JEE MAIN - Mathematics (2025 - 23rd January Morning Shift - No. 1)
Let a curve $y=f(x)$ pass through the points $(0,5)$ and $\left(\log _e 2, k\right)$. If the curve satisfies the differential equation $2(3+y) e^{2 x} d x-\left(7+e^{2 x}\right) d y=0$, then $k$ is equal to
32
8
4
16
Explanation
$$\begin{aligned} & \frac{d y}{d x}=\frac{2(3+y) \cdot e^{2 x}}{7+e^{2 x}} \\ & \frac{d y}{d x}-\frac{2 y \cdot e^{2 x}}{7+e^{2 x}}=\frac{6 \cdot e^{2 x}}{7+e^{2 x}} \\ & \text { I.F. }=e^{-\int \frac{2 e^{2 x}}{7+e^{2 x}}}=\frac{1}{7+e^{2 x}} \end{aligned}$$
$$\begin{aligned} & \therefore y \cdot \frac{1}{7+\mathrm{e}^{2 \mathrm{x}}}=\int \frac{6 \mathrm{e}^{2 \mathrm{x}}}{\left(7+3^{2 x}\right)^2} \mathrm{dx} \\ & \quad \frac{\mathrm{y}}{7+\mathrm{e}^{2 \mathrm{x}}}=\frac{-3}{7+\mathrm{e}^{2 \mathrm{x}}}+C \\ & (0,5) \Rightarrow \frac{5}{8}=\frac{-3}{8}+C \Rightarrow C=1 \\ & \therefore y=-3+7+\mathrm{e}^{2 \mathrm{x}} \\ & y=\mathrm{e}^{2 \mathrm{x}}+4 \\ & \therefore \mathrm{k}=8 \end{aligned}$$
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