Let a curve y=f(x) pass through the points (0 | vedclass

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(B) Given the differential equation: $2(3+y) e^{2x} dx = (7+e^{2x}) dy$. Rearranging the terms,we get: $\frac{dy}{dx} = \frac{2(3+y) e^{2x}}{7+e^{2x}}

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$8$

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(B) Given the differential equation: $2(3+y) e^{2x} dx = (7+e^{2x}) dy$. Rearranging the terms,we get: $\frac{dy}{dx} = \frac{2(3+y) e^{2x}}{7+e^{2x}}$. Separating the variables: $\frac{dy}{3+y} = \frac{2e^{2x}}{7+e^{2x}} dx$. Integrating both sides: $\int \frac{dy}{3+y} = \int \frac{2e^{2x}}{7+e^{2x}} dx$. Let $u = 7+e^{2x}$,then $du = 2e^{2x} dx$. So,$\ln|3+y| = \ln|7+e^{2x}| + C$. This can be written as $3+y = C(7+e^{2x})$. Since the curve passes through $(0,5)$,we substitute $x=0$ and $y=5$: $3+5 = C(7+e^0) \Rightarrow 8 = 8C \Rightarrow C=1$. Thus,the equation of the curve is $3+y = 7+e^{2x}$,which simplifies to $y = e^{2x} + 4$. Now,for the point $(\log_e 2, k)$,we substitute $x = \log_e 2$: $k = e^{2 \log_e 2} + 4 = e^{\log_e 4} + 4 = 4 + 4 = 8$.

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

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