Question

For $x\in R,x\ne 0$ if $y\left(x\right)$ is a differentiable function such that $x\underset{1}{\overset{x}{\int }}y\left(t\right)dt=(x+1)\underset{1}{\overset{x}{\int }}ty\left(t\right)dt,$ then $y\left(x\right)$ equals:

• A. $C{x}^3{e}^{\frac{1}{x}}$
• B. $\frac{C}{x^2}{e}^{-\frac{1}{x}}$
• C. $\frac{C}{x^3}{e}^{-\frac{1}{x}}$
• D. $\frac{C}{x}{e}^{-\frac{1}{x}}$

Answer 1

The correct option is C $\frac{C}{x^3}{e}^{-\frac{1}{x}}$

$x\underset{1}{\overset{x}{\int }}y\left(t\right)dt=(x+1)\underset{1}{\overset{x}{\int }}ty\left(t\right)dt$
Differentiating w.r.t $x$, we get
$\underset{1}{\overset{x}{\int }}y\left(t\right)dt+xy\left(x\right)=\underset{1}{\overset{x}{\int }}ty\left(t\right)dt+(x+1)xy\left(x\right)$
$\Rightarrow x^{2}y\left(x\right)=\underset{1}{\overset{x}{\int }}\left(y\right(t)-ty(t\left)\right)dt$
Again differentiating w.r.t $x$, we get
$2xy\left(x\right)+x^2\frac{dy}{dx}=y\left(x\right)-xy\left(x\right)$
$\Rightarrow \frac{dy}{dx}=\frac{1-3x}{x^2}y$
$\Rightarrow \int \frac{dy}{y}=\int \frac{1-3x}{x^2}dx$
$\Rightarrow \log y=-\frac{1}{x}-3\log x+\log C$
$\Rightarrow \log y=-\frac{1}{x}-\log x^3+\log C$
$\Rightarrow \log \frac{y{x}^3}{C}=-\frac{1}{x}$
$\Rightarrow y=\frac{C}{x^3}{e}^{-\frac{1}{x}}$