# Check my solution for solvin this IVP $y''' + 3 y'' + 4y' + 12 y = 0$

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

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

• Did you try substituting this back in the equation? – Sean Roberson Nov 10 '18 at 19:59
• 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. – Eevee Trainer Nov 10 '18 at 20:01
• Well I did, but it will better if I have an additional set of eyes to verify if it's right! – Itsnhantransitive Nov 10 '18 at 20:06
• I checked looks good to me... – Isham Nov 10 '18 at 20:08