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I am a bit stuck on the following second order differential equation, using the method of undertermined coefficients. $$y'' + 3y' + 2y = xe^{-x}$$

The homogenous solution is easy to find, but I run into some issues with the particular solution. $y = Axe^{-x}$ doesn't seem to be a good enough guess, but I am not sure why? Since it doesn't appear in the homogenous solution I thought it would be independent from it. What is a better particular solution to get started with this?

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4 Answers 4

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Using the substitution $u=y'+2y$, so $u'=y''+2y'$, the equation becomes $$u'+u=x\mathrm{e}^{-x}$$ Using the integrating factor method this becomes $$u'\mathrm{e}^x+u\mathrm{e}^x=\frac{\mathrm{d}}{\mathrm{d}x}\left(u\mathrm{e}^x\right)=x\\u\mathrm{e}^x=\frac12x^2+A\\u=y'+2y=\left(\frac12x^2+A\right)\mathrm{e}^{-x}$$Using the integrating factor method again this becomes $$y'\mathrm{e}^{2x}+2y\mathrm{e}^{2x}=\frac{\mathrm{d}}{\mathrm{d}x}\left(y\mathrm{e}^{2x}\right)=\left(\frac12x^2+A\right)\mathrm{e}^{x}\\y\mathrm{e}^{2x}=\left(\frac12x^2-x+B\right)\mathrm{e}^x+C\\\boxed{y= \left(\frac12x^2-x+B\right)\mathrm{e}^{-x}+C\mathrm{e}^{-2x}}$$Note that $x^2\mathrm{e}^{-x}$ is used as well as $x\mathrm{e}^{-x}$.

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The idea is $e^{-x}$ is a homogeneous solution of the ODE. So if you purely let $y=Axe^{-x}$, then you can see the function with top degree of $y''+3y'+2y$ will vanish itself because the top degree term is just simply keep differentiating $e^{-x}$ (by product rule) (Top degree here means terms with $xe^{-x}$)

So you should increase the degree of your guess by one to prevent this, i.e. $y=Ax^2e^{-x}+Bxe^{-x}$. This is generally true for any $x^me^{-x}$, as long as $e^{-x}$ is a homogeneous solution on any order of ODE.

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  • $\begingroup$ Because $Ax^2e^{-x}$ will give terms with $ke^{-x}$ on its second derivative, but no one in $y$ and $y'$ can eliminate this guy, so we need some help, from the lower degree one. $\endgroup$
    – Angae MT
    Apr 22 at 16:32
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This 2nd order ODE equation $y'' + 3y' + 2y = xe^{-x}$ can be rewritten as

$$ (D^2+3D+2)y_p=(D+1)(D+2)y_p=xe^{-x} $$ which can be solved by

\begin{align} y_p&=\frac{1}{(D+1)(D+2)}xe^{-x}\\ &=\left(\frac{1}{D+1}-\frac{1}{D+2}\right)xe^{-x}\\ &=e^{-x}D^{-1}x-e^{-x}(1-\color{red}{D})x\\ &=e^{-x}\left(\frac{x^2}{2}-x+1\right) \end{align} where $D^{-1}$ stands for integration. Note that the last term $e^{-x}Dx=e^{-x}$ is actually redundant as a particular solution since it is already included in the homogeneous solution $y_h=c_1e^{-x}+c_2e^{-2x}$, but the operator produced one by-product through integration (actually convolution).

For the operator approach to solve a general inhomogenous ODE in any order, the answer (2nd order example) to this post might be useful General Solution of $y''+4y=\frac{3}{\sin(2t)}$.

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$$y'' + 3y' + 2y = xe^{-x}$$

$$y=z\,e^{-x} \quad \implies \quad z''+z'=x$$

Reduction of order makes now things simple.

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