Answer
See below.
Work Step by Step
To make life easier for us, let's start with some substitution.
$a=Acos(3x)$
$b=Bsin(3x)$
We are given:
$y=e^{2x}(Acos(3x)+Bsin(3x))$
Using our substitution:
$y=e^{2x}(a+b)$
Now, we take the first derivative using the product rule ($(f(x)g(x))'=f'(x)g(x)+g'(x)f(x)$):
$y'=2e^{2x}(a+b)+e^{2x}(a'+b')$
$y'=e^{2x}(2a+2b+a'+b')$
Then, the second derivative:
$y''=2e^{2x}(2a+2b+a'+b')+e^{2x}(2a'+2b'+a''+b'')$
$y''=e^{2x}(4a+4b+2a'+2b'+2a'+2b'+a''+b'')$
$y''=e^{2x}(4a+4b+4a'+4b'+a''+b'')$
We are asked to show that the original function satisfies the equation $y''-4y'+13y=0$. We will substitute the equations we found into this equation.
$e^{2x}(4a+4b+4a'+4b'+a''+b'')-4(e^{2x}(2a+2b+a'+b'))+13(e^{2x}(a+b))$
$=e^{2x}(4a+4b+4a'+4b'+a''+b''-8a-8b-4a'-4b'+13a+13b)$
$=e^{2x}(9a+9b+a''+b'')$
We'll now solve for $a''$ and $b''$.
$a=Acos(3x)$
$a'=-3Asin(3x)$
$a'' = -9Acos(3x)$
$b=Bsin(3x)$
$b'=3Bscos3x)$
$b''=-9Bsin(3x)$
Finally, we'll substitute $a$, $b$, $a''$, and $b''$ back into $e^{2x}(9a+9b+a''+b'')$
$=e^{2x}(9a+9b+a''+b'')$
$=e^{2x}(9Acos(3x)+9Bsin(3x)-9Acos(3x)-9Bsin(3x))$
$=0$
Done!
