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Math Help - Method to solve this DE

  1. #1
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    Method to solve this DE

    2y" = 3y^2

    Basically I don't have any idea about what method I should use to solve this DE...I've tried several ways but they don't seem to work
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  2. #2
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    Quote Originally Posted by Naples View Post
    2y" = 3y^2

    Basically I don't have any idea about what method I should use to solve this DE...I've tried several ways but they don't seem to work
    Use the substition

    u=y' \implies \frac{du}{dy}\frac{dy}{dx}=y'' \iff u\frac{du}{dy}=y''

    This reduces the ODE to the first order separable ODE

    u\frac{du}{dy}=3y^2

    Can you finish from here?
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  3. #3
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    So..?
    2u du = 3y^2 dy
    u^2 = y^3 + A
    u = y^(^3^/^2^) + A^(^1^/^2^)
    y' = y^(^3^/^2^) + A^(^1^/^2^)
    A=0
    y' = y^(^3^/^2^)
    y = (2/5)y^(^5^/^2^) + B
    B = 3/5
    y = (2/5)y^(^5^/^2^) + (3/5)
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  4. #4
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    Quote Originally Posted by Naples View Post
    So..?
    2u du = 3y^2 dy
    u^2 = y^3 + A
    u = y^(^3^/^2^) + A^(^1^/^2^)
    y' = y^(^3^/^2^) + A^(^1^/^2^)
    A=0
    y' = y^(^3^/^2^)
    y = (2/5)y^(^5^/^2^) + B
    B = 3/5
    y = (2/5)y^(^5^/^2^) + (3/5)
    Not quite. Unless you have some inital conditons I don't think this will simplify very nicely.
    We have

     2u\frac{du}{dy}=3y^2 \iff \int udu = \int 3y^2dy \implies u^2=y^3+C \implies u=\pm  \sqrt{y^3+C}

    But now we get

    \frac{dy}{dx}= \pm  \sqrt{y^3+C} \implies \int \frac{dy}{\sqrt{y^3+C}}= \pm \int dx \implies x=\pm \int \frac{dy}{\sqrt{y^3+C}}
    Last edited by TheEmptySet; May 8th 2011 at 07:02 PM. Reason: extra 2!
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  5. #5
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    I was given the initial conditions y(0)=1 and y'(0)=1
    Sorry I forgot to put them in the original post
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  6. #6
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    Quote Originally Posted by Naples View Post
    I was given the initial conditions y(0)=1 and y'(0)=1
    Sorry I forgot to put them in the original post
    Yes this is good!

    Now since

    y'=u

    We can use the equation above and the intial condions to get

    u^2=  y^3+C \implies 1^2 = 1^3 +0 \implies C=0

    Now we just have to solve the ODE

    (y')^2=y^3 \implies \frac{dy}{dx} = \pm y^{\frac{3}{2}} \implies y^{-\frac{3}{2}}dy = \pm dx \implies \frac{-2}{\sqrt{y}}= \pm x+D

    Now use the initial conditions and solve for y. The solution to this equation is not unique.
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  8. #8
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    Quote Originally Posted by Naples View Post
    Yes that is correct. You can verify the solution by plugging back into the original ODE.
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