# Partial Fraction expansion

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• December 5th 2010, 06:58 PM
deltafee
Partial Fraction expansion
Hey I need to learn how to solve a problem using Partial fraction expansion theory
the problem is:
Quote:

y''+y=x^2+e^(2x)
Thank you!
• December 5th 2010, 08:39 PM
Prove It
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

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