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Homework Help: Solving Initial value problem

  1. Apr 21, 2007 #1


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    1. The problem statement, all variables and given/known data
    Solve the following initial-value problem
    [tex] \(y'+5y=1,\:y(1)=0\) [/tex]

    3. The attempt at a solution

    y' + 5y = 0 (this is to solve for a general solution)

    y = -(5/2)y^2

    y' + 5y = 1 (this is for the particular solution)

    y = 1/5

    Therefore when you add the 2 equations up the answer should be

    1/5 - (5/2)y^-2

    But I went wrong somewhere.
    My math tutor said that the answer involves e somehow but frankly im kinda lost here. Thanks.
  2. jcsd
  3. Apr 21, 2007 #2


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    Homework Helper

    y' = dy/dt
    So if you go with y = -(5/2)y^2, y' = 0. This would mean that the original differential equation isn't satisfied.

    What methods do you have at your disposal? Are you familiar with separation of variables?
    Write the equation as
    y' + 5y = 1
    dy/dt + 5y = 1
    dy/(1-5y) = dt
    Does that ring any bells?

    EDIT: You could use x instead of t if you're more used to that.
    Last edited: Apr 21, 2007
  4. Apr 21, 2007 #3


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    Olay, how in the world did you get this?
    y is a function of x not y!
    If you meant y= - (5/2)x2, that obviously doesn't work:
    If y= -(5/2)x2, then y'= -5x, not -5 y.
    To solve y'= -5y write it as dy/y= -5dx and integrate both sides.

  5. Apr 21, 2007 #4


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    You got this in before looking at my respons. You want y as a function of x, not y!

    1. Integrate dy/y= -5 dx

    2. Sollve for y.
  6. Jul 23, 2007 #5
    I think the easiest way to do it is like Päällikkö said, separation of variables.
    Just do the following:

    [tex]y' + 5y = 1[/tex]

    [tex]y' = 1 -5y[/tex]

    [tex]\frac{dy}{1 - 5y} = 1 dt[/tex]

    [tex]\int\frac{dy}{1 - 5y} = \int1 dt[/tex]

    [tex]-\frac{1}{5}\ln(1 - 5y) = t + C[/tex]

    [tex]\ln(1 - 5y) = -5t + C[/tex]

    [tex]1 -5y = Ce^{-5t}[/tex]

    [tex]5y = Ce^{-5t} - 1[/tex]

    [tex]y = \frac{1 - Ce^{-5t}}{5}[/tex]

    Now use the initial condition to find C:

    [tex]0 = \frac{1 - Ce^{-5}}{5}[/tex]

    [tex]0 = 1 - Ce^{-5}[/tex]

    [tex]1 = Ce^{-5}[/tex]

    [tex]\frac{1}{e^{-5}} = C[/tex]

    [tex]C = e^{5}[/tex]

    Now put it all together:

    [tex]y = \frac{1 - e^{5}e^{-5t}}{5}[/tex]

    Simplify and you get:

    [tex]y = \frac{1 - e^{5 (1 - t)}}{5}[/tex]
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