JEE Main 2021 — Mathematics Calculus

Given:$\frac{y}{x}\frac{dy}{dx}=[\frac{{y}^{2}}{{x}^{2}}+\frac{\phi (\frac{{y}^{2}}{{x}^{2}})}{{\phi }^{'}(\frac{{y}^{2}}{{x}^{2}})}]....(1)$xy​dxdy​=[x2y2​+ϕ′(x2y2​)ϕ(x2y2​)​]....(1)

Let $\frac{y}{x}=t$xy​=t

$$\Rightarrow y=xt$$⇒y=xt

$$\Rightarrow \frac{dy}{dx}=t+x\cdot \frac{dt}{dx}$$⇒dxdy​=t+x⋅dxdt​

$$\therefore t(t+x\frac{dt}{dx})=({t}^{2}+\frac{\phi ({t}^{2})}{{\phi }^{'}({t}^{2})})$$∴t(t+xdxdt​)=(t2+ϕ′(t2)ϕ(t2)​)

$$\Rightarrow xt\frac{dt}{dx}=\frac{\phi ({t}^{2})}{\phi '({t}^{2})}$$⇒xtdxdt​=ϕ′(t2)ϕ(t2)​

$$\Rightarrow \frac{t\cdot {\phi }^{'}({t}^{2})}{\phi ({t}^{2})}dt=\frac{1}{x}dx$$⇒ϕ(t2)t⋅ϕ′(t2)​dt=x1​dx

Integrating both sides

$$\int \frac{t\cdot {\phi }^{'}({t}^{2})}{\phi ({t}^{2})}dt=\int \frac{1}{x}dx$$∫ϕ(t2)t⋅ϕ′(t2)​dt=∫x1​dx

Let $\phi ({t}^{2})=p$ϕ(t2)=p

$$\Rightarrow {\phi }^{'}({t}^{2}).2t=dp$$⇒ϕ′(t2).2t=dp

$$\Rightarrow \frac{1}{2}\int \frac{1}{p}dp=\int \frac{1}{x}dx$$⇒21​∫p1​dp=∫x1​dx

$$\Rightarrow \frac{1}{2}\mathrm{ln}p=\mathrm{ln}x+C$$⇒21​lnp=lnx+C

$$\Rightarrow \frac{1}{2}\mathrm{ln}\phi ({t}^{2})=\mathrm{ln}x+C$$⇒21​lnϕ(t2)=lnx+C

$$\Rightarrow \frac{1}{2}\mathrm{ln}(\phi (\frac{{y}^{2}}{{x}^{2}}))=\mathrm{ln}x+C...(2)$$⇒21​ln(ϕ(x2y2​))=lnx+C...(2)

If $x=1,y=-1$x=1,y=−1 then $C=\frac{1}{2}\mathrm{ln}(\phi (1))$C=21​ln(ϕ(1))

Substituting value of $C$C in $(2)$(2)

$$\frac{1}{2}\mathrm{ln}(\phi (\frac{{y}^{2}}{{x}^{2}}))=\mathrm{ln}x+\frac{1}{2}\mathrm{ln}(\phi (1))$$21​ln(ϕ(x2y2​))=lnx+21​ln(ϕ(1))

$\Rightarrow$⇒$\mathrm{ln}(\phi (\frac{{y}^{2}}{{x}^{2}}))=\mathrm{ln}{x}^{2}+\mathrm{ln}(\phi (1))$ln(ϕ(x2y2​))=lnx2+ln(ϕ(1))

If $x=2$x=2 then

$$\mathrm{ln}(\phi (\frac{{y}^{2}}{4}))=\mathrm{ln}4+\mathrm{ln}(\phi (1))$$ln(ϕ(4y2​))=ln4+ln(ϕ(1))

SO, $\phi (\frac{{y}^{2}}{4})=4\phi (1)$ϕ(4y2​)=4ϕ(1)