1. ## Differential equation

Use a suitable trial function to find the particular solution of

$\frac{dy}{dx}+6y=3sin(4x)$

$y=Acos(4x)+Bsin(4x)$

$\frac{dy}{dx}=-4Asin(4x)+4Bcos(4x)$

From the equation, $-4Asin(4x)+4Bcos(4x)+6[Acos(4x)+Bsin(4x)]=3sin(4x)$

Now I don't understand the last bit where you have to pick the two parts which correspond to 0 and 3. Could someone explain it for me?

2. Hi

$-4Asin(4x)+4Bcos(4x)+6[Acos(4x)+Bsin(4x)]=3sin(4x)$

Separate cos and sin
$(-4A+6B-3)sin(4x)+(4B+6A)cos(4x)=0$

Therefore
$-4A+6B-3=0$ and $4B+6A=0$