Let y=f(x) be the solution of the differential equation y(x+1) d x-x2 d y=0, y(1)=e. Then displaystyle lim x arrow 0+ f(x) is equal to

Question

Let y=f(x) be the solution of the differential equation y(x+1) d x-x2 d y=0, y(1)=e. Then displaystyle lim x arrow 0+ f(x) is equal to (A) (1/e2) (B) (1/e) (C) 0 (D) e2

Tardigrade Admin · Accepted Answer

(x+1/x2) d x=(d y/y) ln x-(1/x)= ln y+c (1, e ) c =-2 ln x-(1/x)= ln y-2 y=e ln x-(1/x)+2 displaystyle lim x arrow 0+ e ln x-1-(1/x)+2 =e-∞ =0

Q. 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 $x \to 0^{+} lim f (x)$ is equal to

$\frac{1}{e ^{2}}$

$\frac{1}{e}$

0

$e^{2}$

$\frac{x + 1}{x ^{2}} d x = \frac{d y}{y}$
$ln x - \frac{1}{x} = ln y + c$
$(1, e)$
$c = - 2$
$ln x - \frac{1}{x} = ln y - 2$
$y = e^{l n x} - \frac{1}{x} + 2$
$x \to 0^{+} lim e^{l n x - 1} - \frac{1}{x} + 2$
$= e^{- \infty}$
$= 0$

Q. Let y=f(x) be the solution of the differential equation y(x+1)dx−x2dy=0,y(1)=e. Then x→0+lim​f(x) is equal to

e21​

e1​

0

e2

x2x+1​dx=ydy​ lnx−x1​=lny+c (1,e) c=−2 lnx−x1​=lny−2 y=elnx−x1​+2 x→0+lim​elnx−1−x1​+2 =e−∞ =0

Q. 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 $x \to 0^{+} lim f (x)$ is equal to

$\frac{1}{e ^{2}}$

$\frac{1}{e}$

$e^{2}$

$\frac{x + 1}{x ^{2}} d x = \frac{d y}{y}$
$ln x - \frac{1}{x} = ln y + c$
$(1, e)$
$c = - 2$
$ln x - \frac{1}{x} = ln y - 2$
$y = e^{l n x} - \frac{1}{x} + 2$
$x \to 0^{+} lim e^{l n x - 1} - \frac{1}{x} + 2$
$= e^{- \infty}$
$= 0$