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Math Help - Differential Equation

  1. #1
    Junior Member Singular's Avatar
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    Differential Equation

    Solve this Differential Equation

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

    here is my attempt to solve it :

    \begin{array}{l}<br />
 \int {\frac{{2 + y}}{{4 + y^2 }}dy = } \int {dx}  \\ <br />
 \int {\frac{2}{{4 + y^2 }}} {\rm{ }}dy + \int {\frac{y}{{4 + y^2 }}dy}  = x \\ <br />
 arc{\rm{ }}tg{\rm{ }}\frac{{\rm{y}}}{{\rm{2}}} + \frac{1}{2}{\rm{ }}\ln (4 + y^2)  = x \\ <br />
 \end{array}


    Is this right ?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Singular View Post
    Solve this Differential Equation

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

    here is my attempt to solve it :

    \begin{array}{l}<br />
\int {\frac{{2 + y}}{{4 + y^2 }}dy = } \int {dx} \\ <br />
\int {\frac{2}{{4 + y^2 }}} {\rm{ }}dy + \int {\frac{y}{{4 + y^2 }}dy} = x \\ <br />
arc{\rm{ }}tg{\rm{ }}\frac{{\rm{y}}}{{\rm{2}}} + \frac{1}{2}{\rm{ }}\ln (4 + y^2) = x \\ <br />
\end{array}


    Is this right ?
    It looks OK, but you would be better off using the log version for the first integral:

    2\int \frac{1}{4+y^2}\,dy=\frac{1}{2}\ln \left( \frac{2+y}{2-y}\right)

    RonL
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  3. #3
    Junior Member Singular's Avatar
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    Ya...

    It looks more simple
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