JEE Mains · Maths · STD 12 - 9. differential equations
If the solution curve of the differential equation \(\left(y-2 \log _e x\right) d x+\left(x \log _e x^2\right) d y=0, x > 1\) passes through the points \(\left(e, \frac{4}{3}\right)\) and \(\left(e^4, \alpha\right)\), then \(\alpha\) is equal to \(................\).
- A \(2\)
- B \(3\)
- C \(1\)
- D \(6\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
\((y-2 \ln x) d x+(2 x \ln x) d y=0\) \(d y(2 x \ln x)=[(2 \ln x)-y] d x\) \(\frac{d y}{d x}=\frac{1}{x}-\frac{y}{2 x \ln x}\) \(\frac{d y}{d x}+\frac{y}{2 x \ln x}=\frac{1}{x}\) \(\text { I.F }=e^{\int \frac{1}{2 x \ln x} d x}\)…
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