Differential Equation

**Singular** · January 27th 2007, 12:01 AM

Solve this Differential Equation

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

here is my attempt to solve it :

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

Is this right ?

**CaptainBlack** · January 27th 2007, 12:23 AM

Originally Posted by Singular

Solve this Differential Equation

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

here is my attempt to solve it :

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

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

**Singular** · January 27th 2007, 12:30 AM

Ya...

It looks more simple

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