Solveeit Logo
Login

Question

Mathematics Question on Differential equations

If y = y (x) is the solution of the differential equation
(1+e2x)dydx+2(1+y2)ex=0(1 + e^{2x})\frac{dy}{dx} + 2(1 + y^2)e^x = 0
and y(0) = 0, then
6(y′(0)+(log⁡e(3))2)6 \left( y'(0) + \left( \log_e\left(\sqrt{3}\right) \right)^2 \right)
is equal to

A

2

B

-2

C

-4

D

-1

Answer

-4

Explanation

Solution

The correct answer is (C) : -4
(1+e2x)dydx+2(1+y2)ex=0(1 + e^{2x})\frac{dy}{dx} + 2(1 + y^2)e^x = 0
∫dy1+y2=−∫2ex1+e2x dx\int \frac{dy}{1+y^2} = -\int \frac{2e^x}{1+e^{2x}} \,dx exdx=dtex=t\stackrel{e^x = t}{e^xdx=dt}
tan⁡−1(y)=−2∫dt1+t2\tan^{-1}(y) = -2\int \frac{dt}{1+t^2}
tan⁡−1(y)=2tan⁡−1(ex)+c\tan^{-1}(y) = 2\tan^{-1}(e^x) + c
y(0)⇒c=π2y(0) ⇒ c = \frac{π}{2}
tan⁡−1(y)=−2tan⁡−1(ex)+π2\tan^{-1}(y) = -2\tan^{-1}(e^x) + \frac{\pi}{2}
y=cot⁡(2tan⁡−1(ex))y = \cot(2\tan^{-1}(e^x))
dydx=−cosec2(2tan⁡−1(ex)⋅(2ex1+e2x)\frac{dy}{dx} = -cosec^2(2\tan^{-1}(e^x) \cdot (\frac{2e^x}{1+e^{2x}})
y′(0)=dydx∣x=0=−22=−1y'(0) = \left. \frac{dy}{dx} \right|_{x=0} = -\frac{2}{2} = -1
y=cot⁡(2tan⁡−1(ex))y = \cot(2\tan^{-1}(e^x))
y(ln⁡3)=cot⁡(2tan⁡−1(elog⁡e3))y(\ln \sqrt{3}) = \cot(2 \tan^{-1}(e^{\log_e \sqrt{3}}))
=cot⁡(2tan⁡−1(3))=cot⁡(2π3)=−cot⁡(π3)=−13= \cot(2\tan^{-1}(\sqrt{3})) = \cot\left(\frac{2\pi}{3}\right) = -\cot\left(\frac{\pi}{3}\right) = -\frac{1}{\sqrt{3}}
6(y′(0)+(y(ln⁡3))2)=6(−1+(−13)2)=6(−1+13)=−46\left(y'(0) + \left(y(\ln \sqrt{3})\right)^2\right) = 6\left(-1 + \left(-\frac{1}{\sqrt{3}}\right)^2\right) = 6\left(-1 + \frac{1}{3}\right) = -4