Question

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

Answer 1

Given,

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

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

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

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

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

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

$y(1+x^3)=\int (1+x^2)dx$

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