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General solution of ordinary differential equation

  1. Jan 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the general solution of the differential equation,

    y' + y = be^(-λx)

    where b is a real number and λ is a positive constant.

    2. Relevant equations

    y' + P(x)y = Q(x)

    Integrating factor: e^(∫P(x) dx)

    3. The attempt at a solution

    Let P(x) = 1, Q(x) = be^(-λx)

    The equation is already in the form y' + P(x)y = Q(x).

    So, the integrating fator is I(x) = e^(∫1 dx) = e^(x)

    Multiplying both sides by the integrating factor.

    e^(x)y + e^(x)y = be^(-λx)e^(x)

    (e^(x)y)' = be^(-λx)e^(x)

    Now integrating the left hand side,

    e^(x)y = be^(-λx)e^(x)

    Here is my problem. I don't know where to go from here. How do I integrate the right hand side? That's my main problem.


    Any help will be greatly appreciated.
     
  2. jcsd
  3. Jan 18, 2009 #2

    jgens

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    Gold Member

    Why not simplify it? e^(-ax) = 1/(e^(ax)).
     
  4. Jan 18, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Please do not write an equation that you know isn't true! You have integrated the left side but not yet the right side so you know they are not equal. You might write
    [tex]e^xy= \int be^{-\lambda x}e^x dx[/tex]
    to indicate that the right has yet to be integrated.


    [tex]e^{-\lambda x}e^x= e^{(1-\lambda)x}[/tex]
    Can you integrate that?

     
  5. Jan 19, 2009 #4
    I apologize for the integral symbol, simple mistake.

    I integrated e^((1-λ)x) and came up with this,

    ∫e^((1-λ)x) dx

    Let u = ((1-λ)x) so du = (1-λ) dx

    Now,

    ∫e^(u) du/(1-λ)

    1/(1-λ) ∫e^(u) du

    1/(1-λ)*e^(u) + C

    Substituting back in for u,

    1/(1-λ)*e^(x(1-λ)) + C

    Going back to e^(x)y = ∫be^(-λx)e^(x) dx, I obtained,

    y = be^(-x)*e^(x(1-λ))/(1-λ)

    y = [be^(x(1-λ))] / [e^(x)*(1-λ)]
     
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