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Y'' - y' = e^x [2nd order nonhomogenous diff Eq]

  1. Sep 3, 2008 #1
    I have an equation I need to solve by using undetermined coefficients:

    y'' - y' = ex

    The auxiliary equation is:

    r2- r = 0 , so 2 real roots (R1=0, R2 = 1)

    So, yc(x) = C1 + C2ex

    Now for the particular solution:

    I can try Aex but this is already present in the complementary solution. Do I use:

    yp(x) = xAex

    Is this the right move at this point in the problem?
     
  2. jcsd
  3. Sep 3, 2008 #2

    rock.freak667

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    Homework Helper

    Yes, since r=1 is a root of the auxiliary equation, your PI should be Axex
     
  4. Sep 4, 2008 #3

    Mute

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    You could also note that you have a first order ODE in disguise: set v = y' and you have

    v' - v = e^x

    If you multiply by e^(-x) you get

    e^(-x)v' - e^(-x)v = (e^(-x)v)' = 1

    Which means e^(-x)v = x + C => y' = xe^(x) + Ce^(x); integrate once more to get y.
     
  5. Sep 5, 2008 #4
    Hi there!

    there's also another way of finding the particular solution, but it works, iff the inhomogeneous part is of the form e^(ax) where a is any complex number (i.e. it can be also real or pure imaginary)

    so, here's the general formula:
    [tex]y''+py'+qy=be^{\alpha x}[/tex]

    using the differential operator [tex]D=\frac{d}{dx}[/tex], we obtain:

    [tex](D^2+pD+q)y=be^{\alpha x}[/tex]

    considering D^2+pD+q as the characteristic polynomial and rewriting it as p(D):

    [tex]p(D)y=be^{\alpha x}[/tex]

    so far so good: Now every particular solution has the following form:

    [tex]y_p=\frac{be^{\alpha x}}{p(\alpha)}[/tex]

    !!! IF it turns out that [tex]p(\alpha)=0[/tex], we take:

    [tex]y_p=\frac{bxe^{\alpha x}}{p'(\alpha)}[/tex]

    !!! NOW, if also [tex]p'(\alpha)=0[/tex], we take:

    [tex]y_p=\frac{bx^2e^{\alpha x}}{p''(\alpha)}[/tex]
     
  6. Sep 5, 2008 #5
    In this example:

    [tex]y''-y'=e^x[/tex]
    using the differential operator [tex]D=\frac{d}{dx}[/tex], we obtain:
    [tex](D^2-D)y=e^x[/tex]

    Now, consider D^2-D as the characteristic polynomial and write p(D) instead:

    [tex]p(D)y=e^x[/tex]

    we try the particular solution:

    [tex]y_p=\frac{be^{\alpha x}}{p(\alpha)}=\frac{e^x}{p(1)}[/tex]

    but 1 is a root of the char. polynomial(i.e. p(1)=0), so we take:

    [tex]y_p=\frac{bxe^{\alpha x}}{p'(\alpha)}=\frac{xe^x}{p'(1)}[/tex]

    [tex]p'(1)=2(1)-1=1[/tex] and we obtain for our y_p:

    [tex]y_p=xe^x[/tex]

    *** Note that trig functions sinx and cosx could be represented as imaginary or real part of the complex exponential - then this formula would also provide us with the correct y_p :)

    Best regards, Marin
     
  7. Sep 5, 2008 #6
    Great answers all - I didn't realize there were several ways to attack this type of problem.

    Jeff
     
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