Question

Solve : $\frac{dy}{dx}=\frac{2x+3y}{3x+2y}$

Answer 1

Here, $(3x+2y)dy=(2x+3y)dx$

$\Rightarrow 2xdx-2ydy=3(xdy-ydx)\Rightarrow \frac{d(x^2-y^2)}{x^2}=3\frac{xdy-ydx}{x^2}$ (Divided by $x^2$)

$\Rightarrow \frac{d\left(x^2(1-\frac{y^2}{x^2})\right)}{x^2}=3d\left(\frac{y}{x}\right)$

Let $\frac{y}{x}=z$ and $v=x^2$.

ie, $\frac{(1-z^2)dv+vd(1-z^2)}{v}=3dz$ (Product Rule)

Let $1-z^2=u$.

On dividing throughout with $(1-z^2)$,

$\Rightarrow \frac{dv}{v}+\frac{du}{u}=\frac{z}{1-z^2}$

On integrating,

$\int \frac{dv}{v}+\int \frac{du}{u}=\int \frac{dz}{1-z^2}+\ln C$

$\Rightarrow \ln v+\ln u=\ln \sqrt{\frac{1+z}{1-z}}+\ln C$

$\Rightarrow \ln \frac{vu\sqrt{1-z}}{\sqrt{1+z}}=\ln C$

On taking antilog,

$\frac{vu\sqrt{1-z}}{\sqrt{1+z}}=C$

$\Rightarrow \frac{x^2(1-\frac{y^2}{x^2})\sqrt{1-\frac{y}{x}}}{\sqrt{1+\frac{y}{x}}}=\frac{(x^2-y^2)\sqrt{x-y}}{\sqrt{x+y}}=C$

$\Rightarrow \frac{(x^2-y^2)\sqrt{x^2-y^2}}{x+y}=(x-y)\sqrt{x^2-y^2}=C$