JEE Main 2022 — Differential Equations Question with Solution

Given,

$$x\frac{dy}{dx}-y=\sqrt{y^2+16x}$$

$$\Rightarrow x\frac{dy}{dx}=y+\sqrt{y^2+16x}$$

$$\Rightarrow \frac{dy}{dx}=\frac{y}{x}+\sqrt{{\left(\frac{y}{x}\right)}^2+16}$$

This is a homogenous different equation.

Let $$\frac{y}{x}=v$$

$$\Rightarrow y=vx$$

$$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$$

$$\therefore$$ $$v+x\frac{dv}{dx}=v+\sqrt{v^2+16}$$

$$\Rightarrow$$ $$x\frac{dv}{dx}=\sqrt{v^2+16}$$

$$\Rightarrow \frac{dv}{\sqrt{v^2+16}}=\frac{dx}{x}$$

Integrating both sides, we get

$$\int \frac{dv}{\sqrt{v^2+16}}=\int \frac{dx}{x}$$

$$\Rightarrow \ln \left|v+\sqrt{v^2+16}\right|=\ln x+\ln c$$

$$\Rightarrow v+\sqrt{v^2+16}=cx$$

Now putting, $$v=\frac{y}{x}$$, we get

$$\frac{y}{x}+\sqrt{\frac{y^2}{x^2}+16}=cx$$

$$\Rightarrow \frac{y}{x}+\sqrt{\frac{y^2+16x^2}{x^2}}=cx$$

$$\Rightarrow y+\sqrt{y^2+16x^2}=c{x}^2$$ ...... (1)

Given, $$y(1)=3$$

$$\therefore$$ When x = 1 then y = 3.

Putting in equation (1) we get,

$$3+\sqrt{9+16}=c.1$$

$$\Rightarrow c=8$$

$$\therefore$$ Solution of equation,

$$y+\sqrt{y^2+16x^2}=8x^2$$

Now, y(2) means when x = 2 then y = ?

$$\therefore$$ $$y+\sqrt{y^2+16\times 4}=8\times 4$$

$$\Rightarrow y=15$$