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Math Help - Solving a DE

  1. #1
    MHF Contributor arbolis's Avatar
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    Solving a DE

    I must solve (1+x^2)y'+4xy=\frac{1}{(1+x^2)^2}.
    I think it's a first order linear non homogeneous DE which can be solved by the integrating factor method.
    So far I've rewritten the equation as \frac{dy}{dx}+ \underbrace {\left ( \frac{4x}{1+x^2}\right ) }_{p(x)}  y=\underbrace{\frac{1}{(1+x^2)^3}}_{q(x)}.
    I found the integrating factor to be 2 \ln (1+x^2).
    So multiplying the DE by it and integrating with respect to x, I found out that 2 \ln (1+x^2)=2 \int \frac{\ln (1+x^2)}{(1+x^2)^3}dx.
    I thought of using integration by parts to solve the integral, but I'm stuck at finding a primitive of \ln (1+x^2), as stupid as it may sound.
    Am I in the right direction? If so, could you provide a tip to find a primitive of \ln (1+x^2)?
    Thanks in advance.
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  2. #2
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    Quote Originally Posted by arbolis View Post
    I must solve (1+x^2)y'+4xy=\frac{1}{(1+x^2)^2}.
    I think it's a first order linear non homogeneous DE which can be solved by the integrating factor method.
    So far I've rewritten the equation as \frac{dy}{dx}+ \underbrace {\left ( \frac{4x}{1+x^2}\right ) }_{p(x)}  y=\underbrace{\frac{1}{(1+x^2)^3}}_{q(x)}.
    I found the integrating factor to be 2 \ln (1+x^2).
    So multiplying the DE by it and integrating with respect to x, I found out that 2 \ln (1+x^2)=2 \int \frac{\ln (1+x^2)}{(1+x^2)^3}dx.
    I thought of using integration by parts to solve the integral, but I'm stuck at finding a primitive of \ln (1+x^2), as stupid as it may sound.
    Am I in the right direction? If so, could you provide a tip to find a primitive of \ln (1+x^2)?
    Thanks in advance.
    You are definitely on the right track, but you have the wrong integrating factor.

    The integrating factor is actually

    e^{\int{\frac{4x}{1+x^2}\,dx}} = e^{2\ln{(1 + x^2)}} = e^{\ln{(1 + x^2)^2}} = (1 + x^2)^2.


    So multiplying through by the integrating factor:

    (1 + x^2)^2\frac{dy}{dx} + 4x(1 + x^2)y = \frac{1}{1 + x^2}

    \frac{d}{dx}[(1 + x^2)^2y] = \frac{1}{1 + x^2}

    (1 + x^2)^2y = \int{\frac{1}{1 + x^2}\,dx}

    (1 + x^2)^2y = \arctan{x} + C

    y = \frac{\arctan{x} + C}{(1 + x^2)^2}.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Oops, I forgot to take the exponential of "my" integrating factor.
    Thanks for the rest.
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