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Differential Equations

Section: Mathematics

Question : 18 of 23

Marks: +1, -0

Solution:

We have

\frac{dy}{dx} - y\tan x

= 2x sec x

\cos x \frac{dy}{dx} + (-\sin x)y

= 2x

\frac{d}{dx} (y\cos x)

= 2x

y (x) cos x =

x^2 + c

, where c = 0 since y(0) = 0

When x =

\frac{\pi}{4} , y\left(\frac{\pi}{4}\right)

\frac{\pi^2}{8\sqrt{2}}

When x =

\frac{\pi}{3} , y\left(\frac{\pi}{3}\right)

\frac{2\pi^2}{9}

When x =

\frac{\pi}{4}. y'\left(\frac{\pi}{4}\right)

\frac{\pi^2}{8\sqrt{2}}+\frac{\pi}{\sqrt{2}}

When x =

\frac{\pi}{3}, y'\left(\frac{\pi}{3}\right)

\frac{2\pi^2}{3\sqrt{3}}+\frac{4\pi}{3}

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