Question

The solution of $x(x-1)\frac{dy}{dx}-(x-2)y=x^3(2x-1)$ is

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

Answer 1

The correct option is A $y\frac{(x-1)}{x^2}=x^2-x+c$

$x\left(x-1\right)\frac{dy}{dx}-\left(x-2\right)=x^3\left(2x-1\right)$

$\Rightarrow \frac{dy}{dx}-\frac{\left(x-2\right)}{x\left(x-1\right)}y=\frac{x^2\left(2x-1\right)}{x-1}$

$P=\frac{-\left(x-2\right)}{x\left(x-1\right)}Q=\frac{x^2\left(2x-1\right)}{x-1}$

$I.F=e^{\int pdx}=e^{\int \frac{-\left(x-2\right)}{\left(x-1\right)}dx}$

$=e^{\int \left(\frac{1}{x-1}\frac{-2}{x}\right)dx}$

$e^{\log x-1-2\log x}$

$=e^{\log \frac{x-1}{x^2}}$

$=\frac{x-1}{x^2}$

Solution of Given $D.E$ is

$y\times I\cdot F=\int Q.I.Fdx+c$

$\Rightarrow \frac{y\left(x-1\right)}{x^2}=\int \frac{x^2\left(2x-1\right)}{x-1}\times \frac{x-1}{x^2}dx+c$

$\Rightarrow \frac{y\left(x-1\right)}{x^2}=\int \left(2x-1\right)dx=c$

$\Rightarrow \frac{y\left(x-1\right)}{x^2}=x^2-x+c$