴ C.F= yc=c1 cos2 x+¿ c1 sin 2x ¿.And P.I= y p=
x sin xD2+4
y p=I . Pof xe ix
D2+4=I .Pof e ix x
(D+ i)2+4
= I . Pof eix xD2+2Di+3
= I . Pof e ix3 [1+D2+2Di
3 ]−1
x
= I . Pof e ix3 [1−D2+2Di3
..] x = I . Pof e ix3 [1−2Di
3..]x
= I . P of 13
(cos x+ isinx )[ x−i 23 ]
y p = 13 (−2
3cos x+ xsin x).
∴ the general solution is y = yc+ y p
y = c1 cos2x+¿c1 sin2 x+13 (−2
3cos x+xsin x) .¿
CAUCHY’S LINEAR EQUATION:
Its general form is
xnDny + …. +y = X
-----------------------------------------------------------------------------------------------
then to solve this equation put x = ez and convert into ordinary form.
Example 1: x2D2y+xDy+y=1
Example 2: x3D3y+3x2D2y+2xDy+6y=x2
Legendres linear differential equation:
(1). solve (2x-1)3 d3 yd x3 + (2x-1) dydx - 2y = x .
Answer: Given equation is (2x-1)3 d3 yd x3 + (2x-1) dydx - 2y = x ----------(i)
Put 2x-1 ------------------------------------------(ii)
=>2 = dtdx => d ydx =dydt
. dtdx
=2 dydt ---------------(iii)
And d2 yd x2 =22 d2 y
d t2 --------------------(iV)
d3 yd x3 =23 d3 y
d x3 -------------(v)
Now equation (iii) (iv) and (v) substituting in equation (i)
We get 8t3 d3 yd t3
+ 2t dydt - 2y = t+12 ---------------(vi)
Let z =logt => t = ez and t ddt=θ , t3 d3
d t 3=θ (θ−1)(θ−2)
From (vi)
[8θ (θ−1 ) (θ−2 )+2θ−2 ] y = ez+12
=>(8θ3−24θ2+18θ−2)y = ez+12
Now write the auxiliary equation i.e f(m) = 0=> 8m3−24m2+18m−2=0
-----------------------------------------------------------------------------------------------
∴ m = 1 and m = 1 ± √32
Therefore C.F IS yc= c1 ez+c2e
(1+√32 )z
+c3 e(1−√3
2 )z
And
y p1=e z
2(8θ3−24θ2+18θ−2) = −ze z
12 and y p2=
e0 z
2(8θ3−24θ2+18θ−2)=−1
4
∴ y p= y p1+y p2 = −ze z
12 + −14
∴ general solution is y = yc+ y p
y= c1 ez+c2e
(1+√32 )z
+c3 e(1−√3
2 )z+−z ez
12 + −14
where z = log (t) = log (2x -1).
METHOD OF VARIATION OF PARAMETERS:
Its general form is D2y + P Dy + Q = R
where P, Q, R are real valued functions of x.
Let C.F = C1u + C2v
P.I = Au + Bv Where A = −∫ vRdx
u dvdx
−v dudx
and B = ∫uRdx
u dvdx
−v dudx
Example 1: (D2+1)y=Cosecx. Here A=-x, B=log(Sinx)
Example 2: (D2+1)y=Cosx. Here A=Cos2x/4, B=(x+Sin2x)/2
-----------------------------------------------------------------------------------------------
PROBLEMS (1). by using variation of parameters solve d
2 yd x2 +4 y=tan 2 x .
Answer: Given equation can be written as (D2+a2) y=tan ax ----(i)
Write A.E f(m) = 0 => m2+a2=0=¿m=ai ,−aithese roots are complex conjugate numbers
∴ yc=C .F=c1 cosax+c2 sin ax
And y p=P . I=A cosax+B sin ax -------------------(ii)
Where A = −∫vRdx
u dvdx
−v dudx
= −∫ sin ax .cos axa
dx and B = ∫uRdx
u dvdx
−v dudx
Now A = −∫vR
u dvdx
−v dudx
= −∫ sin ax .cos axa dx =−1
a ∫ sin2axcosax
¿−1a∫
1−cos2axcosax
= =−1a [∫ sec axdx−∫ cosax dx ]
A = −1a2 [ log|sec ax+ tan ax|]+ 1
a2 sin ax
And
B = ∫uRdx
u dvdx
−v dudx
= ∫ tan ax .cosaxa
dx=1a∫sinax dx=¿−1
a2 cosax ¿
B = −1a2 cosax
Hence
y p=[−1a2 [ log|sec ax+ tan ax|]+ 1
a2 sinax ]cosax+−1a2 cosax sin ax
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Therefore the general solution is y = yc+ y p
y=c1 cosax+c2sin ax+[−1a2 [ log|sec ax+tan ax|]+ 1
a2 sin ax ]cosax+−1a2 cosax sin ax
It’s the required solution .
(2). solve the method of variation of parameters d2 yd x2 + y=cosec x .
