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Math Help - Solve y'(x)=(4x+3y-1)^2

  1. #1
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    Post Solve y'(x)=(4x+3y-1)^2

    Hello,
    I have solved this differential equation, but I want to know whether all the steps are correct.
    y'(x)=(4x+3y-1)^2
    Let u=4x+3y-1 then du/dx =4+3y'(x)
    This implies y'(x)=(1/3)u'(x)-4/3
    Now
    (1/3)u'(x)-4/3=u^2
    Hence,u'(x)=3u^2+4
    du/(3u^2+4)=dx
    Integrating both the sides,we get
    1/2sqrt(3) arctan(sqrt(3)u)/2=x+c
    So my ans is sqrt(3)*(4x+3y-1)= 2tan(2sqrt(3)*x)+c
    If I am wrong, reply where I am wrong.
    Last edited by Vinod; November 19th 2011 at 04:10 AM.
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  2. #2
    Member sbhatnagar's Avatar
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    Re: Solve y'(x)=(4x+3y-1)^2

    You have made some slight errors. I have corrected them.
    Quote Originally Posted by Vinod View Post
    Hello,
    I have solved this differential equation, but I want to know whether all the steps are correct.
    y'(x)=(4x+3y-1)^2
    Let u=4x+3y-1 then du/dx =4+3y'(x)
    This implies y'(x)=(1/3)u'(x)-4/3
    Now
    (1/3)u'(x)-4/3=u^2
    Hence,u'(x)=3u^2+4
    du/(3u^2+4)=dx
    Integrating both the sides,we get
    1/2sqrt(3) arctan((sqrt(3)u)/2)+k=x+c
    So my ans is sqrt(3)*(4x+3y-1)= 2tan((2sqrt(3)*x)+c-k)
    If I am wrong, reply where I am wrong.
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  3. #3
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    Re: Solve y'(x)=(4x+3y-1)^2

    You don't have to write the constant of integration on both sides and then subtract. That will simply be a single constant.
    Vinod, your solution is correct.
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