Differential Equations Question 5
Question 5 - 29 January - Shift 1
Let $y=f(x)$ be the solution of the differential equation $y(x+1) d x-x^{2} d y=0, y(1)=e$. Then $\lim _{x \to 0^{+}} f(x)$ is equal to
(1) 0
(2) $\frac{1}{e}$
(3) $e^{2}$
(4) $\frac{1}{e^{2}}$
Show Answer
Answer: (1)
Solution:
Formula: Variables separable
$ \frac{x+1}{x^{2}} d x=\frac{d y}{y} $
$\ln x-\frac{1}{x}=\ln y+c$
$(1, e)$
$c=-2$
$\ln x-\frac{1}{x}=\ln y-2$
$y=e^{{\ln x}-\frac{1}{x}+2}$
$\lim e^{{\ln x-1}-\frac{1}{x}+2 }; \text{as} ; ; {x \to} 0^{+}$
$=e^{-\infty}$
$=0$
