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find the general solution of

$$\frac {d^2y}{dx^2} +9y =18$$

I am not sure how to write it in its complementary form because of the roots one being positive and the other negative

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    $\begingroup$ The general solution is $c_1e^{r_1t} + c_2e^{r_2t}$ where $r_1$ and $r_2$ are the roots of the characteristic polynomial. The roots being positive or negative doesn't change anything. $\endgroup$ – Math1000 Apr 29 '16 at 15:55
  • $\begingroup$ The particular solution is $y=2$. $\endgroup$ – velut luna Apr 29 '16 at 15:56
  • $\begingroup$ I don't understand can you give me some more help $\endgroup$ – Tesha Caesar Apr 29 '16 at 15:58
  • $\begingroup$ Note that $\frac{d^2}{dx^2}(y-2)+9(y-2)=0$ - can you solve this? $\endgroup$ – πr8 Apr 29 '16 at 16:04
  • $\begingroup$ as in implicit? $\endgroup$ – Tesha Caesar Apr 29 '16 at 16:07
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Take Laplace Transform, $(s^2+9)Y(s) - c_1 s - c_2 = \frac{18}{s}$

Thus, $Y(s) = \frac{18}{s(s^2+9)} + c_1 \frac{s}{s^2+9} + c_2\frac{1}{s^2+9} = 2\frac{s^2+9-s^2}{s(s^2+9)}+ c_1 \frac{s}{s^2+9} + c_2\frac{1}{s^2+9} = \frac{2}{s} + (c_1-2) \frac{s}{s^2+9} + c_2\frac{1}{s^2+9}$.

Now, do an inverse Laplace transform. $y(x) = 2 + (c_1-2) cos(3x) + \frac{c_2}{3} sin(3x)$, where $c_1 = y(0)$, and $c_2 = y'(0)$

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