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

Question

Aicharya · Accepted Answer

$\begin{aligned} & \frac{d y}{d x}=\frac{2(3+y) \cdot e^{2 x}}{7+e^{2 x}} \ & \frac{d y}{d x}-\frac{2 y e^{2 x}}{7+e^{2 x}}=\frac{6 \cdot e^{2 x}}{7+e^{2 x}} \ & 	ext { I.F. }=e^{-\int \frac{2 e^{2 x}}{7+e^{2 x}} d x} \ & \Rightarrow e^{-\ln \left(7+e^{2 x}ight)} \ & =\frac{1}{7+e^{2 x}} \ & y \cdot \frac{1}{7+e^{2 x}}=\int \frac{6 e^{2 x}}{\left(7+e^{2 x}ight)^2} d x \ & \frac{y}{7+e^{2 x}}=\frac{-3}{7+e^{2 x}+C} \ & 	herefore y(0)=5 \ & \Rightarrow \frac{5}{8}=\frac{-3}{8}+C \ & \Rightarrow C=1 \ & 	herefore y=-3+7+e^{2 x} \ & y=e^{2 x}+4 \ & 	herefore k=8\end{aligned}$

Let a curve $y=f(x)$ pass through the points $(0,5)$ and $\left(\log _e 2, k\right)$. If the curve satisfies…

Solution