Let a curve y=f(x) pass through the points (0,5) and ( _e2,k).

Question

Let a curve $y=f(x)$ pass through the points $(0,5)$ and $({\log }_{e}2,k)$. If the curve satisfies the differential equation $2(3+y){e}^{2x}dx-(7+{e}^{2x})dy=0$, then $k$ is equal to

Accepted Answer

$\begin{gathered} \frac{dy}{dx}=\frac{2(3+y)\cdot {e}^{2x}}{7+{e}^{2x}} \ \frac{dy}{dx}-\frac{2y\cdot {e}^{2x}}{7+{e}^{2x}}=\frac{6\cdot {e}^{2x}}{7+{e}^{2x}} \ ={e}^{-\int \frac{2{e}^{2x}}{7+{e}^{2x}}}=\frac{1}{7+{e}^{2x}} \end{gathered}$

$\begin{gathered} 	herefore y\cdot \frac{1}{7+{e}^{2x}}=\int \frac{6{e}^{2x}}{{(7+{3}^{2x})}^{2}}dx \ \, \frac{y}{7+{e}^{2x}}=\frac{-3}{7+{e}^{2x}}+C \ (0,5)\Rightarrow \frac{5}{8}=\frac{-3}{8}+C\Rightarrow C=1 \ 	herefore y=-3+7+{e}^{2x} \ y={e}^{2x}+4 \ 	herefore k=8 \end{gathered}$

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