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JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=y(x)\) be the solution of the differential equation \(d y=e^{a x+y} d x ; \alpha \in N\). If \(y\left(\log _{e} 2\right)=\log _{e} 2\) and \(y(0)=\log _{e}\left(\frac{1}{2}\right)\), then the value of \(\alpha\) is equal to \(.....\)
- A \(1\)
- B \(2\)
- C \(3\)
- D \(5\)
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
\(\int \mathrm{e}^{-y} \mathrm{~d} \mathrm{y}=\int \mathrm{e}^{\alpha x} \mathrm{~d} x\) \(\Rightarrow-\mathrm{e}^{-y}=\frac{\mathrm{e}^{\alpha \mathrm{x}}}{\alpha}+\mathrm{c}....(i)\) Put \((x, y)=(\ell n 2, \ell n 2)\) \(\frac{-1}{2}=\frac{2^{\alpha}}{\alpha}+C....(ii)\) Put…
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