JEE Mains · Maths · STD 12 - 9. differential equations
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
- A \(4\)
- B \(32\)
- C \(8\)
- D \(16\)
Answer & Solution
Correct Answer
(C) \(8\)
Step-by-step Solution
Detailed explanation
\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}} \\ & \text { I.F. }=e^{-\int \frac{2 e^{2 x}}{7+e^{2 x}} d x} \\ & \Rightarrow e^{-\ln \left(7+e^{2 x}\right)} \\ &…
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