Find A,B such that the given function is a solution to the given differential equation

Question

The equation is: $x(t)=A\sin(t)+B\cos(t), \; x' - 3x = \frac{1}{2}\cos(t)$

I'm honestly just lost from where to start.

I'd really appreciate any help.

Answer: $A = \frac{1}{20}, \; B= -\frac{3}{20}$

Answer 1

It's easier than you think.

$$x(t)= A\sin(t)+B\cos(t)$$

Now differentiate to find $x'(t)$

$$x'(t)= A\cos(t) - B\sin(t) $$

Sub this into your original equation and solve for A and B

$$A\cos(t) - B\sin(t) - 3A\sin(t) - 3B\cos(t) = \frac{\cos(t)}{2}$$

Simplifying to

$$\cos(t)[A - 3B - \frac12] - \sin(t)[B + 3A] = 0$$

Since there is no $\sin(t)$ or $\cos(t)$ on RHS, these components must equal 0.

Hence

$A - 3B - \frac12 = 0 $ and $B + 3 = 0$

Solve to get $A = \frac{1}{20}$ and $B = \frac{-3}{20}$