GATE 2022 IN Q.45: solution and answer

Official answer: CSource: Official answer key, GATE 2022 IN (IIT Kharagpur)

1. Rewrite the equation as $\frac{dy}{dx}=-y\ln y$. It is first order and variables separable. At the initial point $y=e$, so $y\ln y=e\ne 0$ and we may divide: $\frac{dy}{y\ln y}=-dx$.

2. Substitute $u=\ln y$, so $du=\frac{dy}{y}$. The left side becomes $\frac{du}{u}$, and integrating gives $\ln |u|=-x+C$, i.e. $u=A{e}^{-x}$ with $A$ a constant.

3. So the general solution is $\ln y=A{e}^{-x}$.

4. Apply the initial condition $y(0)=e$: $\ln e=A{e}^0$, so $A=1$. Hence $\ln y=e^{-x}$, i.e. $y=e^{e^{-x}}$.

5. At $x=1$: $\ln y(1)=e^{-1}=\frac{1}{e}$, so $y(1)=e^{(1/e)}$.

6. Check: with $y=e^{e^{-x}}$, $\frac{dy}{dx}=-e^{-x}{e}^{e^{-x}}=-(\ln y)y$, so $\frac{dy}{dx}+y\ln y=0$ holds, and $y(0)=e^1=e$. The answer is option (C).