2nd order non-hom. eqn: y'' - 2y' - 3y = x^2 - 1

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f1player

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Feb 25, 2005
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I need to solve the following:

y'' - 2y' - 3y = x^2 - 1

I found the homogeneous solution which was : Ae^(3x) + Be^(-x)

For the particular solution i tried Ax^2 - B since the RHS term is x^2 - 1 but this doesnt seem to work. what else should i try?
 
so if we try: Ax^2 + Bx + C

then find the first and second derivatives and sub into ODE i get:

2A - 4ax -2B - 3Ax^2 -3Bx - 3C = x^2 - 1

??? what now, how do u simplify this??
 
Hello f1player:

The method of undetermined coefficients.

\(\displaystyle \L\\y"-2y'-3y=x^{2}-1\)

The roots of \(\displaystyle \L\\m^{2}-2m-3=0\) are 3 and -1.

So we have the complementary solution:

\(\displaystyle \L\\y_{c}=C_{1}e^{3x}+C_{2}e^{-x}\)

Because the right hand side is a quadratic we can use the form

\(\displaystyle \L\\Ax^{2}+Bx+C\)

\(\displaystyle \L\\y_{p}=Ax^{2}+Bx+C\)

\(\displaystyle \L\\y'_{p}=2Ax+B\)

\(\displaystyle \L\\y''_{p}=2A\)

Sub into the equation:

\(\displaystyle \L\\2A-2(2Ax+B)-3(Ax^{2}+Bx+C)=x^{2}-1\)

\(\displaystyle \L\\-3Ax^{2}-4Ax-3Bx+2A-2B-3C=x^{2}-1\)

Equating coefficients gives the system:

\(\displaystyle \L\\-3A=1\)

\(\displaystyle \L\\-4A-3B=0\)

\(\displaystyle \L\\2A-2B-3C=-1\)

Solving the system gives:

\(\displaystyle \L\\A=\frac{-1}{3}, \;\ B=\frac{4}{9}, \;\ C=\frac{-5}{27}\)

Which gives the particular solution:

\(\displaystyle \L\\y_{p}=\frac{-1}{3}x^{2}+\frac{4}{9}x-\frac{5}{27}\)

Therefore, the general solution is \(\displaystyle y=y_{c}+y_{p}\)

\(\displaystyle \L\\\fbox{y=C_{1}e^{3x}+C_{2}e^{-x}-\frac{1}{3}x^{2}+\frac{4}{9}x-\frac{5}{27}}\)


See?.
 
yes, i understand now. i was on the right track just couldn't see that last step. however im still having some problems with choosing the correct function for the particular solution. for example in another question the term on the RHS equals 4x + 4e^x . I chose Ax + Be^x and got the wrong answer, so im still a bit hazy on this.
 
I believe then you would use \(\displaystyle \L\\Ax+B+Ce^{x}\)
 
f1player said:
so if we try: Ax^2 + Bx + C

then find the first and second derivatives and sub into ODE i get:

2A - 4Ax -2B - 3Ax^2 -3Bx - 3C = x^2 - 1

??? what now, how do u simplify this??
Click to expand...
Compare coefficients of x^2, x and constant terms on each side.

x^2 coefficient: -3A = 1, so A = -1/3,

x coefficient: -4A - 3B = 0, so B = -4A/3 = 4/9,

constant: 2A - 2B - 3C = -1, so C = (1/3)(1 + 2A - 2B) = -5/27

(which agrees with galactus's answer).

f1player said:
... in another question the term on the RHS equals 4x + 4e^x . I chose Ax + Be^x and got the wrong answer, so im still a bit hazy on this.
Click to expand...
Not having seen the problem, I would guess that you should try Ax + B + Ce^x. In general, if you're trying to get a polynomial on the RHS you need to use a general polynomial of that degree.
 
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