Question

Solve the following differential equation :

$(1+y+x^{2}y)dx+(x+x^3)dy=0$

Answer 1

$(1+y+x^{2}y)dx+(x+x^3)dy=0$

$(x+x^3)dy=-(1+y+x^{2}y)dx$

$\frac{dy}{dx}=\frac{-[1+y(1+x^2\left)\right]}{x+x^3}$

$\frac{dy}{dx}=\frac{-[1+y(1+x^2\left)\right]}{x(1+x^2)}$

$\frac{dy}{dx}+\frac{y}{x}=-\frac{1}{x(1+x^2)}$

This is a linear differential equation of the form

$\frac{dy}{dx}+Py=Q$

where, $P=\frac{1}{x}$ and $Q=-\frac{1}{x(1+x^2)}$

Now, $IF=e^{\int Pdx}=e^{\log x}=x$

Hence, its solution is given by

$yx=\int \frac{-1}{x(1+x^2)}.xdx+C$

$yx=\int \frac{-1}{1+x^2}dx+C$

$yx={\cot }^{-1}x+C$

$y=\frac{{\cot }^{-1}x+C}{x}$ which is the required solution.