JEE Main 2022 — Mathematics Calculus

Given,

$$x\frac{dy}{dx}-y=\sqrt{{y}^{2}+16{x}^{2}}$$xdxdy​−y=y2+16x2​

$$\Rightarrow \frac{dy}{dx}=\frac{\sqrt{{y}^{2}+16{x}^{2}}+y}{x}$$⇒dxdy​=xy2+16x2​+y​

Let $y=xt\Rightarrow \frac{dy}{dx}=x\frac{dt}{dx}+t$y=xt⇒dxdy​=xdxdt​+t

$$\Rightarrow x\frac{dt}{dx}+t=t+\sqrt{{t}^{2}+16}$$⇒xdxdt​+t=t+t2+16​

$$\Rightarrow \frac{dt}{\sqrt{{t}^{2}+16}}=\frac{dx}{x}$$⇒t2+16​dt​=xdx​

Now integrating both side we get,

$$\Rightarrow \int \frac{dt}{\sqrt{{t}^{2}+16}}=\int \frac{dx}{x}$$⇒∫t2+16​dt​=∫xdx​

$$\Rightarrow \mathrm{ln}|t+\sqrt{{t}^{2}+16}|=\mathrm{ln}x+\mathrm{ln}c$$⇒ln∣t+t2+16​∣=lnx+lnc

$$\Rightarrow \frac{y}{x}+\frac{\sqrt{{y}^{2}+16{x}^{2}}}{x}=cx$$⇒xy​+xy2+16x2​​=cx

$\because y(1)=3$∵y(1)=3 so $\Rightarrow \frac{3}{1}+\frac{\sqrt{{3}^{2}+16\times {1}^{2}}}{1}=c\times 1\Rightarrow c=8$⇒13​+132+16×12​​=c×1⇒c=8

So solution becomes,

$$\frac{y}{x}+\frac{\sqrt{{y}^{2}+16{x}^{2}}}{x}=8x$$xy​+xy2+16x2​​=8x

Now $y(2)=\frac{y}{2}+\frac{\sqrt{{y}^{2}+16\times {2}^{2}}}{2}=8\times 2$y(2)=2y​+2y2+16×22​​=8×2

$$\Rightarrow y+\sqrt{{y}^{2}+64}=32$$⇒y+y2+64​=32

$$\Rightarrow y=15$$⇒y=15