Answer to Question #212899 in Differential Equations for Saad

Question

solve 2xydy+(3x^2+4xy)dx=0

Answer 1

Solution

New function: z=y/x, y=z*x => $\frac{dy}{dx}=x\frac{dz}{dx}+z$

From DE: $x\frac{dz}{dx}+z=-\frac{3+4z}{2z}$ => $x\frac{dz}{dx}=-\frac{3+4z+{2z}^2}{2z}$

It is a separable differential equation

$\frac{2zdz}{3+4z+{2z}^2}=-\frac{dx}{x}$

$\int\frac{2zdz}{3+4z+{2z}^2}=-\int\frac{dx}{x}$

Left side integral is

$\int\frac{2zdz}{3+4z+2z^2}=\frac{1}{2}\int\frac{\left(4z+4\right)dz}{3+4z+2z^2}-2\int\frac{dz}{3+4z+2z^2}=\frac{1}{2}\int\frac{d\left(3+4z+2z^2\right)}{3+4z+2z^2}-2\int\frac{dz}{3+4z+2z^2}$

Therefore

$\frac{1}{2}ln\left|3+4z+2z^2\right|-2\int\frac{dz}{3+4z+2z^2}=-ln\left|x\right|+C$

Integral in this expression is

$2\int\frac{dz}{3+4z+2z^2}=2\int{\frac{dz}{1+\left(2+4z+2z^2\right)}=}2\int{\frac{dz}{1+2\left(z+1\right)^2}=}$

$\frac{2}{\sqrt2}\int{\frac{d\left[\sqrt2\left(z+1\right)\right]}{1+\left[\sqrt2\left(z+1\right)\right]^2}=\frac{2}{\sqrt2}arctan\left(\sqrt2\left(z+1\right)\right)=\frac{2}{\sqrt2}arctan\left(\frac{2z+2}{\sqrt2}\right)}$

So

$\frac{1}{2}ln\left|3+4z+2z^2\right|-\frac{2}{\sqrt2}arctan\left(\frac{2z+2}{\sqrt2}\right)=-ln\left|x\right|+C$

Returning to function y(x):  

$\frac{1}{2}ln\left|3+4\frac{y}{x}+2\left(\frac{y}{x}\right)^2\right|-\frac{2}{\sqrt2}arctan\left(\frac{2y+2x}{x\sqrt2}\right)=-ln\left|x\right|+C$

$\frac{1}{2}ln\left|3x^2+4xy+2y^2\right|-\frac{2}{\sqrt2}arctan\left(\frac{2y+2x}{x\sqrt2}\right)=C$

Answer

$\frac{1}{2}ln\left|3x^2+4xy+2y^2\right|-\frac{2}{\sqrt2}arctan\left(\frac{2y+2x}{x\sqrt2}\right)=C$