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Mathematics Question on Differential equations

Let the solution curve y=y(x)y = y(x) of the differential equation (1+e2x)(dydx+y)=1(1 + e^{2x})\left(\frac{dy}{dx} + y\right) = 1 pass through the point (0,π2)(0, \frac{\pi}{2}). Then, lim⁡x→∞exy(x)\lim_{{x \to \infty}} e^x y(x) is equal to

A

π4\frac{\pi}{4}

B

3π4\frac{3\pi}{4}

C

π2\frac{\pi}{2}

D

3π2\frac{3\pi}{2}

Answer

3π4\frac{3\pi}{4}

Explanation

Solution

D.E(1+e2x)dydx+y=1D.E (1 + e^{2x})\frac{dy}{dx} + y = 1
⇒⇒ dydx+y=11+e2x\frac{dy}{dx} + y = \frac{1}{1+e^{2x}}

I.F.=e∫1 dx=ex\text{I.F.} = e^{\int 1 \,dx} = e^x

∴$$e^x y(x) = \int \frac{e^x}{1 + e^{2x}} \,dx

⇒$$e^x y(x) = \tan^{-1}(e^x) + C
∵∵ It passes through
(0,π2),C=π2−π4=π4(0, \frac{\pi}{2}), \quad C = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}

∴∴ lim⁡x→∞exy(x)=lim⁡x→∞tan⁡−1(ex)+π4\lim_{{x \to \infty}} e^x y(x) = \lim_{{x \to \infty}} \tan^{-1}(e^x) + \frac{\pi}{4}
=3π4= \frac{3\pi}{4}
So, the correct option is (B): 3π4\frac{3\pi}{4}