Question

$\frac{dy}{dx}+\frac{3x^2}{1+x^3}y=\frac{{\sin }^{2}x}{1+x^3}$.

Answer 1

Given,

$\frac{dy}{dx}+\frac{3x^2}{1+x^3}y=\frac{{\sin }^{2}x}{1+x^3}$

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

$P=\frac{3x^2}{1+x^3}\to I.F=e^{\int Pdx}=e^{\int \frac{3x^2}{1+x^3}dx}=e^{\log (1+x^3)}=1+x^3$

$Q=\frac{{\sin }^{2}x}{1+x^3}$

$y\times I.F=\int Q\times I.Fdx$

$y\times (1+x^3)=\int \frac{{\sin }^{2}x}{1+x^3}\times (1+x^3)dx$

$y(1+x^3)=\int {\sin }^{2}xdx$

$y(1+x^3)=\frac{1}{2}(x-\frac{1}{2}\sin 2x)+c$