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?

  • $\begingroup$ Plug your solution in the equation and see. $\endgroup$
    – user65203
    May 5, 2019 at 20:44

2 Answers 2


$$y_p = A\sin{kx} + B\cos{kx}$$ 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.


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 $-4A-6B=0$,

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

  • $\begingroup$ Edited, I wrote the particular equation y incorrectly, but I think my derivatives are correct? $\endgroup$
    – Jwan622
    May 5, 2019 at 20:51
  • 1
    $\begingroup$ @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 $B = -12/216 = -1/18.$ $\endgroup$ May 5, 2019 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.


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