JEE Mains · Maths · STD 12 - 9. differential equations
If \(x=f(y)\) is the solution of the differential equation
\(\left(1+y^2\right)+\left(x-2 \mathrm{e}^{\tan ^{-1} y}\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=0, y \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
with \(f(0)=1\), then \(f\left(\frac{1}{\sqrt{3}}\right)\) is equal to :
- A \(e^{\pi / 12}\)
- B \(e^{\pi / 4}\)
- C \(e^{\pi / 3}\)
- D \(e^{\pi / 6}\)
Answer & Solution
Correct Answer
(D) \(e^{\pi / 6}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \frac{\mathrm{dx}}{\mathrm{dy}}+\frac{\mathrm{x}}{1+\mathrm{y}^2}=\frac{2 \mathrm{e}^{\tan ^{-1} \mathrm{y}}}{1+\mathrm{y}^2} \\ & \text { I.F. }=\mathrm{e}^{\tan ^{-1} \mathrm{y}} \\ & \mathrm{xe}^{\tan ^{-1} \mathrm{y}}=\int \frac{2\left(\mathrm{e}^{\tan…
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