Hey I need to learn how to solve a problem using Partial fraction expansion theory
the problem is:
Thank you!y''+y=x^2+e^(2x)
Can't you just solve this using inspired guesswork?
$\displaystyle \displaystyle \frac{d^2y}{dx^2} + y= x^2 + e^{2x}$.
First evaluating the homogeneous solution, the characteristic equation is
$\displaystyle \displaystyle m^2 + 1 = 0$
$\displaystyle \displaystyle m^2 = -1$
$\displaystyle \displaystyle m = 0 \pm i$.
Since these solutions are complex conjugates of the form $\displaystyle \displaystyle \alpha + \beta i$, the solution is of the form $\displaystyle \displaystyle Ae^{\alpha x}\cos{(\beta x)} + Be^{\alpha x}\sin{(\beta x)}$. In this case $\displaystyle \displaystyle \alpha = 0$ and $\displaystyle \displaystyle \beta = 1$.
So the homogeneous solution is $\displaystyle y = \displaystyle A\cos{x} + B\sin{x}$.
Now to look for a particular solution, since the RHS of your DE is $\displaystyle \displaystyle x^2 + e^{2x}$, a particular solution might be $\displaystyle \displaystyle y = ax^2 + bx + c + d\,e^{2x}$.
That means $\displaystyle \displaystyle \frac{dy}{dx} = 2ax + b + 2d\,e^{2x}$
and $\displaystyle \displaystyle \frac{d^2y}{dx^2} = 2a + 4d\,e^{2x}$.
So substituting into the DE $\displaystyle \displaystyle \frac{d^2y}{dx^2} + y = x^2 + e^{2x}$gives
$\displaystyle \displaystyle 2a + 4d\,e^{2x} + ax^2 + bx + c + d\,e^{2x} = x^2 + e^{2x}$
$\displaystyle \displaystyle ax^2 + bx + c + 2a + 5d\,e^{2x} = x^2 + 0x + 0 + e^{2x}$.
Equating like coefficients gives $\displaystyle \displaystyle a = 1, b = 0, c + 2a = 0, 5d = 1$.
Therefore $\displaystyle \displaystyle a = 1, b = 0, c = -2, d = \frac{1}{5}$.
So your particular solution is $\displaystyle \displaystyle y = x^2 - 2 + \frac{1}{5}e^{2x}$.
Adding your homogeneous solution to your particular solution gives the general solution, so the general solution is
$\displaystyle \displaystyle y = A\cos{x} + B\sin{x} + x^2 - 2 + \frac{1}{5}e^{2x}$.