Some more Differential Equations

**brd_7** · February 8th 2008, 01:51 PM

The question states to find the solution of the following -

$\frac{d^2y}{dx^2} - 6\frac{dy}{dx} + 10y = 20 - e^2x$

I found the root, it was 3+- i

Im not too sure what to do after this, any help would be appreciated.
Thanks in advance

**Soroban** · February 8th 2008, 04:29 PM

Hello, brd_7!

Find the solution: . $\frac{d^2y}{dx^2} - 6\frac{dy}{dx} + 10y \;= \;20 - e^{2x}$

I found the root, it was $3 \pm i$

I'm not too sure what to do after this.

Those roots give us the homogeneous solution: . $y \;=\;e^{3x}\left(C_1\cos x + C_2\sin x\right)$

Now we must find the particular solution . . .

I conjectured that the solution is of the form: . $y_p \:=\:A + Be^{2x}$

. . and found it to be: . $y_p \;=\;2 - \frac{1}{2}e^{2x}$

The solution is: . $y \;=\;e^{3x}\left(C_1\cos x + C_2\sin x\right) + 2 - \frac{1}{2}e^{2x}$

**mr fantastic** · February 8th 2008, 05:52 PM

Originally Posted by brd_7

[snip]
$\frac{d^2y}{dx^2} - 6\frac{dy}{dx} + 10y = 20 - e^2x$

I found the root, it was 3+- i

Im not too sure what to do after this, any help would be appreciated.
[snip]

Then you need to read this.

**brd_7** · February 9th 2008, 05:26 AM

Ok, i managed to prove it all myself.. I just didnt know what to do because of the extra '20' term on the RHS.. now i know that you just put an A to represent the constant makes it a whole lot easier. Thank you!

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