# Second order differential equation check. Find general solution

I could use a process check. In general since the solution is $$y = y_c + y_p$$ where $$y_c$$ is the complementary solution and $$y_p$$ is the particular, I can break it down.

The equation in question is this:

$$y" - 3y' = sin2x$$

Let's find the complementary solution first:

$$y_c = c_1e^{r_1x} + c_2e^{r_2x}$$

so let's solve the auxiliary equation:

$$r^2 - 3r = 0$$ $$r(r - 3) = 0$$ so the roots are $$r = {0, 3}$$

so the complementary equation is:

$$y = c_1 + c_2e^{3x}$$

Let's find the particular solution second:

so the form of the particular since it's a trig function on the right side is: $$y_p = A\cos{kx} + B\sin{kx}$$

where k is the coefficient for the trig function radian measure.

$$y'p = -2A\sin{2x} + 2B\cos{2x}$$ $$y''p = -4A\cos{2x} - 4B\sin{2x}$$

so substituting:

$$(-4A\cos2x - 4B\sin2x) - 3(-2A\sin2x + 2B\cos2x) = \sin{2x}$$

$$(-4B + 6A)sin2x + (-4A - 6B)cos2x = sin2x$$

so matching coefficients $$(-4B + 6A) = 1$$ and $$-6B - 4A = 0$$ and $$-6B = 4A$$ so $$B = \frac{-2}{3}A$$

solving further:

$$\frac{8}{3}A + 6A = 1$$ so $$\frac{26}{3}A = 1$$ so $$A = \frac{3}{26}$$ and so $$B = \frac{-12}{216}$$

so then the general solution is:

$$y = c_1 + c_2e^{3x} + \frac{3}{26}cos2x - \frac{12}{216}sin2x$$

Is my process right? Did I make an error?

• Plug your solution in the equation and see. – Yves Daoust May 5 at 20:44

$$y_p = A\sin{kx} + B\cos{kx}$$ where k is the coefficient for the trig function radian measure. $$y'p = -2A\sin{2x} + 2B\cos{2x}$$ $$y''p = -4A\cos{2x} - 4B\sin{2x}$$

A problem results here. When taking $$k=2$$, you should see

$$y_p = A \sin 2x + B \cos 2x$$

and then you get the derivatives

$$y_p' = 2A \cos 2x - 2B \sin 2x \;\;\;\;\; y_p'' = -4A \sin 2x - 4B \cos 2x$$

You seem to have forgotten to taken the derivative of the trigonometric functions themselves when getting $$y_p'$$.

This is mostly just an arithmetic error, though, and the overarching idea (and the complimentary solution) are correct.

Edit:

After an edit made to the OP, that error was fixed. The other error that results is finding $$A,B$$, the latter in particular.

$$A=3/26$$ is correct. However, an arithmetic error seems to have resulted in finding $$B$$. Using substitution into $$-4A-6B=0$$,

$$-4\left( \frac 3 {26} \right) - 6B = 0 \implies B = \frac{-1}{6} \cdot 4\left( \frac 3 {26} \right) = \frac{-12}{156} = \frac{-1}{13} \ne \frac{-12}{216} = \frac{-1}{18}$$

• Edited, I wrote the particular equation y incorrectly, but I think my derivatives are correct? – Jwan622 May 5 at 20:51
• @Jwan622 Okay, made an addendum to my original post with a new error. Everything with $y_p$ seems to be correct, you seem to have somehow solved for $B$ wrong though. Details are above, but you basically should have gotten $B = -1/13$ instead of $B = -12/216 = -1/18.$ – Eevee Trainer May 5 at 21:03

Your process is correct. Your answer is wrong because you have a mistake in finding your constants $$A$$ and $$B$$.

The correct values for $$A$$ and $$B$$ are $$A=\frac {3}{26}$$ and $$B=\frac {-1}{13}$$

Fix this minor miscalculation and your solution is as good as gold.