Question

The solution of the equation $x\underset{0}{\overset{x}{\int }}y\left(t\right)dt=(x+1)\underset{0}{\overset{x}{\int }}ty\left(t\right)dt,x>0$ is
(where $c$ is integration constant)

• A. $|c{y}^3|=\frac{e^{\frac{1}{x}}}{|x|}$
• B. $|c{y}^3|=\frac{e^{-\frac{1}{x}}}{|x|}$
• C. $|cy|=\frac{e^{\frac{1}{x}}}{|x^3|}$
• D. $|cy|=\frac{e^{-\frac{1}{x}}}{|x^3|}$

Answer 1

The correct option is D $|cy|=\frac{e^{-\frac{1}{x}}}{|x^3|}$

$x\underset{0}{\overset{x}{\int }}y\left(t\right)dt=(x+1)\underset{0}{\overset{x}{\int }}ty\left(t\right)dt,x>0$
Differentiate w.r.t. $x,$ we get
$xy\left(x\right)+\underset{0}{\overset{x}{\int }}y\left(t\right)dt=(x+1)xy\left(x\right)+\underset{0}{\overset{x}{\int }}ty\left(t\right)dt$
Again differentiating w.r.t. $x,$ we get
$x\frac{dy}{dx}+y\left(x\right)+y\left(x\right)=(2x+1)y\left(x\right)+(x^2+x)\frac{dy}{dx}+xy\left(x\right)$
$\Rightarrow \frac{1-3x}{x^2}dx=\frac{dy}{y}$
Integrating both sides we get,
$-\frac{1}{x}-3\ln |x|=\ln |y|+\ln |c|\Rightarrow |cy|=\frac{e^{-\frac{1}{x}}}{|x^3|}$