Question

solve the given differential equation
(1 + x)y = xy^{2}\frac{dy}{dx}

Explanation:

Step1: Rearrange the equation

First, rewrite the given differential equation $\frac{dy}{dx}-\frac{(1 + x)}{x}y=xy^{2}$ in the Bernoulli - form $y^{-2}\frac{dy}{dx}-\frac{(1 + x)}{x}y^{-1}=x$. Let $z = y^{-1}$, then $\frac{dz}{dx}=-y^{-2}\frac{dy}{dx}$. The equation becomes $-\frac{dz}{dx}-\frac{(1 + x)}{x}z=x$, or $\frac{dz}{dx}+\frac{(1 + x)}{x}z=-x$.

Step2: Find the integrating factor

The integrating factor $I.F.=e^{\int\frac{1 + x}{x}dx}=e^{\int(\frac{1}{x}+ 1)dx}=e^{\ln x+x}=xe^{x}$.

Step3: Multiply the equation by the integrating factor

Multiply $\frac{dz}{dx}+\frac{(1 + x)}{x}z=-x$ by $xe^{x}$, we get $xe^{x}\frac{dz}{dx}+(e^{x}+xe^{x})z=-x^{2}e^{x}$. The left - hand side is the derivative of the product $z(xe^{x})$ by the product rule. So, $\frac{d}{dx}(zxe^{x})=-x^{2}e^{x}$.

Step4: Integrate both sides

Integrate $\frac{d}{dx}(zxe^{x})=-x^{2}e^{x}$ with respect to $x$. Using integration by parts $\int u dv=uv-\int v du$. Let $u = - x^{2}$, $dv=e^{x}dx$, then $du=-2xdx$, $v = e^{x}$. $\int - x^{2}e^{x}dx=-x^{2}e^{x}+2\int xe^{x}dx$. For $\int xe^{x}dx$, let $u = x$, $dv=e^{x}dx$, then $du = dx$, $v = e^{x}$, and $\int xe^{x}dx=xe^{x}-e^{x}+C$. So, $\int - x^{2}e^{x}dx=-x^{2}e^{x}+2(xe^{x}-e^{x})+C=-e^{x}(x^{2}-2x + 2)+C$. Then $zxe^{x}=-e^{x}(x^{2}-2x + 2)+C$.

Step5: Solve for $z$ and then $y$

Since $z = y^{-1}$, we have $y^{-1}=\frac{- (x^{2}-2x + 2)}{x}+\frac{C}{xe^{x}}$. So, $y=\frac{xe^{x}}{C-(x^{2}-2x + 2)e^{x}}$.

Answer:

$y=\frac{xe^{x}}{C-(x^{2}-2x + 2)e^{x}}$