Nonhomogeneous differential equation: y"+2y'+y=xe-x

jman2807

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Sep 4, 2006
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I need to solve this equation :

y'' + 2y' + y = xe^-x

I found the complimentary solution which i will dub yc(x) and i know that the solution is the complimentary plus the particular ( yc(x) + yp(x) ) but I am having trouble finding the particular. I know it will have the form

y = (Ax + B)*e^-x
y' = e^-x * (A - Ax + B)
y'' = -e^-x * (2A - Ax + B)

right?

Then when i plug back in everything cancels out except 2B * e^-x = xe^-x

This can't be right. Any help appreciated.
 
You need to go to higher powers: \(\displaystyle y_p = (Ax^3+Bx^2)e^{-x}\). This gives \(\displaystyle B=0,A=1/6\).
 
Let's use Variation of Parameters:

\(\displaystyle \L\\y''+2y'+y=xe^{-x}\)

\(\displaystyle m^{2}+m+1=(m+1)^{2}\)

m=-1

Therefore,

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

\(\displaystyle \L\\y_{1}=e^{-x}; \;\ y_{2}=xe^{-x}\)

Wronskian:

\(\displaystyle \L\\W(e^{-x},xe^{-x})=\begin{vmatrix}e^{-x}&xe^{-x}\\{-}e^{-x}&(1-x)e^{-x}\end{vmatrix}=e^{-2x}\)

Since the coefficient of y'' is 1 we can use \(\displaystyle f(x)=xe^{-x}\)

\(\displaystyle \L\\W_{1}=\begin{vmatrix}0&xe^{-x}\\xe^{-x}&(1-x)e^{-x}\end{vmatrix}={-}x^{2}e^{-2x}\)

\(\displaystyle \L\\W_{2}=\begin{vmatrix}e^{-x}&0\\{-}e^{-x}&xe^{-x}\end{vmatrix}=xe^{-2x}\)

\(\displaystyle \L\\U_{1}^{'}=\frac{-x^{2}e^{-2x}}{e^{-2x}}={-}x^{2}\)

\(\displaystyle \L\\U_{2}^{'}=\frac{xe^{-2x}}{e^{-2x}}=x\)

Hence \(\displaystyle \L\\U_{1}=\frac{-x^{3}}{3}\)

\(\displaystyle \L\\U_{2}=\frac{x^{2}}{2}\)

So, \(\displaystyle \L\\y_{p}=\frac{-x^{3}}{3}e^{-x}+\frac{x^{2}}{2}xe^{-x}\)

\(\displaystyle \H\\y=y_{c}+y_{p}=e^{-x}\left(C_{1}+xC_{2}+\frac{x^{3}}{6}\right)\)
 
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