ma 108 - ordinary differential equationsdey/diffeqn_spring14/lecture14_d2_prt.pdf · suppose emx is...

MA 108 - Ordinary Differential Equations

Santanu Dey

Department of Mathematics,Indian Institute of Technology Bombay,

Powai, Mumbai [email protected]

March 27, 2014

Santanu Dey Lectures 14

Outline of the lecture

Second order vs nth order

Method of variation of parameters

Method of undetermined coefficients


Abel’s formula

We have seen that the Wronskian of any two solutions y1(x), y2(x)of y ′′ + p(x)y ′ + q(x)y = 0 is given by

W (y1, y2)(x) = W (y1, y2)(x0)e−

∫ xx0

p(t)dt,

for any x0, x ∈ I . Note that the formula depends only on p(x) andnot on q(x). What should be the n-th order analogue?

Theorem

If y1, y2, . . . , yn are solutions of

y (n) + p1(x)y (n−1) + . . .+ pn(x)y = 0,

then,

W (y1, . . . , yn)(x) = W (y1, . . . , yn)(x0)e−

∫ xx0

p1(t)dt .


Proof

Proceeding as in the proof for second order case, we need to show

W ′ = −p1(x)W .

Notice that the derivative of∣∣∣∣ y1 y2y ′1 y ′2

∣∣∣∣is ∣∣∣∣ y ′1 y ′2

y ′1 y ′2

∣∣∣∣+

∣∣∣∣ y1 y2y ′′1 y ′′2

∣∣∣∣ ,which is ∣∣∣∣ y1 y2

y ′′1 y ′′2

∣∣∣∣ .In this, we had substituted y ′′i = −p1(x)y ′i − p2(x)yi to concludethe claim.


Proof

The derivative of W (y1, y2, . . . , yn) is the sum of n determinantswith derivative being taken in the first row in the first one, in thesecond row in the second one, etc. Except the last one, all vanishbecause two rows are identical. In the last one, make thesubstitution

y(n)i = −p1(x)y

(n−1)i − p2(x)y

(n−2)i − . . .− pn(x)yi .

Expand this into sum of n determinants. Once again, all but onevanish. The non-vanishing one gives

−p1(x) ·W (y1, y2, . . . , yn).

Thus, for solutions of linear homogeneous DE’s, the Wronskian iseither never zero or identically zero on I !


Basis of solutions & General solution (2nd order) - RECALL

If p(x) and q(x) are continuous on an open interval I , then

y ′′ + p(x)y ′ + q(x)y = 0

has a basis of solutions y1, y2 on I .

y(x) = C1y1(x) + C2y2(x)

is the general solution of the DE.That is, every solution y = Y (x) of the DE has the form

Y (x) = C1y1(x) + C2y2(x),

where C1 and C2 are arbitrary constants.


Basis of solutions & General solution (nth order)

If p1(x), . . . , pn(x) are continuous on an open interval I , then

y (n) + p1(x)y (n−1) + . . .+ pn(x)y = 0

has a basis of solutions y1, . . . , yn on I .

y(x) = C1y1(x) + C2y2(x) + · · ·+ Cnyn(x)

is the general solution of the DE.That is, every solution y = Y (x) of the DE has the form

Y (x) = C1y1(x) + · · ·+ Cnyn(x),

where C1, · · · ,Cn are arbitrary constants. (Prove this ! )


Non-homogeneous 2nd order Linear ODE’s - RECALL

Consider the non-homogeneous DE

y ′′ + p(x)y ′ + q(x)y = r(x)

where p(x), q(x), r(x) are continuous functions on an interval I .Let yp(x) be any solution of

y ′′ + p(x)y ′ + q(x)y = r(x)

and y1(x), y2(x) be a basis of the solution space of the correspondinghomogeneous DE.Then the set of solutions of the non-homogeneous DE is

{c1y1(x) + c2y2(x) + yp(x) | c1, c2 ∈ R}.

Summary:

In order to find the general solution of a non-homogeneous DE, we need to

get the general solution of the corresponding homogeneous DE.

get one particular solution of the non-homogeneous DE


Non-homogeneous nth order Linear ODE’s

Consider the non-homogeneous DE

y (n) + p1(x)y (n−1) + . . .+ pn(x)y = r(x)

where p1(x), · · · , pn(x), r(x) are continuous functions on an interval I .Let yp(x) be any solution of

y (n) + p1(x)y (n−1) + . . .+ pn(x)y = r(x)

and y1(x), · · · , yn(x) be a basis of the solution space of thecorresponding homogeneous DE.Then the set of solutions of the non-homogeneous DE is

{c1y1(x) + · · ·+ cnyn(x) + yp(x) | c1, · · · , cn ∈ R}.

