Answer to Question #245835 in Differential Equations for Gabriel

Question

(1+x)dy/dx-xy=x+x^2

Answer 1

Answer:-

$y'+P(x)y=Q(x)$

$P(x)=-\frac{x}{x+1}$ , $Q(x)=x$

$y'+P(x)y=0$

$\frac{dy}{y}=-P(x)dx$

$\int\frac{dy}{y}=-\int P(x)dx$

$ln|y|=-\int P(x)dx$

$y=\pm e^{-\int P(x)dx}$

$\int P(x)dx=\int(-\frac{x}{x+1})dx=-x+ln(x+1)+c$

$y=\frac{ce^x}{x+1}$

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

$\frac{d}{dx}c(x)=Q(x)e^{\int P(x)dx}$

$\frac{d}{dx}c(x)=(x^2+x)e^{-x}$

$\int (x^2+x)e^{-x}dx=(-x^2-3x-3)e^{-x}+c$

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