JEE Main 2023 — Differential Equations Question with Solution

Given,

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

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

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

Integrating both sides, we get

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

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

Given $$y(1)=e$$

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

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

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

$$\Rightarrow 0-1=1+C$$

$$\Rightarrow C=-2$$

$$\therefore$$ Equation (1) becomes,

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

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

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

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

Now,

$$\underset{x\to 0^{+}}{\lim }y$$

$$=\underset{x\to 0^{+}}{\lim }\left(x.e^{2-\frac{1}{x}}\right)$$

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

$$=0$$