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Math Help - Differential Equation 1

  1. #1
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    Differential Equation 1

    Find the solution:

     xy  dx - (x^{2}+y^{2}) dy =0

    So far I have:

    y=vx
    dy=vdx+xdv

    Plug the  y and  dy into the equation to get:

     x(vx) dx -(x^{2}+(vx)^{2}) (vdx+xdv)=0

     vx^{2} dx -[vx^{2} dx+ x^{3} dv+ v^{3}x^{2}dx+v^{2}x^{3}dv]=0

    No matter what I do, I'm just going in circles. I can't seem to get the variables separated.
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  2. #2
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    Quote Originally Posted by ahhh View Post
    Find the solution:

     xy dx - (x^{2}+y^{2}) dy =0

    So far I have:

    y=vx
    dy=vdx+xdv

    Plug the  y and  dy into the equation to get:

     x(vx) dx -(x^{2}+(vx)^{2}) (vdx+xdv)=0

     vx^{2} dx -[vx^{2} dx+ x^{3} dv+ v^{3}x^{2}dx+v^{2}x^{3}dv]=0

    No matter what I do, I'm just going in circles. I can't seem to get the variables separated.
    \frac{\partial }{\partial y}(xy)=x and

    \frac{\partial }{\partial x}(-(x^2+y^2))=-2x

    \frac{-2x-x}{xy}=-\frac{3}{y}

    So the integrating factor is e^{\int -\frac{3}{y}dy}=y^{-3}

    xy^{-2}dx-(x^2y^{-3}+y^{-1})dy=0

    This is now an exact equation so we get

    \frac{\partial f}{\partial x}=xy^{-2} \implies f(x,y)=\frac{1}{2}x^2y^{-2}+g(y)

    Now taking the partial with respect to y we get

    \frac{\partial f}{\partial y}=-x^2y^{-3}+g'(y)

    but this must equal

    -(x^2y^{-3}+y^{-1})=-x^2y^{-3}+g'(y)

    -y^{-1}=g'(y) \implies g(y)=-\ln(y)

    so \frac{1}{2}x^2y^{-2}-\ln(y)=c is an implicit solution
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  3. #3
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    I'm sorry, but I don't understand the process. In my class, the only way we have solved differentials so far is to use  y=vx and dy=vdx+xdv.
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  4. #4
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    Quote Originally Posted by ahhh View Post
    Find the solution:

     xy dx - (x^{2}+y^{2}) dy =0

    So far I have:

    y=vx
    dy=vdx+xdv

    Plug the  y and  dy into the equation to get:

     x(vx) dx -(x^{2}+(vx)^{2}) (vdx+xdv)=0

     vx^{2} dx -[vx^{2} dx+ x^{3} dv+ v^{3}x^{2}dx+v^{2}x^{3}dv]=0

    No matter what I do, I'm just going in circles. I can't seem to get the variables separated.

    ummm sorry lets try this again...

     vx^{2} dx -[vx^{2} dx+ x^{3} dv+ v^{3}x^{2}dx+v^{2}x^{3}dv]=0

    simplifying

    -x^3dv-v^{3}x^2dx-v^{2}x^3dv=0

    -x^3(1+v^2)dv=v^{3}x^2dx \iff \frac{1+v^2}{v^{3}}dv=\frac{x^2}{-x^3}dx

    \left( \frac{1}{v^3}+\frac{1}{v} \right) dv =-\frac{1}{x}dx<br />

    I hope this helps...
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post
    ummm sorry lets try this again...

     vx^{2} dx -[vx^{2} dx+ x^{3} dv+ v^{3}x^{2}dx+v^{2}x^{3}dv]=0

    simplifying

    -x^3dv-v^{3}x^2dx-v^{2}x^3dv=0

    -x^3(1+v^2)dv=v^{3}x^2dx \iff \frac{1+v^2}{v^{3}}dv=\frac{x^2}{-x^3}dx

    \left( \frac{1}{v^3}+\frac{1}{v} \right) dv =-\frac{1}{x}dx<br />

    I hope this helps...
    Thank you so much! I couldn't seem to get the variables separated.
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  6. #6
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    So it should be  \frac {[\frac {y}{x}]^{-2}}{-2} + \ln \frac {y}{x} = \ln \frac {1}{x}+c

    I'm not worried about simplifying it because I am liable to mess up doing algebra.
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  7. #7
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    Quote Originally Posted by ahhh View Post
    So it should be  \frac {[\frac {y}{x}]^{-2}}{-2} + \ln \frac {y}{x} = \ln \frac {1}{x}+c

    I'm not worried about simplifying it because I am liable to mess up doing algebra.
    Just for fun and to note if we do simplify a little we can show that this is the same as the solution we got solving the exact equation

    -\frac{1}{2}x^2y^{-2}+\ln(y)-\ln(x)=\ln(1)-\ln(x)+c

    \frac{1}{2}x^2y^{-2}-\ln(y)=c

    Yay!!
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  8. #8
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    Oh wait...dumb question, but how did you simplify
    \frac {[\frac {y}{x}]^{-2}}{-2} <br />

    into this

    \frac{1}{2}x^2y^{-2}
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  9. #9
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    Quote Originally Posted by ahhh View Post
    Oh wait...dumb question, but how did you simplify
    \frac {[\frac {y}{x}]^{-2}}{-2} <br />

    into this

    \frac{1}{2}x^2y^{-2}
    \frac {[\frac {y}{x}]^{-2}}{-2}=\frac{\left( \frac{x}{y} \right)^2}{\frac{-2}{1}}=\left( \frac{x}{y} \right)^2 \left( \frac{1}{-2} \right) =\frac{x^2}{-2y^2}=-\frac{1}{2}x^2y^{-2}

    Then I multiplied the whole equation by -1 to change it from a negative to a +.
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