1. ## Differential Equation

Solve this Differential Equation

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

here is my attempt to solve it :

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

Is this right ?

2. Originally Posted by Singular
Solve this Differential Equation

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

here is my attempt to solve it :

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

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

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

RonL

3. Ya...

It looks more simple