Question

Let $y=f\left(x\right)$ be a real-valued differentiable function on the set of all real numbers $R$ such that $f\left(1\right)=1$. If $f\left(x\right)$ satisfies $x{f}^{\prime }\left(x\right)=x^2+f\left(x\right)-2$, then the area enclosed by $y=f\left(x\right)$ with $x$-axis between ordinates $x=0$ and $x=3$ is

Answer 1

$x{f}^{\prime }\left(x\right)=x^2+f\left(x\right)-2$
$y=f\left(x\right),$ so $\frac{dy}{dx}=f^{\prime }\left(x\right)$
Now, $\frac{dy}{dx}+\left(\frac{-1}{x}\right)y=x-\frac{2}{x}$
which is a linear differential equation.
$I.F.=\exp \left(\int \frac{-1}{x}dx\right)=e^{-\ln x}=\frac{1}{x}$

The general solution is
$y\left(\frac{1}{x}\right)=\int (x-\frac{2}{x})\frac{1}{x}dx$
$\therefore \frac{y}{x}=x+\frac{2}{x}+C$
As $y\left(1\right)=1\Rightarrow C=-2$
$\therefore \frac{y}{x}=x+\frac{2}{x}-2$
$\Rightarrow f\left(x\right)=x^2-2x+2=(x-1{)}^2+1$


$\therefore$ Required area
$={\int }_0^3((x-1{)}^2+1)dx$
$={\left[\frac{1}{3}\right(x-1{)}^3]}_0^3+[x{]}_0^3$
$=(\frac{8}{3}+\frac{1}{3})+3$
$=3+3=6$