2y'(x) + y''(x) = x : How do I find particular solution?

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Apr 28, 2006
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Homogeneous solution:

\(\displaystyle \L\\2y'(x)+y''(x)=x\)

\(\displaystyle \L\\r^{2}+2r=0\)
\(\displaystyle \L\\r(r+2)=0\)
\(\displaystyle \L\\r_{1}=0\)
\(\displaystyle \L\\r_{2}=-2\)

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

How do I find particular solution?
 
You already have \(\displaystyle \L\\y_{c}=C_{1}e^{-2x}+C_{2}\)

Since we had a quadratic, perhaps we can assume a particular solution in the form \(\displaystyle \L\\y_{p}=Ax^{2}+Bx+C\)

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

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

Sub them into your equation and equate coefficients:

\(\displaystyle \L\\4Ax+2A+2B=x\)

This gives:

\(\displaystyle \L\\4A=1\)

\(\displaystyle \L\\2A+2B=0\)

\(\displaystyle \L\\A=\frac{1}{4}, \;\ and \;\ B=\frac{-1}{4}\)

So, the particular solution is:

\(\displaystyle \L\\C_{1}e^{-2x}+C_{2}+\frac{1}{4}x^{2}-\frac{1}{4}x\)

Check it out and make sure I didn't flub up somewhere.
 
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