Partial Fraction expansion

Can't you just solve this using inspired guesswork?

$\displaystyle \frac{d^2y}{dx^2} + y= x^2 + e^{2x}$ .

First evaluating the homogeneous solution, the characteristic equation is

$\displaystyle m^2 + 1 = 0$

$\displaystyle m^2 = -1$

$\displaystyle m = 0 \pm i$ .

Since these solutions are complex conjugates of the form $\displaystyle \alpha + \beta i$ , the solution is of the form $\displaystyle Ae^{\alpha x}\cos{(\beta x)} + Be^{\alpha x}\sin{(\beta x)}$ . In this case $\displaystyle \alpha = 0$ and $\displaystyle \beta = 1$ .

So the homogeneous solution is $y = \displaystyle A\cos{x} + B\sin{x}$ .

Now to look for a particular solution, since the RHS of your DE is $\displaystyle x^2 + e^{2x}$ , a particular solution might be $\displaystyle y = ax^2 + bx + c + d\,e^{2x}$ .

That means $\displaystyle \frac{dy}{dx} = 2ax + b + 2d\,e^{2x}$

and $\displaystyle \frac{d^2y}{dx^2} = 2a + 4d\,e^{2x}$ .

So substituting into the DE $\displaystyle \frac{d^2y}{dx^2} + y = x^2 + e^{2x}$ gives

$\displaystyle 2a + 4d\,e^{2x} + ax^2 + bx + c + d\,e^{2x} = x^2 + e^{2x}$

$\displaystyle ax^2 + bx + c + 2a + 5d\,e^{2x} = x^2 + 0x + 0 + e^{2x}$ .

Equating like coefficients gives $\displaystyle a = 1, b = 0, c + 2a = 0, 5d = 1$ .

Therefore $\displaystyle a = 1, b = 0, c = -2, d = \frac{1}{5}$ .

So your particular solution is $\displaystyle y = x^2 - 2 + \frac{1}{5}e^{2x}$ .

Adding your homogeneous solution to your particular solution gives the general solution, so the general solution is

$\displaystyle y = A\cos{x} + B\sin{x} + x^2 - 2 + \frac{1}{5}e^{2x}$ .