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Math Help - A second grade differential equation

  1. #1
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    A second grade differential equation

    Hey,

    I'm stuck on this one and it's due tomorrow so I'd love some help.

    Solve y''-y'+y=x+e^x
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  2. #2
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    Re: A second grade differential equation

    Ok first take the characteristic equition x^2-x+1=0 and solve that first.
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: A second grade differential equation

    Another approach would be the annihilator method.

    We are given:

    (1) y''(x)-y'(x)+y(x)=x+e^x

    Observe that D^2 annihilates x and (D-1) annihilates e^x and so define:

    A\equiv D^2(D-1)

    Applying A to both sides of the ODE we get:

    A\left[y''-y'+y \right]=A\left[x+e^x \right]

    (2) D^2(D-1)(D^2-D+1)=0

    Thus, the auxiliary equation is:

    r^2(r-1)(r^2-r+1)=0

    The roots are:

    r=0\text{ multiplicity 2},\,1,\,\frac{1\pm i\sqrt{3}}{2}

    Thus, a general solution to (1) is:

    (3) y(x)=c_1+c_2x+c_3e^x+c_4e^{ \frac{x}{2}}\cos\left( \frac{ \sqrt{3}}{2}x \right)+c_5e^{ \frac{x}{2}}\sin\left( \frac{\sqrt{3}}{2}x \right)

    Now, recall that a general solution to (1) is of the form y(x)=y_h(x)+y_p(x). Since every solution to (1) is also a solution to (2), then y(x) must have the form displayed on the right-hand side of (3). But, we recognize that:

    y_h(x)=c_6e^{ \frac{x}{2}}\cos\left( \frac{ \sqrt{3}}{2}x \right)+c_7e^{ \frac{x}{2}}\sin\left( \frac{\sqrt{3}}{2}x \right)

    and so there must exist a particular solution of the form:

    y_p(x)=c_1+c_2x+c_3e^x

    In order to substitute this into (1), we must first compute:

    y_p'(x)=c_2+c_3e^x

    y_p''(x)=c_3e^x

    and so we find:

    (c_3e^x)-(c_2+c_3e^x)+(c_1+c_2x+c_3e^x)=x+e^x

    (c_1-c_2)+(c_2)x+(c_3)e^x=0+1\cdot x+1\cdot e^2

    Equating coefficients, we find:

    c_1-c_2=0

    c_2=1

    c_3=1

    and so:

    y_p(x)=1+x+e^x and we have the solution:

    y(x)=1+x+e^x+c_1e^{ \frac{x}{2}}\cos\left( \frac{ \sqrt{3}}{2}x \right)+c_2e^{ \frac{x}{2}}\sin\left( \frac {\sqrt{3}}{2}x \right)
    Last edited by MarkFL; January 8th 2013 at 01:57 PM.
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