Summary:

In order to find the general solution of a non-homogeneous DE, we need to

get the general solution of the corresponding homogeneous DE.

get one particular solution of the non-homogeneous DE


General Solution of Homogeneous Equations withConstant Coefficients

Considery (n) + p1y

(n−1) + . . .+ pny = 0

where p1, · · · , pn are in R; that is, an nth order homogeneous linearODE with constant coefficients.Suppose emx is a solution of this equation. Then,

mnemx + p1mn−1emx + · · ·+ pne

mx = 0,

and this implies

mn + p1mn−1 + · · ·+ pn = 0.

This is called the characteristic equation or auxiliary equation ofthe linear homogeneous ODE with constant coefficients.


This polynomial of degree n has n zeros say m1, · · · ,mn, some ofwhich may be equal and hence the characteristic polynomial canbe written in the form

(m −m1)(m −m2) · · · (m −mn).

Depending on the the nature of these zeros (real & unequal, real &equal, complex ), we write down the general solution of thehomogeneous DE in a similar way as we did for 2nd order equations.


Example 1

Find a basis of solutions and general solution of the DE:

y (3) − 7y ′ + 6y = 0.

1 The auxiliary equation is

m3 − 7m + 6 = 0.

2 The roots are 1, 2,−3.

3 Hence a basis for solutions is {ex , e2x , e−3x} Why?.

4 Thus, the general solution is of the form

c1ex + c2e

2x + c3e−3x ,

where c1, c2, c3 ∈ R.


Example 2

Find the general solution of the DE:

L(y) = (D3 − D2 − 8D + 12)(y) = 0.

1 The auxiliary equation is

m3 −m2 − 8m + 12 = (m − 2)2(m + 3) = 0.

2 The roots are 2, 2,−3.

3 Hence a basis for solutions is {e2x , xe2x , e−3x}. Why?

4 Thus, the general solution is of the form

c1e2x + c2xe

2x + c3e−3x ,

where c1, c2, c3 ∈ R.


Examples 3 & 4

Find the general solution of the DE:

L(y) = (D6 + 2D5 − 2D3 − D2)(y) = 0.

The general solution is

c1 + c2x + c3ex + c4e

−x + c5xe−x + c6x

2e−x ,

with ci ∈ R.Find the general solution of

y (5) − 9y (4) + 34y (3) − 66y (2) + 65y ′ − 25y = 0.

The characteristic polynomial is

(m − 1)(m2 − 4m + 5)2.

The roots are1, 2± ı, 2± ı.

Hence, the general solution is

y = c1ex + c2e

2x cos x + c3xe2x cos x + c4e

2x sin x + c5xe2x sin x ,

where ci ∈ R.Santanu Dey Lectures 14

Variation of Parameters- finding yp

Recall the case n = 2: We replaces c1, c2 in the general solution ofthe associated homogeneous equation by functions v1(x), v2(x), sothat

y(x) = v1(x)y1(x) + v2(x)y2(x)

is a solution of

y ′′ + p(x)y ′ + q(x)y = r(x).

Now,y ′ = v1y

′1 + v2y

′2 + v ′1y1 + v ′2y2.

We also demandedv ′1y1 + v ′2y2 = 0.


Variation of Parameters

Then we substituted y , y ′, y ′′ in the given non-homogeneous ODE,and rearranged terms, to obtain:

v ′1y′1 + v ′2y

′2 = r(x).

Together with,v ′1y1 + v ′2y2 = 0,

this yielded :

v ′1 =

∣∣∣∣ 0 y2r(x) y ′2

∣∣∣∣W (y1, y2)

, v ′2 =

∣∣∣∣ y1 0y ′1 r(x)

∣∣∣∣W (y1, y2)

,

and hence

v1 = −∫

y2r(x)

W (y1, y2)dx , v2 =

∫y1r(x)

W (y1, y2)dx .

Thus,

y = v1y1 + v2y2 = y2

∫y1r(x)

W (y1, y2)dx − y1

∫y2r(x)

W (y1, y2)dx .



For general n, we proceed exactly the same way.

c1y1 + c2y2 → c1y1 + c2y2 + . . .+ cnyn

y = v1y1 + v2y2 → y = v1y1 + v2y2 + . . .+ vnyn

y ′ = v1y′1 + v2y

′2 + v ′1y1 + v ′2y2

→

y ′ = v1y′1 + . . .+ vny

′n + v ′1y1 + . . .+ v ′nyn



Demandv ′1y1 + v ′2y2 = 0

→

Demandv ′1y1 + . . .+ v ′nyn = 0.

Takey ′′ = v1y

′′1 + . . .+ vny

′′n + v ′1y

′1 + . . .+ v ′ny

′n

and demandv ′1y′1 + . . .+ v ′ny

′n = 0.

. . . . . .



