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Given differential eqn of,

$y'' + 16y = \cos(4x) - 8e^{3x}$

Initial conditions $y=0$ and $dy/dx = 0$ when $x = 0$

So therefor,

$r^2 + 16 = 0$

where $r$ will be $4i$

so, $Y = C\cos(4x) + D\sin(4x)$

The RHS is the part I have problem with...

This is what I've done so far

$Y_p = Ae^{3x} + B\cos(4x)$

$Y_p' = 3Ae^{3x} - 4B\sin(4x)$

$Y_p'' = 9Ae^{3x} - 16B\cos(4x)$

Sub into eqn...

$9Ae^{3x} - 16B\cos(4x) + 16(Ae^{3x} + B\cos(4x)) = \cos(4x) - 8e^{3x}$

$9Ae^{3x} - 16B\cos(4x) + 16Ae^{3x} + 16B\cos(4x) = \cos(4x) - 8e^{3x}$

$25Ae^{3x} = \cos(4x) - 8e^{3x}$

Taking coeff of $e^{3x}$,

$25A = -8$

$A = -8/25$

From here on Im stuck, not to sure how to find $B$.

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    You have written sometimes e^3x and sometimes e^-3x in $Y_p$ and its derivatives. Did you mean $e^{3x}$ everywhere?2012-05-19
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    BTW including $\cos 4x$ into $Y_p$ does not seem to be very good, since $\cos 4x$ is a solution of homogeneous equation.2012-05-19
  • 0
    @MartinSleziak oh yes, my mistake. I meant to type e^3x for all2012-05-19

3 Answers 3