LECTURE 25

NON-HOMOGENEOUS SECOND ORDER ODEs

To solve the non-homogeneous second order constant coefficient ODE

ay + by + cy = r(x)

we first solve the homogeneous problem ay + by + cy = 0 to obtain a homogenous solution yh.We then find a particular solution yp by using the method of undetermined coefficients. Thefinal solution is then y = yh + yp.

We now deal with the constant coefficient second order ODEs of the previous lectureexcept that we will allow nice functions to appear on the RHS. This makes the ODEnon-homogeneous. We begin by finding the homogeneous solution yh by using the methodsof the previous lecture. We then find a particular solution yp by essentially guessing ananswer. The final solution is then y = yh + yp. It should be noted that this technique isprone to fail and will only work when the RHS is uncomplicated.

Example 1 Solve y 5y + 6y = 6x+ 23 .

y = Ae2x +Be3x x+ 3

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Why does this work?

The crucial question is of course how do we know what to guess for yp? The generalrule is that you guess an arbitrary representation of the RHS. The following table givessome RHSs and their associated guesses for yp .

RHS Choice of yP3e4x Ce4x

x3 7 x3 + x2 + x+ 3 sin(4x) sin(4x) + cos(4x)

5e7x cos(2x) e7x( sin(2x) + cos(2x))9x2e3x e3x(x2 + x+ )

Note that if the RHS is too weird then no amount of guessing will save you.

Example 2 Solve the initial value problem

y + y = 55e2x + 3x2 + 14

where y(0) = 20 and y(0) = 28.

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y = 6 sin x+ cosx+ 11e2x + 3x2 + 8

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VARIATION OF PARAMETERS

If the method of undetermined coefficients fails to produce a particular solution ypit is possible that technique of variation of parameters will still work. This is a highlyspecialised process:

Variation of Parameters

Suppose that the second order ODE

y + p(x)y + q(x)y = f(x)

has homogeneous solution yh = Ay1(x) +By2(x). Then a particular solution is given by

yP (x) = y1(x)

y2(x)f(x)

W (x)dx+ y2(x)

y1(x)f(x)

W (x)dx

where W (x) is called the Wronskian and is given by W (x) = det

(y1(x) y2(x)

y1(x) y

2(x)

).

A full proof of this formula is found in your printed notes.

Example 3 Use Variation of Parameters to solve y + y = sec x.

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y = A cosx+B sin x+ cosx ln(| cosx|) + x sin x

25You can now do Q85 a and b, Q88

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