In each of the following differential equation show that it is homogeneous and solve it. $2 x y d x+\left(x^{2}+2 y^{2}\right) d y=0$

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In each of the following differential equation show that it is homogeneous and solve it. $2 x y d x+\left(x^{2}+2 y^{2}\right) d y=0$

Solution:

$\begin{array}{l} 2 x y d x+\left(x^{2}+2 y^{2}\right) d y=0 \\ \Rightarrow \frac{d y}{d x}=-\frac{2 x y}{x^{2}+2 y^{2}}=-\frac{2}{\left(\frac{y}{x}\right)^{-1}+2^{\frac{y}{x}}} \\ \Rightarrow \frac{d y}{d x}=f\left(\frac{y}{x}\right) \end{array}$
$\Rightarrow$ the given differential eq. is a homogenous eq.
Solution of the given differential eq. is:
$\begin{array}{l} \text { Put } y=v x \\ \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\ \Rightarrow v+x \frac{d v}{d x}=-\frac{2}{\left(\frac{v x}{x}\right)^{-1}+2 \frac{v x}{x}}=-\frac{2 v}{1+2 v^{2}} \\ \Rightarrow x \frac{d v}{d x}=-\frac{2 v}{1+2 v^{2}}-v=-\frac{2 v+v+2 v^{3}}{1+2 v^{2}}=-\frac{3 v+2 v^{3}}{1+2 v^{2}} \\ \Rightarrow \frac{1+2 v^{2}}{3 v+2 v^{3}} d v=-\frac{d x}{x} \end{array}$
On integrating both sides we get:
$\begin{array}{l} \int \frac{1+2 v^{2}}{3 v+2 v^{3}} d v=\int \frac{d x}{x}+c^{\prime} \\ \Rightarrow \frac{\ln \left|3 v+2 v^{3}\right|}{3}=\ln |x|+c^{\prime} \end{array}$
By resubstituting the value of $y=v x$ we have
$\begin{array}{l} \Rightarrow \frac{\ln \left|3^{\frac{y}{x}}+2\left(\frac{y}{x}\right)^{3}\right|}{3}=\ln |x|+c^{\prime} \\ \Rightarrow 3 x^{2} y+2 y^{3}=C \end{array}$