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Differential Equations question

2019 · 9 Apr · Shift 2 · Q25

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JEE Main
Mathematics
Differential Equations
2019 · 9 Apr · Shift 2 · Q25

Differential Equations question

2019 · 9 Apr · Shift 2 · Q25

JEE MainMathematicsDifferential EquationsMCQ+4 / −1

If

\cos x{{dy} \over {dx}} - y\sin x = 6x

, (0 < x <

{\pi \over 2}

) and

y\left( {{\pi \over 3}} \right)

= 0 then

y\left( {{\pi \over 6}} \right)

is equal to :-

A
$- {{{\pi ^2}} \over {2 }}$
B
$- {{{\pi ^2}} \over {4\sqrt 3 }}$
C
${{{\pi ^2}} \over {2\sqrt 3 }}$
D
$- {{{\pi ^2}} \over {2\sqrt 3 }}$

View written solutionFree

Correct answer: D

Given differential equation

$\cos x\,\frac{dy}{dx}-y\sin x=6x, \qquad 0<x<\frac\pi2$

with condition

$y\left(\frac\pi3\right)=0.$
Recognize the left-hand side

Since

$\frac{d}{dx}(y\cos x)=\cos x\frac{dy}{dx}-y\sin x,$

the differential equation becomes

$\frac{d}{dx}(y\cos x)=6x.$
Integrate both sides

$y\cos x=\int 6x\,dx=3x^2+C.$

Hence,

$y=\frac{3x^2+C}{\cos x}.$
Use the given condition

At $x=\frac\pi3$ , we have $y=0$ :

$0=\frac{3\left(\frac\pi3\right)^2+C}{\cos\frac\pi3}.$

Since $\cos\frac\pi3=\frac12\neq 0$ , we get

$3\cdot \frac{\pi^2}{9}+C=0$ $\frac{\pi^2}{3}+C=0$ $C=-\frac{\pi^2}{3}.$

Therefore,

$y=\frac{3x^2-\frac{\pi^2}{3}}{\cos x}.$
Find $y\left(\frac\pi6\right)$

$y\left(\frac\pi6\right)=\frac{3\left(\frac\pi6\right)^2-\frac{\pi^2}{3}}{\cos\frac\pi6}.$

First simplify the numerator:

$3\left(\frac{\pi^2}{36}\right)-\frac{\pi^2}{3}=\frac{\pi^2}{12}-\frac{4\pi^2}{12}=-\frac{3\pi^2}{12}=-\frac{\pi^2}{4}.$

Also,

$\cos\frac\pi6=\frac{\sqrt3}{2}.$

So,

$y\left(\frac\pi6\right)=\frac{-\frac{\pi^2}{4}}{\frac{\sqrt3}{2}}=-\frac{\pi^2}{4}\cdot \frac{2}{\sqrt3}=-\frac{\pi^2}{2\sqrt3}.$
Match with options

$-\frac{\pi^2}{2\sqrt3}$

This corresponds to Option D.

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