Question

Solve the differential equation:

$\frac{dy}{dx}$=$\frac{x^2-y^2}{2xy}$

Answer 1

$(x^2-y^2)dx=2xydy$

or, $\frac{dy}{dx}=\frac{x^2-y^2}{2xy}$ ............ $\left(1\right)$

Let $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$\therefore$ From $\left(1\right),v+x\frac{dv}{dx}=\frac{1-v^2}{2v}$

or, $x\frac{dv}{dx}=\frac{1-v^2}{2v}-v=\frac{1-3v^2}{2v}$
or, $\frac{v}{1-3v^2}dv=\frac{dx}{x}$

Integrating both sides,

$\int \frac{vdv}{1-3v^2}=\int \frac{dx}{x}+c$ ......... $\left(2\right)$ $c=$ constant of integrating.

Let $1-3v^2=z\Rightarrow -6vdv=d\Rightarrow vdv=-\frac{1}{6}dz$

$\therefore$ $\left(2\right)$

$\int -\frac{1}{6}\frac{dz}{z}=\int \frac{dx}{x}+c$

or, $-\frac{1}{6}\log z=\log x+c$

or, $-\log (1-3v^2)=6\log x+6c$

or, $\log x^6+\log (1-3v^2)=-6c$

or, $\log x^6(1-3v^2)=-6c$

or, $x^6(1-3v^2)=e^{-6c}=k$ $k=$ constant of integration

or, $x^6(1-\frac{3y^2}{x^2})=k$

$\therefore x^4(x^2-3y^2)=4\Rightarrow$ Required general solution.