# nonhomogeneous 2nd order linear eqn

• Dec 17th 2009, 03:59 PM
harveyo
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!
• Dec 17th 2009, 05:19 PM
kjchauhan
Use method of variation of parameters,
You get,
$y=\left( \frac{2}{9}\left(e^{-3x}-3e^{-x}\right)+c_1\right)e^x-\left(\frac{1}{3}\left(e^{2x}-x\right)+c_2\right)e^{-2x}$
• Dec 18th 2009, 05:09 AM
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 $Y_p(x)= A+ Bxe^{-2x}$.
• Dec 18th 2009, 09:19 AM
kjchauhan
Quote:

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..
• Jan 4th 2010, 06:52 PM
harveyo
could somebody show me the next line combining yp to yc
• Jan 4th 2010, 08:14 PM
harveyo
i am now at 10 posts.
• Jan 4th 2010, 09:33 PM
harveyo
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

????
• Jan 5th 2010, 04:22 AM
HallsofIvy
Quote:

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 $y_p'= 0+ Be^{-2x}- 2Bxe^{-2x}$

Quote:

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

Quote:

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

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

????
• Jan 6th 2010, 01:26 PM
harveyo
makes complete sense now. (Doh)