Question ID: #226
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
- (1) 16
- (2) 8
- (3) 32
- (4) 4
Solution:
$$2(3+y)e^{2x}dx = (7+e^{2x})dy$$
Separating variables:
$$\frac{2e^{2x}}{7+e^{2x}} dx = \frac{dy}{y+3}$$
Integrating both sides:
$$\ln(7+e^{2x}) = \ln(y+3) + \ln C$$
$$7+e^{2x} = C(y+3)$$
Passes through $(0, 5)$:
$$7+e^0 = C(5+3) \Rightarrow 8 = 8C \Rightarrow C = 1$$
Curve Equation: $7+e^{2x} = y+3 \Rightarrow y = e^{2x} + 4$
Passes through $(\ln 2, k)$:
$$k = e^{2\ln 2} + 4 = e^{\ln 4} + 4 = 4 + 4 = 8$$
Ans.(2)
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