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Math Help - Initial Value Problem General Question

  1. #1
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    Initial Value Problem General Question

    Given an initial value problem with separable differential equation \frac{dy}{dx}=\frac{g(x)}{h(y)} and initial condition (x_0,y_0) is it true that \int_{y_0}^y h(t)dt=\int_{x_0}^x g(t)dt and is that the same as solving by using indefinite integrals and then using the initial condition to solve for the constant? If yes, how? I'm having trouble understanding. Thank you!
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  2. #2
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    Quote Originally Posted by rualin View Post
    Given an initial value problem with separable differential equation \frac{dy}{dx}=\frac{g(x)}{h(y)} and initial condition (x_0,y_0) is it true that \int_{y_0}^y h(t)dt=\int_{x_0}^x g(t)dt and is that the same as solving by using indefinite integrals and then using the initial condition to solve for the constant? If yes, how? I'm having trouble understanding. Thank you!
    Yes, they are the same.

    Let F(x) be an anti-derivative of f(x) and G(t) be an anti-derivative of g(t).
    Then \int_{y_0}^y f(t)dt= F(y)- F(y_0) and \int_{x_0}^x g(t)dt= G(x)- G(x_0). Your solution becomes F(y)- F(y_0)= G(x)- G(x_0) which can be written F(y)= G(x)+ (F(y_0)- G(x_0)).

    Now, suppose you used indefinite integrals: \int f(y)dy= F(y)+ C_1 and \int g(x)dx= G(x)+ C_2 so you would have F(y)+ C_1= G(x)+ C_2 or just F(y)= G(x)+ C where C= C_2-C_1. Now put y= y_0, x= x_0 into that: F(y_0)= G(x_0)+ C so C= F(y_0)- G(x_0). That gives F(y)= G(x)+ (F(y_0)- G(x_0)), exactly the same answer as above.
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  3. #3
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    Nice explanation! Thanks!
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