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Thread: ode differiential equation

  1. #1
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    ode differiential equation

    Could someone solve these:
    xyy'+y^2=sinx by letting y^2=v
    y'=2/(x+2y-3) by letting x+2y-3=v
    2y'y''=1+(y')^2 by letting y'=v
    or explain one and if they are similar then i'd be fine?
    Last edited by Monster32432421; Aug 12th 2010 at 07:37 AM.
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Monster32432421 View Post
    Could someone solve these:

    xyy'+y^2=sinx

    or explain one and if they are similar then i'd be fine?
    Hint:

    Multiply both sides of the equation by $\displaystyle 2x$

    $\displaystyle 2x^2yy'+2xy^2=2x\sin{x}$

    $\displaystyle (x^2y^2)'=2x\sin{x}$

    $\displaystyle x^2y^2=2\int{x\sin{x}\,dx}$
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Monster32432421 View Post
    Could someone solve these:

    2y'y''=1+(y')^2 by letting y'=v

    or explain one and if they are similar then i'd be fine?
    Hint:

    $\displaystyle 2y'y''=1+(y')^2$

    $\displaystyle \dfrac{2y'y''}{1+(y')^2}=1$

    $\displaystyle \Bigl(\ln(1+(y')^2)\Bigl)'=1$

    $\displaystyle \ln(1+(y')^2)=x+C$

    $\displaystyle 1+(y')^2=e^{x+C}=C_1e^x$

    $\displaystyle (y')^2=C_1e^x-1$

    $\displaystyle y'=\pm\sqrt{C_1e^x-1}$

    $\displaystyle y=\pm\int\!\sqrt{C_1e^x-1}\,dx$
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  4. #4
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    Quote Originally Posted by DeMath View Post
    Hint:

    Multiply both sides of the equation by $\displaystyle 2x$

    $\displaystyle 2x^2yy'+2xy^2=2x\sin{x}$

    $\displaystyle (x^2y^2)'=2x\sin{x}$

    $\displaystyle x^2y^2=2\int{x\sin{x}\,dx}$
    I actually forgot to write the substitute for it... sorry... it's y^2=v
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  5. #5
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    Quote Originally Posted by DeMath View Post
    Hint:

    $\displaystyle 2y'y''=1+(y')^2$

    $\displaystyle \dfrac{2y'y''}{1+(y')^2}=1$

    $\displaystyle \Bigl(\ln(1+(y')^2)\Bigl)'=1$

    $\displaystyle \ln(1+(y')^2)=x+C$

    $\displaystyle 1+(y')^2=e^{x+C}=C_1e^x$

    $\displaystyle (y')^2=C_1e^x-1$

    $\displaystyle y'=\pm\sqrt{C_1e^x-1}$

    $\displaystyle y=\pm\int\!\sqrt{C_1e^x-1}\,dx$
    how did you get from... $\displaystyle \dfrac{2y'y''}{1+(y')^2}=1$ to $\displaystyle \Bigl(\ln(1+(y')^2)\Bigl)'=1$
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  6. #6
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    If you let $\displaystyle y' = v$ then $\displaystyle y'' = v' $ and you get

    $\displaystyle 2v \dfrac{dv}{dx} = 1 + v^2$ (separable).
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