"Let $y=f(x)$ be the solution of the differential equation $y(x+1)dx-x^2dy=0,y(1)=e$. Then $\mathop {\lim }\limits_{x \to {0^ + }} f(x)$ is equal to"

Question

Accepted Answer

"

Given,

$y(x + 1)dx - {x^2}dy = 0$

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

$\Rightarrow {1 \over x}dx + {{dx} \over {{x^2}}} = {{dy} \over y}$

Integrating both sides, we get

$\int {{{dx} \over x} + \int {{{dx} \over {{x^2}}} = \int {{{dy} \over y}} } }$

$\Rightarrow \ln |x| - {1 \over x} = \ln |y| + C$ ..... (1)

Given $y(1) = e$

$\therefore$ $x = 1$ and $y = e$

Putting value of x and y in equation (1), we get

$\ln |1| - {1 \over 1} = \ln |e| + C$

$\Rightarrow 0 - 1 = 1 + C$

$\Rightarrow C = - 2$

$\therefore$ Equation (1) becomes,

$\ln |x| = - {1 \over x} = \ln |y| - 2$

$\Rightarrow \ln |y| = \ln |x| - {1 \over x} + 2$

$\Rightarrow y = {e^{\ln |x|}}\,.\,{e^{2 - {1 \over x}}}$

$\Rightarrow y = x\,.\,{e^{2 - {1 \over x}}}$

Now,

$\mathop {\lim }\limits_{x \to {0^ + }} y$

$$ = \mathop {\lim }\limits_{x \to {0^ + }} \left( {x\,.\,{e^{2 - {1 \over x}}}} \right)$$

$= 0\,.\,{e^{ - \alpha }}$

$= 0$

"

Let $y=f(x)$ be the solution of the differential equation $y(x+1)dx-x^2dy=0,y(1)=e$. Then $\mathop {\lim }\limits_{x \to {0^ + }} f(x)$ is equal to