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Math Help - making equation exact

  1. #1
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    making equation exact

    Solve (x^{2}+y^{2}-5)dx=(y+xy)dy
    y(0)=1

    I have tried finding a factor to make this equation exact, and I get (x+1)^{-3}. I then integrate by parts with respect to x and get \,{\frac {-{x}^{2}-{y}^{2}+5}{2 \left( x+1 \right) ^{2}}}+\ln <br />
 \left( x+1 \right) + \left( x+1 \right) ^{-1}+\rho \left( y \right) =<br />
{\frac {{x}^{2}+{y}^{2}-5}{ \left( x+1 \right) ^{3}}}<br /> <br />
    I feel like I'm not getting anywhere because nothing cancels like it should. Can someone please help me? Thanks!
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  2. #2
    Super Member General's Avatar
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    Rewrite your differential as
    (x^{2}+y^{2}-5)dx-y(1+x)dy=0

    Multiplying by the integrating factor

    <br />
\displaystyle \dfrac{x^2+y^2-5}{(x+1)^3} dx + \left( \dfrac{-y}{(x+1)^2} \right) dy = 0<br />

    The last equation is exact. Its solution of the form F(x,y)=C where :

    F_x=\dfrac{x^2+y^2-5}{(x+1)^3} ... (1)

    F_y=\dfrac{-y}{(x+1)^2} ... (2)

    I think you integrate (1) with respect to x. This is too complicated .

    Try to integrate (2) with respect to y and put p(x) as the constant of integration.
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  3. #3
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    \frac{\partial{f}}{\partial{y}}={\frac {-{y}{\it }}{ \left( x+1 \right) ^{2}}}dy Now I integrate w.r.t.y

    f( x,y)  ={\frac {-{y}^{2}}{ \left2( x+1 \right) ^{2}}<br />
}+p \left( x \right) Now I take partial w.r.t.x

    \frac{\partial{f}}{\partial{x}}={\frac {-{y}^{2}}{4(x+1)}}+\mbox p'  \left( x<br />
 \right) ={\frac { \left( {x}^{2}+{y}^{2}-5 \right) {\it }}{ \left( <br />
x+1 \right) ^{3}}} I thought things would cancel if it was exact. Can anyone tell me what I'm doing wrong? Thanks!
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  4. #4
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    {{\rm e}^{\int \!{\frac {M_{{y}}-N_{{x}}}{N}}{dx}}}

    This is how I get (x+1)^{-3}

    How do you get x+1?
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by duaneg37 View Post
    {{\rm e}^{\int \!{\frac {M_{{y}}-N_{{x}}}{N}}{dx}}}

    This is how I get (x+1)^{-3}

    How do you get x+1?
    Okay, I see my mistake. Hang on.

    -Dan
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by General View Post
    Rewrite your differential as
    (x^{2}+y^{2}-5)dx-y(1+x)dy=0

    Multiplying by the integrating factor

    <br />
\displaystyle \dfrac{x^2+y^2-5}{(x+1)^3} dx + \left( \dfrac{-y}{(x+1)^2} \right) dy = 0<br />

    The last equation is exact. Its solution of the form F(x,y)=C where :

    F_x=\dfrac{x^2+y^2-5}{(x+1)^3} ... (1)

    F_y=\dfrac{-y}{(x+1)^2} ... (2)

    I think you integrate (1) with respect to x. This is too complicated .

    Try to integrate (2) with respect to y and put p(x) as the constant of integration.
    Okay, now that I've fixed my little problem with the minus signs...
    (1) isn't difficult to integrate, but it is tedious. You should get
    F(x, y) = ln|x + 1| + \frac{4x -y^2 + 8}{2(x + 1)^2} + \phi (y)

    What I've always done from here is to integrate (2). This is an easy one. Then just match up common terms. It saves from having to work with the derivative of what you called p(x).

    -Dan
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  7. #7
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    f(x,y)=\ln  \left( x+1 \right) +\,{\frac {4\,x-{y}^{2}+8}{ 2\left( x+1<br />
 \right) ^{2}}}-\frac{7}{2}=1

    Does this look right? What technique did you use to integrate F_{{x}}?
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  8. #8
    Forum Admin topsquark's Avatar
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    F(x, y) = \int \frac{x^2 + y^2 - 5}{(x + 1)^3} dx + \phi(y)

    So we need to evaluate the integral
    \int \frac{x^2 + y^2 - 5}{(x + 1)^3} dx

    = \int \frac{x^2}{(x + 1)^3} dx + \int \frac{y^2 - 5}{(x + 1)^3}~dx

    The second integral we can evaluate directly:
    = \int \frac{x^2}{(x + 1)^3} dx - \frac{y^2 - 5}{2(x + 1)^2}

    For the first integral use a = x + 1:

    = \int \frac{(a - 1)^2}{a^3}da - \frac{y^2 - 5}{2(x + 1)^2}

    = \int \left ( \frac{1}{a} - \frac{2}{a^2} + \frac{1}{a^3} \right ) da - \frac{y^2 - 5}{2(x + 1)^2}

    You can fill in the rest from there.

    -Dan
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by duaneg37 View Post
    f(x,y)=\ln  \left( x+1 \right) +\,{\frac {4\,x-{y}^{2}+8}{ 2\left( x+1<br />
 \right) ^{2}}}-\frac{7}{2}=1

    Does this look right? What technique did you use to integrate F_{{x}}?
    What did you get when you integrated that second function?
    F(x, y) = -\int \dfrac{y}{(x + 1)^2}dy + \omega (x) = -\dfrac{1}{(x + 1)^2} \int y~dy + \omega (x)

    = -\dfrac{y^2}{2(x + 1)^2} + \omega (x)

    Now just compare the two forms.

    By the way, please note the distinction between ln(x + 1) and the correct ln|x + 1|

    -Dan
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  10. #10
    Super Member General's Avatar
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    Quote Originally Posted by duaneg37 View Post
    \frac{\partial{f}}{\partial{y}}={\frac {-{y}{\it }}{ \left( x+1 \right) ^{2}}}dy Now I integrate w.r.t.y

    f( x,y)  ={\frac {-{y}^{2}}{ \left2( x+1 \right) ^{2}}<br />
}+p \left( x \right) Now I take partial w.r.t.x

    \frac{\partial{f}}{\partial{x}}={\frac {-{y}^{2}}{4(x+1)}}+\mbox p'  \left( x<br />
 \right) ={\frac { \left( {x}^{2}+{y}^{2}-5 \right) {\it }}{ \left( <br />
x+1 \right) ^{3}}} I thought things would cancel if it was exact. Can anyone tell me what I'm doing wrong? Thanks!

    Your \displaystyle \frac{\partial{f}}{\partial{x}} is wrong.

    You should get : \displaystyle \dfrac{y^2}{(x+1)^3}+p'(x) ... (1)

    And , from the original equation since M= F_x we have \displaystyle \frac{\partial{f}}{\partial{x}}=\dfrac{x^2+y^2-5}{(x+1)^3}

    Or : \displaystyle \frac{\partial{f}}{\partial{x}}=\dfrac{y^2}{(x+1)^  3}+\dfrac{x^2-5}{(x+1)^3} ... (2)

    Comparing (1) & (2) gives : \displaystyle p'(x)=\dfrac{x^2-5}{(x+1)^3}

    I think you can continue ...
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