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Consider this IVP:

$$y''' + 3 y'' + 4y' + 12 y = 0$$

$y(0) = a, y'(0) = b, y''(0) = c $

where $a,b,$ and $c$ are real numbers

So let's start by finding the characteristic equations

$$r^3 + 3r^2 + 4r + 12 = 0$$ $$(r + 3) (r^2 + 4) = 0 $$ $$ r = -3, \pm 2i $$

So the general solution is

$$\implies y = c_1 e^{-3x} + c_2 \cos 2x + c_3 \sin 2x$$ $$ y' = -3c_1 e^{-3x} - 2c_2 \sin 2x + 2c_3 \cos2x $$ $$ y'' = 9c_1 e^{-3x} - 4c_2 \cos 2x - 4c_3 \sin 2x $$

Initial conditions

$$ c_1 + c_2 = a \\ - 3c_1 + 2c_3 = b \\ 9c_1 - 4c_2 = c$$

So $$ c_1 = \frac{1}{13} (4a + c) $$ $$ c_2 = \frac{1}{13} (9a - c) $$ $$ c_3 = \frac{1}{26} ( 12 a + 13 + 3c) $$

Therefore

$$ y = \frac{1}{13} \left [ (4a + c) e^{-3x} + (9a - c) \cos 2x + \frac{1}{2} (12a + 13b + 3c) \sin 2x \right ] $$

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    $\begingroup$ Did you try substituting this back in the equation? $\endgroup$ – Sean Roberson Nov 10 '18 at 19:59
  • $\begingroup$ You could just check by differentiating $y$ yourself. It looks like it'd be a mess because of the coefficients of your exponential, cosine, and sine functions, but you could just replace them by intermediate variables - like let $(4a+c)/13 = \alpha, (9a-c)/13=\beta, (12a+13b+3c)/26 =\gamma$ - until you've finished differentiating and doing all the other algebra. $\endgroup$ – Eevee Trainer Nov 10 '18 at 20:01
  • $\begingroup$ Well I did, but it will better if I have an additional set of eyes to verify if it's right! $\endgroup$ – Itsnhantransitive Nov 10 '18 at 20:06
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    $\begingroup$ I checked looks good to me... $\endgroup$ – Isham Nov 10 '18 at 20:08
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It should be easy for you to check the conditions directly, remembering that the cosine term drops out of the first derivative and the sine term drops out of the other two conditions.

I do get the same result as yours.

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