Question

Solve the differential equation: $x(x-1)\frac{dy}{dx}-y=x^2{(x-1)}^2$

• A. $\frac{y}{x-1}=\frac{x^2}{3}+c.$
• B. $\frac{yx}{x-1}=\frac{x^3}{3}+c.$
• C. $\frac{yx}{x-1}=\frac{x^3}{3}-\frac{c}{x}.$
• D. $\frac{yx}{x-1}=\frac{x^3}{3}+\frac{c}{x}.$

Answer 1

The correct option is B $\frac{yx}{x-1}=\frac{x^3}{3}+c.$

$x(x-1)\frac{dy}{dx}-y=x^2{(x-1)}^2\Rightarrow \frac{dy}{dx}-\frac{y}{x(x-1)}=x(x-1)$ ...(1)
Here $P=\frac{-1}{x(x-1)}=(\frac{1}{x}-\frac{1}{x-1})\Rightarrow \int Pdx=\int (\frac{1}{x}-\frac{1}{x-1})dx$
$=\log x-\log (x-1)=\log \left(\frac{x}{x-1}\right)$
$\therefore I.F.=e^{\log \frac{x}{x-1}}=\frac{x}{x-1}$
Multiplying (1) by $I.F.$ we get
$\frac{x}{x-1}\frac{dy}{dx}-\frac{y}{{(x-1)}^2}=x^2$
Integrating both sides we get
$\frac{xy}{x-1}=\int x^{2}dx=\frac{x^3}{3}+c$