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Math Help - variables separable ODE

  1. #1
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    variables separable ODE

    When solving a variables separable ODE such as:

    dy/dx=f(x) => intergral dy = integral f(x) dx

    what limits does one integrate between??????
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  2. #2
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    Quote Originally Posted by johnbarkwith View Post
    When solving a variables separable ODE such as:

    dy/dx=f(x) => intergral dy = integral f(x) dx

    what limits does one integrate between??????
    If the boundary condition f(x_1) = y_1 is given, then

    y = \int_{x_1}^{x} f(w) \, dw + y_1.


    If the boundary condition f(x_1) = y_1 is given and the the value of y when x = x_2 is required, then

    y = \int_{x_1}^{x_2} f(w) \, dw + y_1.


    Otherwise there are no limits:

    y = \int f(x) \, dx

    and the right hand side is an indefinite integral. Of course, this approach can also be taken when boundary conditions are given - see the next reply.
    Last edited by mr fantastic; April 7th 2008 at 05:24 AM. Reason: Added the "Of course ..." bit.
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  3. #3
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    Usually you don't use limits on the integral. You evaluate the indefinite integral and add in a constant of integration. Then if you have an initial value for your differential equation you use it to find the value of the constant.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Maybe

    this sample example will help \frac{dy}{dx}=9y^4...seperating we get \frac{dy}{y^4}=9dx then by integrating we get \frac{-1}{3y^3}=9x+C multiplying both sides by -3 we get \frac{1}{y^3}=-27x+C_1 and using the rules of ratios we can switch the means to give us y^3=\frac{1}{-27x+c_1} finally by taking the cuberoot of each side we arrive at our answer y=\frac{1}{[-27x+C_1]^{<br />
\frac{1}{3}}}
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