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I have the DE

$y''-4y'+4y=e^{2x}$

The general solution to the corresponding homogeneous equation is,

$y_h(x)=e^{2x}(A+Bx)$

How do I find a particular solution $y_p(x)$ that fits with this?

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    $\begingroup$ Before we do this, please check whether you have the right DE. If you do, the solution of the homogeneous equation is not quite right. $\endgroup$ – André Nicolas Oct 15 '11 at 16:55
  • $\begingroup$ the solution of the homogeneous equation is ok $\endgroup$ – Peđa Terzić Oct 15 '11 at 17:13
  • $\begingroup$ sosmath.com/diffeq/second/variation/variation.html $\endgroup$ – Peđa Terzić Oct 15 '11 at 17:14
  • $\begingroup$ @pedja, you really believe $x\mapsto\mathrm e^{2x}$ solves the homogenous equation? Wow. $\endgroup$ – Did Oct 15 '11 at 17:42
  • $\begingroup$ @AndréNicolas You're right, the equation was $y''-4y'+4y=e^{2x}$ , not $y''+4y'+4y=e^{2x}$ . Sorry about that. $\endgroup$ – JaHei Oct 15 '11 at 18:07
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There's a method called "the method of undetermined coefficients" that deals with such equations. Normally, when you see $e^{2x}$ on the right-hand side you try $y_p = Ae^{2x}$, plug it in, then solve for $A$. However, since $2$ is a double root of the characteristic equation $r^2 - 4r + 4 = 0$, you have to try $y_p = Ax^2 e^{2x}$ instead. If you plug this in, and solve for $A$, you obtain $A = {1 \over 2}$ and therefore $y_p = {1 \over 2}x^2 e^{2x}$.

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$y^{\prime \prime} -4y^{\prime} + 4y = e^{2x} \quad \Rightarrow \quad (D^2 - 4D + 4)y = e^{2x} \quad \Rightarrow$

$(D -2)^2y = e^{2x} \quad \Rightarrow \quad y = \frac{1}{(D - 2)^2}e^{2x} = e^{2x}\frac{1}{D^2}\cdot 1 = \frac{x^2}{2}e^{2x}$

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