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Thread: initial value problem

  1. #1
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    initial value problem

    I'm in a numerical analysis class and I'm trying to approximate an initial value problem. I get the approximation using Forward Euler and other methods, but I'm not sure what the true answer is. I haven't done ODE's in a while.

    Does anyone know how to solve the problem?
    y'(t) = y - t^2 + 1, t in [0,2]
    y(0) = 2

    Thanks for any help in advance.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by PvtBillPilgrim View Post
    I'm in a numerical analysis class and I'm trying to approximate an initial value problem. I get the approximation using Forward Euler and other methods, but I'm not sure what the true answer is. I haven't done ODE's in a while.

    Does anyone know how to solve the problem?
    y'(t) = y - t^2 + 1, t in [0,2]
    y(0) = 2

    Thanks for any help in advance.
    This is the same thing as $\displaystyle \frac{\,dy}{\,dt}-y=1-t^2$

    Now apply the method of the integrating factor, since this equation is linear.

    Thus, our integrating factor is $\displaystyle \varrho=e^{-\int\,dt}=e^{-t}$

    Thus, we now have $\displaystyle \frac{d}{\,dt}\left[e^{-t}y\right]=e^{-t}-t^{2}e^{-t}$

    Now, integrate both sides:

    $\displaystyle e^{-t}y=-e^{-t}-\int e^{-t}t^2\,dt$

    [I leave you to evaluate the integral on the right side of the equation. You will need to apply integration by parts twice.]

    Thus, $\displaystyle y=-e^t\cdot\left[\int e^{-t}t^2\,dt\right]+C$

    Then apply the initial condition $\displaystyle y(0)=2$ to find $\displaystyle C$.

    Can you take it from here?

    --Chris
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