JEE Main 2023 — Mathematics Calculus

Given equation is:

$$({x}^{2}–3{y}^{2})\mathrm{dx}+3\mathrm{xydy}=0$$(x2–3y2)dx+3xydy=0

We can re-write equation as

$$\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{({x}^{2}-3{y}^{2})}{3\mathrm{xy}}$$dxdy​=−3xy(x2−3y2)​

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{y}{x}-\frac{1}{3}\frac{x}{y}$⇒dxdy​=xy​−31​yx​ ....(1)

Put $y=\mathrm{vx}$y=vx

$$\frac{\mathrm{dy}}{\mathrm{dx}}=v+x\frac{\mathrm{dv}}{\mathrm{dx}}$$dxdy​=v+xdxdv​

Equation (1) can be written as

$$v+x\frac{\mathrm{dv}}{\mathrm{dx}}=v-\frac{1}{3}\frac{1}{v} \Rightarrow \mathrm{vdv}=-\frac{1}{3x}$$v+xdxdv​=v−31​v1​⇒vdv=−3x1​

Integrating both sides, we get

$$\frac{{v}^{2}}{2}=-\frac{1}{3}\mathrm{ln}(x)+c$$2v2​=−31​ln(x)+c

$\Rightarrow \frac{{y}^{2}}{2{x}^{2}}=\frac{-1}{3}\mathrm{ln}(x)+c$⇒2x2y2​=3−1​ln(x)+c .....(2)

$\because y(1)=1$∵y(1)=1 (given)

$\therefore c=\frac{1}{2}$∴c=21​ (from equation 2)

Equation (2) can be written as

$$\frac{{y}^{2}}{2{x}^{2}}=\frac{-1}{3}\mathrm{ln}(x)+\frac{1}{2}$$2x2y2​=3−1​ln(x)+21​

$$\Rightarrow {y}^{2}=\frac{-2}{3}{x}^{2}\mathrm{ln}(x)+{x}^{2}$$⇒y2=3−2​x2ln(x)+x2

Now ${y}^{2}(e)=\frac{-2}{3}{e}^{2}\mathrm{ln}(e)+{e}^{2}=\frac{-2}{3}{e}^{2}+{e}^{2}=\frac{{e}^{2}}{3}$y2(e)=3−2​e2ln(e)+e2=3−2​e2+e2=3e2​

$$\Rightarrow 6{y}^{2}(e)=2{e}^{2}$$⇒6y2(e)=2e2