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Math Help - Partial Fraction expansion

  1. #1
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    Partial Fraction expansion

    Hey I need to learn how to solve a problem using Partial fraction expansion theory
    the problem is:
    y''+y=x^2+e^(2x)
    Thank you!
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  2. #2
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    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}.
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