Answer to Question #2009 in Differential Equations for fady

Question

y''+2y'+24y=16-(x+2)e^4x

Answer 1

Let's solve& the homogeneous equation. The roots of parametric equation are

<img src="/cgi-bin/mimetex.cgi?\lambda^2 + 2\lambda +24 =0 \\ \lambda = -1 \pm i\sqrt{23}" title="\lambda^2 + 2\lambda +24 =0 \\ \lambda = -1 \pm i\sqrt{23}" />

So
<img src="https://latex.codecogs.com/gif.latex?CF = e^{-x}(A \cos{\sqrt{23}} + B \sin{\sqrt{23}})" title="CF = e^{-x}(A \cos{\sqrt{23}} + B \sin{\sqrt{23}})" />

Let's find the particular integral. We choose it in the form

<img src="https://latex.codecogs.com/gif.latex?PI = e^{4x}(Cx+D)+ E" title="PI = e^{4x}(Cx+D)+ E" />

<img src="https://latex.codecogs.com/gif.latex?y' = e^{4x}(4Cx + 4D + C)\\ y'' = e^{4x}(16Cx + 16D + 8C)" title="y' = e^{4x}(4Cx + 4D + C)\\ y'' = e^{4x}(16Cx + 16D + 8C)" />
Thus

<img src="https://latex.codecogs.com/gif.latex?e^{4x}(16Cx + 16D +8C) + e^{4x}(8Cx + 8D +2C)+ \\ + e^{4x}(24Cx+24D)+24E = 16 - (x+2)e^{4x}" title="e^{4x}(16Cx + 16D +8C) + e^{4x}(8Cx + 8D +2C)+ \\ + e^{4x}(24Cx+24D)+24E = 16 - (x+2)e^{4x}" />.

<img src="https://latex.codecogs.com/gif.latex?C = -\frac{1}{48} \hspace{3mm} D = -\frac{43}{24} \hspace{3mm} E = \frac{2}{3}" title="C = -\frac{1}{48} \hspace{3mm} D = -\frac{43}{24} \hspace{3mm} E = \frac{2}{3}" />

hence

<img src="https://latex.codecogs.com/gif.latex?y(x) = e^{-x}(A\cos{\sqrt{23}}+ B \sin{\sqrt{23}})+e^{4x}(-x/48 - 43/24) + 2/3" title="y(x) = e^{-x}(A\cos{\sqrt{23}}+ B \sin{\sqrt{23}})+e^{4x}(-x/48 - 43/24) + 2/3" />