# Thread: nonhomogeneous 2nd order linear eqn

1. ## nonhomogeneous 2nd order linear eqn

y''+y' -2y=2-2e^-2x ...... find general soln

ay''+by'+cy= G(x)

y''+y'-2y=0

r^2+r-2=(r-1)(r+2)=0

roots = 1 & -2

Yc = c1 e ^x + c2 e^-2x

Yp???? stuck!

2. Use method of variation of parameters,
You get,
$y=\left( \frac{2}{9}\left(e^{-3x}-3e^{-x}\right)+c_1\right)e^x-\left(\frac{1}{3}\left(e^{2x}-x\right)+c_2\right)e^{-2x}$

3. "Variation of parameters" always works but is typically much harder than "undetermined coefficients" when that works- which it will as long as the right hand side consists of the kind of functions we get for homogeneous linear equations with constant coefficients, polynomials, exponentials, sine and cosine, or products of those.

Here, your right hand side is $2- 2e^{-2x}$. Because of the constant "2", try some constant A. Because of the " $-2e^{-2x}$", your first thought should be " $Be^{-2x}$". However, $e^{-2x}$ is already a solution to the associated homogeneous equation so try " $Bxe^{-2x}$" instead.

In other words, try $Y_p(x)= A+ Bxe^{-2x}$.

4. Originally Posted by HallsofIvy
"Variation of parameters" always works but is typically much harder than "undetermined coefficients" when that works- which it will as long as the right hand side consists of the kind of functions we get for homogeneous linear equations with constant coefficients, polynomials, exponentials, sine and cosine, or products of those.

Here, your right hand side is $2- 2e^{-2x}$. Because of the constant "2", try some constant A. Because of the " $-2e^{-2x}$", your first thought should be " $Be^{-2x}$". However, $e^{-2x}$ is already a solution to the associated homogeneous equation so try " $Bxe^{-2x}$" instead.

In other words, try $Y_p(x)= A+ Bxe^{-2x}$.
Ok..That also works..

5. could somebody show me the next line combining yp to yc

6. i am now at 10 posts.

7. yp'= -2(A+Bxe^-2x)
yp''= 4(A+Bxe^-2x)

therefore from orginal equation
y''+y' -2y

4(A+Bxe^-2x) + -2(A+Bxe^-2x) - 2*(A+Bxe^-2x) = 2-2e-2x

????

8. Originally Posted by harveyo
yp'= -2(A+Bxe^-2x)
What are you taking as yp? I suggested $y_p= A+ Bxe^{-2x}$ but that certainly is not its derivative.

If $y_p= A+ Bxe^{-2x}$, then $y_p'= 0+ Be^{-2x}- 2Bxe^{-2x}$

yp''= 4(A+Bxe^-2x)
Even with " $y_p'= -2(A+ Bxe^{-2x})$" this would not be the second derivative. You are aware that the derivative of -2A is 0, aren't you?

You should have $y_p"= -4Be^{-2x}+ 4Bxe^{-2x}$

therefore from orginal equation
y''+y' -2y

4(A+Bxe^-2x) + -2(A+Bxe^-2x) - 2*(A+Bxe^-2x) = 2-2e-2x

????

9. makes complete sense now.