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Math Help - problem with first-order nonlinear ordinary differential equation

  1. #1
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    problem with first-order nonlinear ordinary differential equation

    i have problem to find the solution for : (3x^3y+2xy+y^3)+(x^2+y^2)dy/dx=0

    i have tried the exact equation method :

    (3x^3y+2xy+y^3)dx+(x^2+y^2)dy=0

    thus M(x,y)= (3x^3y+2xy+y^3)

    and N(x,y)= (x^2+y^2)

    then deltaM/deltay= 3x^3+2x+3y^2
    and deltaN/deltax= 2x

    Since deltaM/deltay does not equal to deltaN/deltax, this imply that the eqution is not exact

    thus, finding/searching for integrating factor :

    1. 1/N(deltaM/deltay-deltaN/deltax)= (3x^3+3y^3)/(3x^3y+2xy+y^3)<br />
    y cannot be eliminated . thus, this is a function of both x and y, not just x

    2. 1/M(deltaN/deltax-deltaM/deltay)= (3x^3+3y^2)/(x^2+y^2)

    x cannot be eliminated . thus, this is a function of both x and y, not just y


    thus i cannot find the integrating factor in order to solve the DE. where i'm gone wrong? can somebody point it out? i guess may be in algebra...

    is there anyway for me to solve the de? please help me....
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  2. #2
    Member kjchauhan's Avatar
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    Your equation will be,

    \frac{dy}{dx}=-\frac{3x^3y+2xy+y^3}{x^2+y^2}

    Now take y=vx

    Then \frac{dy}{dx}=v+x\frac{dv}{dx}

    Now substitute it in the above,

    you will get,

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

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

    Now integrate....


    Result will be,

    e^{3x}(y^3+3x^2y)-1=0
    Last edited by kjchauhan; January 22nd 2010 at 06:14 PM.
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  3. #3
    MHF Contributor
    Jester's Avatar
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    Quote Originally Posted by kjchauhan View Post
    Your equation will be,

    \frac{dy}{dx}=-\frac{3x^3y+2xy+y^3}{x^2+y^2}

    Now take y=vx

    Then \frac{dy}{dx}=v+x\frac{dv}{dx}

    Now substitute it in the above,

    you will get,

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

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

    Now integrate....


    Result will be,

    e^{3x}(y^3+3x^2y)-1=0
    I only got your's to work out if you start with (with the red term)

    \frac{dy}{dx}=-\frac{3x^3y+2xy+{\color{red}{x}} y^3}{x^2+y^2}
    Last edited by Jester; January 23rd 2010 at 03:16 PM. Reason: fixed typo
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