Question

Let \(y=y(x)\) be the solution of the differential equation \(\dfrac{dy}{dx}+y\tan x=2x+x^2\tan x\), \(x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), such that \(y(0)=1\). Then :

• A. \(y\left(\frac{\pi}{4}\right)-y\left(-\frac{\pi}{4}\right)=\sqrt2\)
• B. \(y^{\prime}\left(\frac{\pi}{4}\right)-y^{\prime}\left(-\frac{\pi}{4}\right)=\pi-\sqrt2\)
• C. \(y\left(\frac{\pi}{4}\right)+y\left(-\frac{\pi}{4}\right)=\frac{\pi^2}{2}+2\)
• D. \(y^{\prime}\left(\frac{\pi}{4}\right)+y^{\prime}\left(-\frac{\pi}{4}\right)=-\sqrt2\)

Hint:

Linear ODE with integrating factor \(\sec x\); the right side times \(\sec x\) is the derivative of \(x^2\sec x\).

Answer 1

The Correct Option is B

Integrating factor \(=e^{\int\tan x\,dx}=\sec x\).
\(\frac{d}{dx}(y\sec x)=(2x+x^2\tan x)\sec x=2x\sec x+x^2\sec x\tan x=\frac{d}{dx}\left(x^2\sec x\right)\).
So \(y\sec x=x^2\sec x+C\Rightarrow y=x^2+C\cos x\). \(y(0)=1\Rightarrow C=1\), hence \(y=x^2+\cos x\).
\(y^{\prime}=2x-\sin x\). \(y^{\prime}\left(\frac{\pi}{4}\right)=\frac{\pi}{2}-\frac{\sqrt2}{2}\), \(y^{\prime}\left(-\frac{\pi}{4}\right)=-\frac{\pi}{2}+\frac{\sqrt2}{2}\).
Difference \(=\pi-\sqrt2\) (option 2). Check others: \(y\) is even so \(y(\pi/4)-y(-\pi/4)=0\ne\sqrt2\); \(y(\pi/4)+y(-\pi/4)=\frac{\pi^2}{8}+\sqrt2\ne\frac{\pi^2}{2}+2\); \(y^{\prime}\) is odd so \(y^{\prime}(\pi/4)+y^{\prime}(-\pi/4)=0\ne-\sqrt2\).
Answer: (2) \(y^{\prime}\left(\frac{\pi}{4}\right)-y^{\prime}\left(-\frac{\pi}{4}\right)=\pi-\sqrt2\)