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Math Help - Find particular solution

  1. #1
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    Find particular solution

    Find the particular solution of the differential equation

    \frac{y}{x}\frac{dy}{dx}=\frac{y^2+1}{x^2+1} y=0 when x=1

    I solved the differential equation by first separating the variables:

    \int \frac{y}{y^2+1}dy=\int \frac{x}{x^2+1}dx

    \frac{1}{2}\ln |y^2+1|=\frac{1}{2}\ln |x^2+1|+A where A=the arbitrary constant.

    Now i don't know how to proceed.
    Thanks
    Last edited by mr fantastic; September 20th 2009 at 11:35 PM. Reason: Added some spacing to make it easier to read.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by arze View Post
    Find the particular solution of the differential equation
    \frac{y}{x}\frac{dy}{dx}=\frac{y^2+1}{x^2+1} y=0 when x=1
    I solved the differential equation by first separating the variables
    \int \frac{y}{y^2+1}dy=\int \frac{x}{x^2+1}dx
    \frac{1}{2}\ln |y^2+1|=\frac{1}{2}\ln |x^2+1|+A where A=the arbitrary constant.
    Now i don't know how to proceed.
    Thanks
    Absolute value bars aren't necessary since both x^2+1>0 and y^2+1>0\,\forall x,y\in\mathbb{R}.

    From there, multiply both sides by 2 to get \ln\left(y^2+1\right)=\ln\left(x^2+1\right)+A (note that 2A is a new constant [why not rename it A?])

    Now, apply the exponential functions to both sides to get y^2+1=e^{\ln\left(x^2+1\right)+A}=Ce^{\ln\left(x^2  +1\right)}=C\left(x^2+1\right).

    Now solve for y^2 to get y^2=C\left(x^2+1\right)-1

    Its best to leave this in implicit form to avoid the plus/minus issue.
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