JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=y(x), x>1\), be the solution of the differential equation \((x-1) \frac{d y}{d x}+2 x y=\frac{1}{x-1}\), with \(y(2)=\frac{1+e^{4}}{2 e^{4}}\). If \(y(3)=\frac{e^{\alpha}+1}{\beta e^{\alpha}}\). then the value of \(\alpha+\beta\) is equal to
- A \(-14\)
- B \(14\)
- C \(-24\)
- D \(24\)
Answer & Solution
Correct Answer
(B) \(14\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}+\frac{2 x}{x-1} \cdot y=\frac{1}{(x-1)^{2}}\) \(y=\frac{1}{(x-1)^{2}}\left[\frac{e^{2 x}+1}{2 e^{2 x}}\right]\) \(y(3)=\frac{e^{6}+1}{8 e^{6}}\) \(\alpha+\beta=14\)
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