Give a differential equation to which the 2-parameter family is a family of solutions The given 2-parameter family is$$ y= e^{5x}(A+B\sin(2x))$$
Here's what I have so far:
$$y'=e^{5x}(5B\sin(2x)+2B\cos(2x)+5A)$$
$$ y''=e^{5x}(21B\sin(2x)+20B\cos(2x)+25A)$$
$$10\cdot 2Be^{5x}\cos(2x)=10y'-10e^{5x}(5B\sin(2x)+5A)=10y'-50y$$
I'm trying to do the same with $y''$ but I get stuck because they have no common factor:
$$20Be^{5x}\cos(2x)=y''-e^{5x}(21B\sin(2x)+25A)$$
How do I proceed?
 A: I would do this in chunks.
First define $w(x)=\exp(-5x)y(x)=A+B\sin (2x)$. Differentiating $w$ twice and dividing the first derivative by the second derivative gives
$\dfrac{d^2w/dx^2}{dw/dx}=-2\tan(2x)$
Clearing the fraction and reinserting $w(x)=\exp(-5x)y(x)$ then yields
$\dfrac{d^2(\exp(-5x)w)}{dx^2}=-2\tan(2x)\dfrac{d(\exp(-5x)w)}{dx} (*)$
We now need the productcrule for both first and second derivatives. The rule for the first derivative should be well known:
$\dfrac{d(uv)}{dx}=\dfrac{du}{dx}v+u\dfrac{dv}{dx}$
To get the second derivative rull, simply differentiate again, treating each term on the right as a product of its own:
$\dfrac{d^2(uv)}{dx^2}=\dfrac{d}{dx}\left(\dfrac{du}{dx}v\right)+\dfrac{d}{dx}\left(u\dfrac{dv}{dx}\right)$
$=\left(\dfrac{d^2u}{dx}v+\dfrac{du}{dx}\dfrac{dv}{dx}\right)+\left(\dfrac{du}{dx}\dfrac{dv}{dx}+u\dfrac{d^2v}{dx^2}\right)$
$\dfrac{d^2(uv)}{dx^2}=\dfrac{d^2u}{dx}v+2\dfrac{du}{dx}\dfrac{dv}{dx}+u\dfrac{d^2v}{dx^2}$
We now substitute both product rules into the $(*)$ equation above and should find we can cancel a factor of $\exp(-5x)$ to get a final answer.
I ultimately get

$\dfrac{d^2y}{dx^2}-(10-2\tan(2x))\dfrac{dy}{dx}+(25-10\tan(2x))y=0.$

A: You have basis solutions $y_1(x)=e^{5x}$ and $y_2(x)=e^{5x}\sin(2x)$. The general homogeneous second order linear DE for $y$ given two basis solutions is
$$
0=\det\pmatrix{y(x)&y_1(x)&y_2(x)\\y'(x)&y_1'(x)&y_2'(x)\\y''(x)&y_1''(x)&y_2''(x)}
$$
