If $y=y(x)$ is the solution of the differential equation $\left(\frac{5+\mathrm{e}^x}{2+y}\right)…

If $y=y(x)$ is the solution of the differential equation $\left(\frac{5+\mathrm{e}^x}{2+y}\right) \frac{\mathrm{d} y}{\mathrm{~d} x}+\mathrm{e}^x=0$ satisfying $y(0)=1$, then a value of $y(\log 13)$ is
  1. $\quad-1$
  2. 0
  3. 1
  4. 2

Solution

$\begin{aligned} & \left(\frac{5+\mathrm{e}^x}{2+y}\right) \frac{\mathrm{d} y}{\mathrm{~d} x}+\mathrm{e}^x=0 \\ & \Rightarrow \frac{\mathrm{~d} y}{2+y}=\frac{-\mathrm{e}^x}{5+\mathrm{e}^x} \mathrm{~d} x \end{aligned}$
Integrating on both sides, we get $\begin{aligned} & \log |2+y|=-\log \left|5+\mathrm{e}^x\right|+\log |\mathrm{c}| \\ & \Rightarrow \log |2+y|=\log \left|\frac{\mathrm{c}}{5+\mathrm{e}^x}\right| \end{aligned}$
Since $y(0)=1$ i.e., $y=1$ when $x=0$ $\begin{aligned} & \therefore \quad \log 3=\log \left|\frac{\mathrm{c}}{6}\right| \\ & \quad \Rightarrow 3=\frac{\mathrm{c}}{6} \\ & \Rightarrow \quad \mathrm{c}=18 \\ & \therefore \quad \log |2+y|=\log \left|\frac{18}{5+\mathrm{e}^x}\right| \quad \ldots[\text { From (i) }] \\ & \Rightarrow 2+y=\frac{18}{5+\mathrm{e}^x} \\ & \Rightarrow y=\frac{18}{5+\mathrm{e}^x}-2\end{aligned}$ $\begin{aligned} \Rightarrow y(\log 13) & =\frac{18}{5+\mathrm{e}^{\log 13}}-2 \\ & =\frac{18}{5+13}-2 \\ & =-1\end{aligned}$

Asked in: MHT CET 2024 (02 May Shift 2)

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