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Math Help - square root equations

  1. #1
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    square root equations

    A.
    xdy-ydx=\sqrt{x^{2}+y^{2}}dx
    B.
    \sqrt{y^{2}+1}dx=xydy

    how to solve these
    Last edited by transgalactic; November 13th 2010 at 05:12 AM.
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    B.
    \sqrt{y^{2}+1}dx=xydy

    how to solve these
    \sqrt{y^{2}+1}~dx=xy~dy

    \frac{\sqrt{y^{2}+1}}{x}~dx=y~dy

    \frac{dx}{x}=\frac{y}{\sqrt{y^{2}+1}}~dy

    \int \frac{dx}{x}=\int \frac{y}{\sqrt{y^{2}+1}}~dy

    Use a substitution u = y^{2}+1 on the RHS.
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  3. #3
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    Quote Originally Posted by transgalactic View Post
    A.
    xdy-ydx=\sqrt{x^{2}+y^{2}}
    B.
    \sqrt{y^{2}+1}dx=xydy

    how to solve these
    I believe that you've stated A incorrectly. Please check.
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  4. #4
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    you are right its
    xdy-ydx=\sqrt{x^{2}+y^{2}}dx

    how to solve it?
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  5. #5
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    This is homogeneous. Try letting y = xu.
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  6. #6
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    how did you know its homogeneous and you need to put
    y/x=u

    ??
    dy=xdu
    x^2du-xudx=\sqrt{x^2+x^2u^2}dx
    what now?
    i cant put on one side dy on the other dx
    but its not looking like something solveable
    ?
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  7. #7
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    No y = xu and dy = x du + u dx. Substitute these and simplify. You should get something separable.
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  8. #8
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    i did this substitution
    i got
    x^2 du=\sqrt{x^2+x^2u^2}dx
    how to solve it
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  9. #9
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    still i got a non saparable equation

    ??
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  10. #10
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    Quote Originally Posted by transgalactic View Post
    i did this substitution
    i got
    x^2 du=\sqrt{x^2+x^2u^2}dx
    how to solve it
    You need to factor

    x^2 du=\sqrt{x^2+x^2u^2}dx

    becomes

    x^2 du=x\sqrt{1+u^2}dx

    Now separate.
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  11. #11
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    Quote Originally Posted by transgalactic View Post
    i did this substitution
    i got
    x^2 du=\sqrt{x^2+x^2u^2}dx
    how to solve it
    x^2 du= \sqrt{x^2(1+ u^2)}dx
    That certainly is separable.
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  12. #12
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    thanks i see that now
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