Question

find a particular solution $y_p$ of the following equation using the method of undetermined coefficients. primes denote the derivatives with respect to x.
$y - y - 6y = 28sin3x$
a particular solution is $y_p(x)=$

Explanation:

Step1: Assume the form of particular solution

Assume $y_p = A\sin3x + B\cos3x$.

Step2: Calculate first - derivative

$y_p'=3A\cos3x - 3B\sin3x$.

Step3: Calculate second - derivative

$y_p''=- 9A\sin3x - 9B\cos3x$.

Step4: Substitute into the given equation

Substitute $y_p$, $y_p'$ and $y_p''$ into $y''-y'-6y = 28\sin3x$.
We get $(-9A\sin3x - 9B\cos3x)-(3A\cos3x - 3B\sin3x)-6(A\sin3x + B\cos3x)=28\sin3x$.
Group the $\sin3x$ and $\cos3x$ terms:
$(-9A + 3B-6A)\sin3x+(-9B - 3A-6B)\cos3x=28\sin3x+0\cos3x$.
So we have the system of equations:

$$\begin{cases}-15A + 3B=28\\-3A-15B = 0\end{cases}$$

From the second equation $A=- 5B$.
Substitute $A = - 5B$ into the first equation:
$-15(-5B)+3B=28$,
$75B + 3B=28$,
$78B=28$, $B=\frac{14}{39}$.
Then $A=-\frac{70}{39}$.

Answer:

$y_p(x)=-\frac{70}{39}\sin3x+\frac{14}{39}\cos3x$