Let a curve y=f(x) pass through the points (0,5) and (log e 2, k ) . If the curve satisfies the differential equation 2(3+y) e 2 x d x- (7+e 2 x ) d y=0 , then k is equal to

Mathematics · Differential Equations

JEE Main 2025 — 23 January, Morning Shift — Question 5

Let a curve $y=f(x)$ pass through the points $(0,5)$ and $\left(\log _{\mathrm{e}} 2, k\right)$ . 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

$\frac{d y}{d x}=\frac{2(3+y) \cdot e^{2 x}}{7+e^{2 x}}$

$\frac{d y}{d x}-\frac{2 y \cdot e^{2 x}}{7+e^{2 x}}=\frac{6 \cdot e^{2 x}}{7+e^{2 x}}$

I.F. $=\mathrm{e}^{-\int \frac{2 \mathrm{e}^{2 \mathrm{x}}}{7+\mathrm{e}^{2 \mathrm{x}}} \mathrm{dx}}=\frac{1}{7+\mathrm{e}^{2 \mathrm{x}}}$

$\therefore y \cdot \frac{1}{7+e^{2 x}}=\int \frac{6 e^{2 x}}{\left(7+3^{2 x}\right)^{2}} d x$

\frac{y}{7+e^{2 x}}=\frac{-3}{7+e^{2 x}}+C

$(0,5) \Rightarrow \frac{5}{8}=\frac{-3}{8}+\mathrm{C} \Rightarrow \mathrm{C}=1$

$\therefore \mathrm{y}=-3+7+\mathrm{e}^{2 \mathrm{x}}$

$y=e^{2 x}+4$

$\therefore \mathrm{k}=8$

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.