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

We need to solve $$2(3+y)e^{2x}dx - (7+e^{2x})dy = 0$$ with $$y(0) = 5$$, and find $$k = y(\log_e 2)$$.

$$\frac{2e^{2x}}{7+e^{2x}}dx = \frac{dy}{3+y}$$

Left side: Let $$u = 7 + e^{2x}$$, then $$du = 2e^{2x}dx$$:

$$\int \frac{du}{u} = \ln|7+e^{2x}| + C_1$$

Right side: $$\int \frac{dy}{3+y} = \ln|3+y| + C_2$$

$$\ln(7 + e^{2x}) = \ln(3 + y) + C$$

$$7 + e^{2x} = A(3 + y)$$

$$7 + 1 = A(3 + 5) \implies 8 = 8A \implies A = 1$$

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

$$y = 4 + e^{2\ln 2} = 4 + e^{\ln 4} = 4 + 4 = 8$$

The correct answer is Option 3: 8.