Let y=f(x) be the solution of the differential equation y(x+1)dx-x^2dy=0,y(1)=e. Then undersetx→ 0^+ limf(x) is equal to

Given,

$$y(x+1)dx-{x}^{2}dy=0$$y(x+1)dx−x2dy=0

$$\Rightarrow \frac{x+1}{{x}^{2}}dx=\frac{dy}{y}$$⇒x2x+1​dx=ydy​

$$\Rightarrow (\frac{1}{x}+\frac{1}{{x}^{2}})dx=\frac{dy}{y}$$⇒(x1​+x21​)dx=ydy​

Now integrating both side we get,

$${\mathrm{log}}_{e}x-\frac{1}{x}={\mathrm{log}}_{e}y+c$$loge​x−x1​=loge​y+c

Now on using $y(1)=e$y(1)=e we get, $c=-2$c=−2

So, the equation of curve becomes ${\mathrm{log}}_{e}x-\frac{1}{x}={\mathrm{log}}_{e}y-2$loge​x−x1​=loge​y−2

$$\Rightarrow y={e}^{\mathrm{ln}x}-\frac{1}{x}+2$$⇒y=elnx−x1​+2

Hence, $\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}{e}^{\mathrm{ln}x-1}-\frac{1}{x}+2$x→0+lim​elnx−1−x1​+2$={e}^{-\infty }$=e−∞$=0$=0.