Take

y (n−1) = v1y(n−1)1 + . . .+ vny

(n−1)n + v ′1y

(n−2)1 + . . .+ v ′ny

(n−2)n

and demandv ′1y

(n−2)1 + . . .+ v ′ny

(n−2)n = 0.



Substituting y , y ′, y ′′, → Substituting y , y ′, . . . , y (n),

rearranging → rearranging

and using L(y1) = L(y2) = 0 → and using L(y1) = . . . = L(yn) = 0

get v ′1y′1 + v ′2y

′2 = r(x) → get

v ′1y(n−1)1 + . . .+ v ′ny

(n−1)n = r(x).



Thus, [y1 y2y ′1 y ′2

] [v ′1v ′2

]=

[0

r(x)

]takes the shape

y1 y2 · yny ′1 y ′2 · y ′n· · · ·

y(n−1)1 y

(n−1)2 · y

(n−1)n

v ′1v ′2·v ′n

=

00·

r(x)

.Use Cramer’s rule to solve for

v ′1, v′2, . . . , v

′n,

and thus getv1, v2, . . . , vn,

and formy = v1y1 + v2y2 + . . .+ vnyn.


Example 5

Solvey (3) − y (2) − y (1) + y = r(x).

Characteristic polynomial for the homogeneous equation is

m3 −m2 −m + 1 = (m − 1)2(m + 1).

Hence, a basis of solutions is

{ex , xex , e−x}.

We need to calculate W (x). Use Abel’s formula:

W (x) = W (0)e−∫ x0 p1(t)dt = W (0) · ex .


Now,

W (x) =

∣∣∣∣∣∣ex xex e−x

ex ex + xex −e−xex 2ex + xex e−x

∣∣∣∣∣∣ .Thus,

W (0) =

∣∣∣∣∣∣1 0 11 1 −11 2 1

∣∣∣∣∣∣ = 4.

Hence,W (x) = 4ex .



Let

W1(x) =

∣∣∣∣∣∣0 xex e−x

0 ex + xex −e−xr(x) 2ex + xex e−x

∣∣∣∣∣∣ = −r(x)(2x + 1).

Similarly,W2(x) = 2r(x), W3(x) = r(x)e2x .

Therefore,

y(x) = ex∫ x

0

−r(t)(2t + 1)

4etdt+xex

∫ x

0

2r(t)

4etdt+e−x

∫ x

0

r(t)e2t

4etdt.


Undetermined Coefficients - Example 6

Find a particular solution of

y (4) + 2y ′′ + y = 3 sin x − 5 cos x .

What’s the general solution of the homogeneous equation?The auxiliary equation is

m4 + 2m2 + 1 = (m2 + 1)2,

and therefore a basis of Ker L is

{cos x , sin x , x cos x , x sin x}.So the candidate is

y = x2(a cos x + b sin x).

Substituting, we get:

−8a cos x − 8b sin x = 3 sin x − 5 cos x .

Thus,

y(x) = x2(5

8cos x − 3

8sin x).


Example 7

Find the candidate solution for

y (4) − y (3) − y ′′ + y ′ = x2 + 4 + x sin x .

Note that the auxiliary/ characteristic equation is

m4 −m3 −m2 + m = m(m − 1)2(m + 1).

Thus a basis of Ker L is {1, ex , xex , e−x}. What’s the candidatefor solution?x2 + 4 on the right suggests ax2 + bx + c , but we should modifythis to

x(ax2 + bx + c),

since a constant is a solution of the homogeneous equation.x sin x suggests

(αx + β) cos x + (γx + δ) sin x .

Do we need to modify this? No. So our final candidate is

x(ax2 + bx + c) + (αx + β) cos x + (γx + δ) sin x .


Example 8

Find the candidate solution for

y (4) − 2y ′′ + y = x2ex + e2x .

Note that the AE is

m4 − 2m2 + 1 = (m − 1)2(m + 1)2.

So a basis of Ker L is {ex , xex , e−x , xe−x}.e2x on the right suggests

ae2x ,

and no modification required.x2ex initially suggests (bx2 + cx + d)ex , but this should bemodified to

x2(bx2 + cx + d)ex .

So final candidate would be

ae2x + x2(bx2 + cx + d)ex .


Review Question 1

For the non-homogeneous equation y ′′ − 5y ′ + 6y = x cos x , theform of yp using the method of undetermined coefficients is————-.(a) A1x cos x + B1x sin x(b) (A0 + A1x) cos x + (B0 + B1x) sin x(c) x((A0 + A1x) cos x + (B0 + B1x) sin x)(d) None of the above.

Solution : 4 (b)


ma 108 - ordinary differential equationsdey/diffeqn_spring14/lecture14_d2_prt.pdf · suppose emx is...

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