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Solve DE: $$y'' + 2y' + 5y = x + 4$$

I have the correct general solution $$y(x) = c_1\,e^{-x}\cos(2x) + c_2\,e^{-x}\sin(2x)\ .$$

So I take my 'guess' , take a couple of derivatives, and plug in for the equation $$2A +5(Ax + B) = x + 4$$

But at this point I'm stuck on solving for the constants. I attempted in several times but I'm just getting a jumbled mess. I realize that this is probably simple algebra I'm struggling with here

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  • $\begingroup$ Do you have some initial conditions? $\endgroup$ – Emilio Novati Feb 27 '15 at 21:05
  • $\begingroup$ What is your guess? $\endgroup$ – Git Gud Feb 27 '15 at 21:08
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You need your equation to hold for all $x$, right? Then the parts that depend on $x$ must be equal ($5Ax = x$) and the parts that do not depend on $x$ must be equal; thus $A$ and $B$ must satisfy

$$ 5A = 1 \\ 2A + 5B = 4. $$

Can you solve for $A$ and $B$ from here?

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The ansatz for $y_p = A x + B$ put in the DE gives $2A + 5(Ax + B) = x + 4$ where you compare coefficients for $x^k$, here $x^1=x$ and $x^0=1$, on both sides of the expanded equation. Finally $y = y_h + y_p$.

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