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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?

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  • $\begingroup$ Plug your solution in the equation and see. $\endgroup$ – Yves Daoust May 5 at 20:44
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$$y_p = A\sin{kx} + B\cos{kx}$$ 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 $-4A-6B=0$,

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

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  • $\begingroup$ Edited, I wrote the particular equation y incorrectly, but I think my derivatives are correct? $\endgroup$ – Jwan622 May 5 at 20:51
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    $\begingroup$ @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 $B = -12/216 = -1/18.$ $\endgroup$ – Eevee Trainer May 5 at 21:03
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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.

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