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Math Help - Solve (x^2 + 4y^2+4xy)dy=(2x+4y+1)dx

  1. #1
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    Solve (x^2 + 4y^2+4xy)dy=(2x+4y+1)dx

    (x^2 + 4y^2+4xy)dy=(2x+4y+1)dx

    How do I solve the xy part? Where will it go?
    I hav tried substitution.. But its of no use..
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Solve this differential equation

    What I noticed first is that the DE can be written as:
    (x+2y)^2dy=[2(x+2y)+1]dx

    So the two sides have something in common, but I'm not entirely sure if you can do something with this.
    Last edited by Siron; August 14th 2011 at 09:01 AM.
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  3. #3
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    Re: Solve this differential equation

    Excellent! And, yes, you can make use of that. Let u= x+2y so that du= dx+ 2dy or dy= (1/2)du- (1/2)dx. That makes the differential equation u^2((1/2)du- (1/2)dx)= (2u+1)dx so (1/2)u^2du= (2u+ 1/2)dx or u^2du= (4u+1)dx. That's a separable equation.
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  4. #4
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    Re: Solve (x^2 + 4y^2+4xy)dy=(2x+4y+1)dx

    Sorry I get separable

    u^2du= (u^2+4u+2)dx \; .
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  5. #5
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    Re: Solve this differential equation

    Quote Originally Posted by HallsofIvy View Post
    Excellent! And, yes, you can make use of that. Let u= x+2y so that du= dx+ 2dy or dy= (1/2)du- (1/2)dx. That makes the differential equation u^2((1/2)du- (1/2)dx)= (2u+1)dx so (1/2)u^2du= (2u+ 1/2)dx or u^2du= (4u+1)dx. That's a separable equation.
    From u^2((1/2)du- (1/2)dx)=(2u+1)dx, I get:

    u^2du- u^2dx=(4u+2)dx

    u^2du=(u^2+4u+2)dx

    Which is certainly separable.
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  6. #6
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    Re: Solve (x^2 + 4y^2+4xy)dy=(2x+4y+1)dx

    Thanx a lot..! i just wanted to get rid of the 'xy' term.. !
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  7. #7
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    Re: Solve (x^2 + 4y^2+4xy)dy=(2x+4y+1)dx

    \textbf{x}(u)=u - 2\, \ln\!\left(u + \sqrt{2} + 2\right) - 2\, \ln\!\left(u - \sqrt{2} + 2\right) - \frac{3\, \sqrt{2}\, \ln\!\left(u + \sqrt{2} + 2\right)}{2} + \frac{3\, \sqrt{2}\, \ln\!\left(u - \sqrt{2} + 2\right)}{2}+C


    \textbf{y}(u) = \frac{1}{2}(u-x)=\ln\!\left(u + \sqrt{2} + 2\right) + \ln\!\left(u - \sqrt{2} + 2\right) + \frac{3\, \sqrt{2}\, \ln\!\left(u + \sqrt{2} + 2\right)}{4} - \frac{3\, \sqrt{2}\, \ln\!\left(u - \sqrt{2} + 2\right)}{4}-\frac{C}{2}
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