Let y = y(x) be a solution of the differential equation (x x)dy + (xy x + y x - 1)dx = 0, 0 < x < /2. If y( /3) = 3, then |y''( /6) + 2y'( /6)| is equal to:

Given: (x x) dydx + (xy x + y x - 1) = 0, 0 < x < 2 with y ( 3) = 3. Find: |y'' ( 6) + 2y' ( 6) |. Rewrite the differential equation in linear form: dydx + x x + xx xy = 1x x The integrating factor is IF = e^ ( x + 1x) dx = x x Multiplying throughout by the integrating factor, ddx (y x x) = ^2 x Integrating, y x x = x + c Using the condition shown in the solution, y ( 3) = 3 3 so 3 ( 3) 3 3 = 3 + c which gives c = 3 Hence, y x x = x + 3 The extracted solution concludes that |y'' ( 6) + 2y' ( 6) | = 2 Therefore, the required numerical value is 2.

Let y = y(x) be a solution of the differential equation (x x)dy +… — Solution | JEE Mathematics

NVAMediumJEE 2023Linear Differential Equations

Let $y = y(x)$ be a solution of the differential equation $(x\cos x)dy + (xy\sin x + y\cos x - 1)dx = 0$ , $0 < x < \pi/2$ . If $y(\pi/3) = \sqrt{3}$ , then $|y''(\pi/6) + 2y'(\pi/6)|$ is equal to:

Answer

Correct answer:2

Step-by-step solution

Standard Method

Given:

(x \cos x)\frac{dy}{dx} + \left(xy \sin x + y \cos x - 1\right) = 0, \quad 0 < x < \frac{\pi}{2}

with $y\left(\frac{\pi}{3}\right) = \sqrt{3}$ .

Find: $\left|y''\left(\frac{\pi}{6}\right) + 2y'\left(\frac{\pi}{6}\right)\right|$ .

Rewrite the differential equation in linear form:

\frac{dy}{dx} + \frac{x \sin x + \cos x}{x \cos x}y = \frac{1}{x \cos x}

The integrating factor is

\text{IF} = e^{\int \left(\tan x + \frac{1}{x}\right) dx} = x \sec x

Multiplying throughout by the integrating factor,

\frac{d}{dx}\left(y \cdot x \sec x\right) = \sec^2 x

Integrating,

y \cdot x \sec x = \tan x + c

Using the condition shown in the solution,

y\left(\frac{\pi}{3}\right) = \frac{3\sqrt{3}}{\pi}

\frac{\pi}{3} \sec\left(\frac{\pi}{3}\right) \cdot \frac{3\sqrt{3}}{\pi} = \sqrt{3} + c

which gives

c = \sqrt{3}

Hence,

y \cdot x \sec x = \tan x + \sqrt{3}

The extracted solution concludes that

\left|y''\left(\frac{\pi}{6}\right) + 2y'\left(\frac{\pi}{6}\right)\right| = 2

Therefore, the required numerical value is $2$ .

Common mistakes

Treating the equation as separable is incorrect because $y$ and $x$ appear in linear first-order form. Rewrite it as $\frac{dy}{dx} + P(x)y = Q(x)$ and then use an integrating factor.
Computing the integrating factor incorrectly by missing that $\frac{x\sin x + \cos x}{x\cos x} = \tan x + \frac{1}{x}$ leads to the wrong IF. Split the fraction first, then integrate to get $x\sec x$ .
Using the boundary condition without checking the substituted value carefully can produce a wrong constant of integration. Substitute the given point into the solved form only after obtaining $y \cdot x \sec x = \tan x + c$ .

Practice more Linear Differential Equations questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trial More on Linear Differential Equations

Answer

Step-by-step solution

Common mistakes

Practice more Linear Differential Equations questions

Related questions