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I am trying to solve for the ODE $$y'' + 2y' + y = e^{-x}$$

I tried the ansatz of the form $$ y = Ae^{-x} $$ But the coefficients cancel out leaving me with $ 0 = e^{-x}$.

I'm guessing I have to be smarter with the ansatz? Maybe include a function like $y = Af(x)e^{-x}$ but that seems like it'll just result in another DE.

Any hint or suggestions on better ways to solve this would be appreciated.

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For a problem like this, you would include a function like $y=Af(x)e^{−x}$

In this case though, it is important to note that the solution to the homogeneous equation has a repeated root, so the roots will take the form $y=C_1e^{−x}$ and $y=C_2f(x)e^{−x}$.

The particular solution must then have a form that is different than both of the homogeneous solutions. Thus if your homogeneous solution takes the form $y=C_1e^{-x} + C_2xe^{-x}$, then your particular solution must have a form with a higher order of x.

When you take the derivatives of your particular solution and plug them into the original equation, you solve for a constant that eliminates the differential equations. Your final solution adds together the two homogeneous solutions and the particular solution.

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Hint

try $Ax^2e^{-x}$ because $c_1e^{-x} ,\, c_2xe^{-x}$ are already part of the homogeneous solution

If you substitute in the equation $y=ze^{-x}$ $$y'' + 2y' + y = e^{-x}$$ Then you got this: $$z''e^{-x}=e^{-x}$$ $$z''=1$$ Integrate $$z'=x+K_1$$ Integrate again $$z=\frac 12x^2+K_1x+K_2$$ Susbstitute back $y=ze^{-x} \implies z=ye^x$

$$ye^x=\frac 12x^2+K_1x+K_2$$ $$\boxed{y=e^{-x}\left(\frac 12x^2+K_1x+K_2\right)}$$

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