Answer: Given equation can be written as (D2+1 ) y=cosec x ----(i)
Write A.E f(m) = 0 => m2+1=0=¿m=i ,−ithese roots are complex conjugate numbers
∴ yc=C .F=c1 cos x+c2sin x
And y p=P . I=A cos x+B sin x -------------------(ii)
Where A = −∫vRdx
u dvdx
−v dud x
and B = ∫uRdx
u dvdx
−v dudx
Let u = cos x and v = sin x
¿> dudx
=−sin x , dvdx
=cos x
Now A = −∫vR
u dvdx
−v dudx
= −∫sin x . cosec x dx =−∫dx=−x
And
B = ∫uRdx
u dvdx
−v dudx
= ∫cos x . cosec xdx=∫ cot x dx=log|sinx|
B = log|sinx|
Hence
y p= [−x ] cos x+ log|sinx|sin x
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Therefore the general solution is y = yc+ y p
y=c1 cos x+c2sin x−x cos x+ log|sinx|sin x
It’s the required solution .
Simple harmonic motion :
(1) Abody weighing 10 kgs is hung from a spring. A pull of 20 kgs will stretch the spring to 10 cms .The body is pulled down to 20 cms below the static equilibrium position and then released . Find the displacement of the body from its equilibrium position at time ‘ t ‘ seconds , the maximum velocity and the period of oscillation.
Answer: Let O be the fixed end and A is the other end of the spring .
Since load of 20 kgs attached to A stretches the spring by 0.1m.
Let e = 0.1 = AB be the elongation produced by the mass m and hanging in equilibrium.
If k be the restoring force per unit stretch of the spring due to elasticity, then for the equilibrium at B .
=> mg =ke => 20 = T 0=K ×0.1 => K = 200kg/m.Let B be the equilibrium position when 10 kg. weight is attached to A
Then 10 = T B= K×AB => AB = 10200
=0.05M
Now the weight is pulled down to C where BC =0.2mAfter any time t of its release from C Let the weight be at P , Where BP = x
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Then the tension T = k × AP = 200(0.05 +x ) = 10 +200x∴ the equation of motion of the body is
wgd2 xd t 2 =w−T
Where g =9.8m/sec2
=> 109.8
d2 xd t 2 =10−(10+200 x ) => d
2xd t 2 =−μ2 x where μ = 14
This shows that the motion of the body in simple harmonic about B as centre and the period of oscillation = 2μμ =0.45 sec.
Also the amplitude of motion being BC = 0.2 m, the displacement of the body from B at time t is given by x = 0.2 cosμt = 0.2cos14t m.And the maximum velocity = μ =14 ×0.2 = 2.8 m/sec.(2) A particle is executing S.H.M,with amplitude 5 meters and time 4 seconds.Find the time required by the particle in passing between points which are atdistances 4 and 2 meters from the centre of force and are on the same side of it.
Answer: the equation of the S.H.M is d2xd t 2 =−μ2 x ---------------(i)
Also given time period = 2πμ
=4=¿μ=π2
The required solution of (i) is x = a cosμt
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Now a = 5,μ= π2 => x = 5 cosπ2 t --------------------------(ii)
Consider the times when the particle is at distance of 4 meters and 2 meters from the centere of motion respectively be t 1 sec∧t 2 sec .
4 = 5 cosπ2 t 1=¿ t1=2π
cos−1( 45 )
2 = 5 cosπ2 t 2 => t 2=2π
cos−1( 25 )
Now t 2−t 1=0.33 sec .
Differentiating equation (ii) w.r.t ‘t’
dxdt = −5π
2 sinπ2 t = −5π
2 √ 25−x2
25 since sin2x + cos2x = 1
dxdt = −π
2 √25−x2
∴when x = 4 => V = π2 √25−42 = 4.71 m/sec.
When x = 2 => V = π2 √21m/sec.
(3). Abody weighing 10 kgs is hung from a spring. A pull of 20 kgs will stretch the spring to 10 cms .The body is pulled down to 20 cms below the static equilibrium position and then released . Find the displacement of the body from its equilibrium position at time ‘ t ‘ seconds , the maximum velocity and the period of oscillation.
Answer: Let O be the fixed end and A is the other end of the spring .
Since load of 20 kgs attached to A stretches the spring by 0.1m.
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Let e = 0.1 = AB be the elongation produced by the mass m and hanging in equilibrium.
If k be the restoring force per unit stretch of the spring due to elasticity, then for the equilibrium at B .
=> mg =ke => 20 = T 0=K ×0.1 => K = 200kg/m.Let B be the equilibrium position when 10 kg. weight is attached to A
Then 10 = T B= K×AB => AB = 10200
=0.05M
Now the weight is pulled down to C where BC =0.2mAfter any time t of its release from C Let the weight be at P , Where BP = xThen the tension T = k × AP = 200(0.05 +x ) = 10 +200x∴ the equation of motion of the body is
wgd2 xd t 2 =w−T
Where g =9.8m/sec2
=> 109.8
d2 xd t 2 =10−(10+200 x ) => d
2xd t 2 =−μ2 x where μ = 14
This shows that the motion of the body in simple harmonic about B as centre and the period of oscillation = 2μμ =0.45 sec.
Also the amplitude of motion being BC = 0.2 m, the displacement of the body from B at time t is given by x = 0.2 cosμt = 0.2cos14t m.-----------------------------------------------------------------------------------------------
And the maximum velocity = μ =14 ×0.2 = 2.8 m/sec.
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