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Mathematics Question on Differential equations

Let y=y(x)y = y(x) be the solution of the differential equation (1+y2)etan⁡xdx+cos⁡2x(1+e2tan⁡x)dy=0(1 + y^2)e^{\tan x} dx + \cos^2 x (1 + e^{2\tan x}) dy = 0, y(0)=1y(0) = 1. Then y(π4)y\left(\frac{\pi}{4}\right) is equal to:

A

2e\frac{2}{e}

B

1e2\frac{1}{e^2}

C

1e\frac{1}{e}

D

2e2\frac{2}{e^2}

Answer

1e\frac{1}{e}

Explanation

Solution

The given differential equation is:
(1+y2)etan⁡−1xdx+cos⁡2x(1+e2tan⁡−1x)dy=0.(1 + y^2)e^{\tan^{-1}x}dx + \cos^2x(1 + e^{2\tan^{-1}x})dy = 0.
Separate the variables:
sec⁡2x⋅etan⁡−1xdx1+e2tan⁡−1x+dy1+y2=0.\frac{\sec^2x \cdot e^{\tan^{-1}x}dx}{1 + e^{2\tan^{-1}x}} + \frac{dy}{1 + y^2} = 0.
Integrating both sides:
tan⁡−1(etan⁡−1x)+tan⁡−1(y)=C.\tan^{-1}(e^{\tan^{-1}x}) + \tan^{-1}(y) = C.
Using the initial condition y(0)=1y(0) = 1:
tan⁡−1(etan⁡−1(0))+tan⁡−1(1)=C.\tan^{-1}(e^{\tan^{-1}(0)}) + \tan^{-1}(1) = C.
Simplify:
tan⁡−1(e0)+tan⁡−1(1)=C  ⟹  tan⁡−1(1)+tan⁡−1(1)=C  ⟹  C=π2.\tan^{-1}(e^0) + \tan^{-1}(1) = C \implies \tan^{-1}(1) + \tan^{-1}(1) = C \implies C = \frac{\pi}{2}.
The solution becomes:
tan⁡−1(etan⁡−1x)+tan⁡−1(y)=π2.\tan^{-1}(e^{\tan^{-1}x}) + \tan^{-1}(y) = \frac{\pi}{2}.
At x=π4x = \frac{\pi}{4}, substitute into the solution:
tan⁡−1(etan⁡−1(π4))+tan⁡−1(y)=π2.\tan^{-1}(e^{\tan^{-1}(\frac{\pi}{4})}) + \tan^{-1}(y) = \frac{\pi}{2}.
Rearrange:
tan⁡−1(y)=π2−tan⁡−1(etan⁡−1(π4)).\tan^{-1}(y) = \frac{\pi}{2} - \tan^{-1}(e^{\tan^{-1}(\frac{\pi}{4})}).
From the properties of tan⁡−1\tan^{-1}, substitute:
tan⁡−1(y)=cot⁡−1(e).\tan^{-1}(y) = \cot^{-1}(e).
Simplify:
y=1e.y = \frac{1}{e}.
Final Answer: 1e\frac{1}{e}.