Question

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

• A. 2
• B. -2
• C. -4
• D. -1

Answer 1

The Correct Option is C

The correct answer is (C) : -4
\((1 + e^{2x})\frac{dy}{dx} + 2(1 + y^2)e^x = 0\)
\(\int \frac{dy}{1+y^2} = -\int \frac{2e^x}{1+e^{2x}} \,dx\)    \(\stackrel{e^x = t}{e^xdx=dt} \)
\(\tan^{-1}(y) = -2\int \frac{dt}{1+t^2}\)
\(\tan^{-1}(y) = 2\tan^{-1}(e^x) + c\)
\(y(0) ⇒ c = \frac{π}{2}\)
\(\tan^{-1}(y) = -2\tan^{-1}(e^x) + \frac{\pi}{2}\)
\(y = \cot(2\tan^{-1}(e^x))\)
\(\frac{dy}{dx} = -cosec^2(2\tan^{-1}(e^x) \cdot (\frac{2e^x}{1+e^{2x}})\)
\(y'(0) = \left. \frac{dy}{dx} \right|_{x=0} = -\frac{2}{2} = -1\)
\(y = \cot(2\tan^{-1}(e^x))\)
\(y(\ln \sqrt{3}) = \cot(2 \tan^{-1}(e^{\log_e \sqrt{3}}))\)
\(= \cot(2\tan^{-1}(\sqrt{3})) = \cot\left(\frac{2\pi}{3}\right) = -\cot\left(\frac{\pi}{3}\right) = -\frac{1}{\sqrt{3}}\)
\(6\left(y'(0) + \left(y(\ln \sqrt{3})\right)^2\right) = 6\left(-1 + \left(-\frac{1}{\sqrt{3}}\right)^2\right) = 6\left(-1 + \frac{1}{3}\right) = -4\)