Solve the following differential equation: 2xy dx+(x^2+2y^2)dy=0

Question

Solve the following differential equation:
$2\text{xy dx}+(\text{x}^2+2\text{y}^2)\text{dy}=0$

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Answer

Here, $2\text{xy dx}+(\text{x}^2+2\text{y}^2)\text{dy}=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{2\text{xy}}{\text{x}^2-2\text{y}^2}$
It is a homogeneous equation.
Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{2\text{xvx}}{\text{x}^2+2\text{v}^2\text{x}^2}$
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{2\text{v}}{1+2\text{v}^2}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{2\text{v}}{1-2\text{v}^2}-\text{v}$
$=\frac{2\text{v}-\text{v}+2\text{v}^3}{1+2\text{v}^2}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}-2\text{v}^3}{1+2\text{v}^2}$
$\int\frac{1+2\text{v}^2}{\text{v}-2\text{v}^3}\text{dv}=\int\frac{\text{dx}}{\text{x}}\ \dots(\text{i})$
$\frac{1+2\text{v}^2}{\text{v}-2\text{v}^3}=\frac{1+2\text{v}^2}{\text{v}(1-2\text{v}^2)}$
$\frac{1+2\text{v}^2}{\text{v}(1-2\text{v}^2)}=\frac{\text{A}}{\text{v}}+\frac{\text{Bv + C}}{1-2\text{v}^2}$
$\frac{1+2\text{v}^2}{\text{v}(1-2\text{v}^2)}=\frac{\text{A}(1-2\text{v}^2)+(\text{Bv + C)}\text{v}}{\text{v}(1-2\text{v}^2)}$
$1+2\text{v}^2=\text{A}-2\text{Av}^2+\text{Bv}^2+\text{Cv}$
$1+2\text{v}^2=\text{v}^2(-2\text{A + B})+\text{Cv + A}$
Comparing the co-efficients of like powers of v,
A = 1
C = 0
-2A + B = 2
-2 + B = 0
B = 4
$\frac{1+2\text{v}^2}{\text{v}-2\text{v}^3}=\frac{1}{\text{v}}+\frac{4\text{v}}{1-2\text{v}^2}$
$\frac{1+2\text{v}^2}{\text{v}-2\text{v}^3}=\frac{1}{\text{v}}-\frac{(-4\text{v})}{(1-2\text{v}^2)}$

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