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Thread: Change of Variables

  1. #1
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    Change of Variables

    I'm having trouble rewriting differential equations in the form F(v). That is, changing a function to make it's variables have the form "y/x" so I can perform the substitution v=y/x. Here's a sample problem:

    Solve the differential equation:

    yy'= sqrt(x^2+y^2)-x


    Is there a general way to go about transforming this into the form f(v)?

    Any help would be much appreciated.

    Thank!
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by divinelogos View Post
    I'm having trouble rewriting differential equations in the form F(v). That is, changing a function to make it's variables have the form "y/x" so I can perform the substitution v=y/x. Here's a sample problem:

    Solve the differential equation:

    yy'= sqrt(x^2+y^2)-x


    Is there a general way to go about transforming this into the form f(v)?

    Any help would be much appreciated.

    Thank!
    \displaystyle yy'=\sqrt{x^2+y^2}-x

    \displaystyle y'=\frac{\sqrt{x^2+y^2}}y}-\frac{x}{y}

    \displaystyle y'=\sqrt{\frac{x^2}{y^2}+1}-\frac{x}{y}
    Last edited by alexmahone; Feb 16th 2011 at 07:22 PM.
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    How did you get from the second step to the third step?
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by divinelogos View Post
    How did you get from the second step to the third step?
    \displaystyle \frac{\sqrt{x^2+y^2}}{y}=\sqrt\frac{x^2+y^2}{y^2}=  \sqrt{\frac{x^2}{y^2}+1}
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  5. #5
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    THAT'S what it was. Some sneaky Algebra right there...Thank you my man!
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  6. #6
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    One more question. That gets it in x/y but I need y/x...
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    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by divinelogos View Post
    One more question. That gets it in x/y but I need y/x...
    You can still substitute \displaystyle v=\frac{y}{x} so that \displaystyle \frac{x}{y}=\frac{1}{v}.
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  8. #8
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    That works, but it leads to an integral that can only be solved in terms of other integrals. Do you know how they got this?



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    Forum Admin topsquark's Avatar
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    \displaystyle y'=\frac{\sqrt{x^2+y^2}}{y}

    \displaystyle y'= \frac{\sqrt{x^2+y^2}}{ \sqrt{y^2} } = \sqrt{ \frac{x^2  + y^2}{y^2}}

    \displaystyle = \sqrt{\frac{x^2}{y^2} + \frac{y^2}{y^2}}

    etc.

    -Dan
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  10. #10
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    That gets me to:

    y'=sqrt((x^2/y^2)) - x/y

    I still don't see how they got to that second step :/
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