JEE Mathematics 2023 Question with Solution

Given:

• $2(y+2)\log(y+2)\,dx+(x+4-2\log(y+2))\,dy=0$
• $x(e^{4}-2)=1$

Find: $x(e^{9}-2)$

Rewrite the equation as a differential equation in $x$ with respect to $y$:

$$\frac{dx}{dy}=-\frac{x+4-2\log(y+2)}{2(y+2)\log(y+2)}$$

So,

$$\frac{dx}{dy}+\frac{1}{2(y+2)\log(y+2)}\,x=\frac{2\log(y+2)-4}{2(y+2)\log(y+2)}$$

This is a linear differential equation of the form $\frac{dx}{dy}+P(y)x=Q(y)$ with

$$P(y)=\frac{1}{2(y+2)\log(y+2)}$$

The integrating factor is

$$I.F.=e^{\int P(y)\,dy}=e^{\frac12\int \frac{dy}{(y+2)\log(y+2)}}$$

Let $t=\log(y+2)$, so that $dt=\frac{dy}{y+2}$. Then

$$I.F.=e^{\frac12\log t}=\sqrt{\log(y+2)}$$

Multiplying the differential equation by the integrating factor,

$$\frac{d}{dy}\!\left(x\sqrt{\log(y+2)}\right)=\sqrt{\log(y+2)}\cdot \frac{2\log(y+2)-4}{2(y+2)\log(y+2)}$$

which simplifies to

$$\frac{d}{dy}\!\left(x\sqrt{\log(y+2)}\right)=\frac{\sqrt{\log(y+2)}}{y+2}-\frac{2}{(y+2)\sqrt{\log(y+2)}}$$

Integrating both sides and again using $t=\log(y+2)$,

$$x\sqrt{\log(y+2)}=\frac{2}{3}(\log(y+2))^{3/2}-4(\log(y+2))^{1/2}+C$$

Hence,

$$x=\frac{2}{3}\log(y+2)-4+\frac{C}{\sqrt{\log(y+2)}}$$

Now use the condition $x(e^{4}-2)=1$. Here $\log(y+2)=4$, so

$$1=\frac{2}{3}\cdot 4-4+\frac{C}{2}$$

Thus,

$$1=\frac{8}{3}-4+\frac{C}{2}\Rightarrow \frac{C}{2}=\frac{7}{3}\Rightarrow C=\frac{14}{3}$$

For $y=e^{9}-2$, we have $\log(y+2)=9$. Therefore,

$$x=\frac{2}{3}\cdot 9-4+\frac{\tfrac{14}{3}}{3}=6-4+\frac{14}{9}=\frac{32}{9}$$

Therefore, the correct option is D.

Given: the differential equation is already linear in $x$ once written as $\frac{dx}{dy}+P(y)x=Q(y)$.

Find: $x(e^{9}-2)$

The key observation is to treat $y$ as the independent variable and identify

$$P(y)=\frac{1}{2(y+2)\log(y+2)}$$

Using $t=\log(y+2)$, the integrating factor becomes

$$I.F.=e^{\frac12\int \frac{dt}{t}}=\sqrt{\log(y+2)}$$

This immediately gives the standard integrated form

$$x=\frac{2}{3}\log(y+2)-4+\frac{C}{\sqrt{\log(y+2)}}$$

Now apply $x(e^{4}-2)=1$ to get $C=\frac{14}{3}$, and then substitute $\log(y+2)=9$:

$$x=\frac{2}{3}\cdot 9-4+\frac{14}{9}=\frac{32}{9}$$

Therefore, the correct option is D.