Question

The solution of differential equation

$\frac{dy}{dx}=\frac{1}{xy[x^2{\sin }^{2}y+1]}$ is

Answer 1

$\frac{dy}{dx}=\frac{1}{xy[x^2\sin y^2+1]}\frac{dx}{dy}=xy[x^2\sin y^2+1]\frac{1}{x^3}\frac{dx}{dy}=\frac{xy}{x^3}[x^2\sin y^2+1]\frac{1}{x^3}\frac{dx}{dy}=\frac{y}{x^2}[x^2\sin y^2+1]\frac{1}{x^3}\frac{dx}{dy}=y\sin y^2+\frac{y}{x^2}\frac{1}{x^3}\frac{dx}{dy}-\frac{y}{x^2}=y\sin y^{2}Let\frac{-1}{x^2}=z-\frac{(-2)}{x^3}\frac{dx}{dy}=\frac{dz}{dy}\frac{2}{x^3}\frac{dx}{dy}=\frac{dz}{dy}\frac{1}{x^3}\frac{dx}{dy}=\frac{1}{2}\frac{dz}{dy}\therefore \frac{1}{2}\frac{dz}{dy}+yz=y\sin y^2\frac{dz}{dy}+2yz=y\sin y^{2}I.F.=e^{\int 2ydy}=e^{2\int ydy}=e^{2\cdot y^2/2}=e^{y^2}$