The function y = f(x) is the solution of the differential equation (dy)/(dx) + xy x ^ 2 - 1 = x ^ 4 + 2x 1 - x ^ 2 in (- 1 , 1) satisfying f(0) = 0 . Then ∫ _ - 3 2 ^ 3 2 f(x)dx is

Correct answer: B. (π)/(3) - 3 4 (dy)/(dx) + x x ^ 2 - 1 y = x ^ 4 + 2x 1 - x ^ 2 This is a linear differential equation I.F. = e ^ ∫ x x ^ 2 - 1 dx = e ^ (1)/(2)In x ^ 2 - 1 = 1 - x ^ 2 ⇒ solutionis y 1 - x ^ 2 = ∫ x( x ^ 3 + 2) 1 - x ^ 2 . 1 - x ^ 2 dx Or y 1 - x ^ 2 = ∫ ( x ^ 4 + 2x)dx = x ^ 5 5 + x ^ 2 + c f(0) = 0 ⇒ c = 0 ⇒ f(x) 1 - x ^ 2 = x ^ 5 5 + x ^ 2 Now, ∫ _ - 3 2 ^ 3 2 f(x)dx = ∫ _ - 3 2 ^ 3 2 x ^ 2 1 - x ^ 2 dx (Using property) = 2 ∫ _ 0 ^ 3 2 x ^ 2 1 - x ^ 2 dx = 2 ∫ _ 0 ^ (π)/(3) ^ 2 θ cos θ cos θ (Taking x = sin θ ) = 2 ∫ _ 0 ^ (π)/(3) ^ 2 θ = 2 [(θ)/(2) - ( 2 θ)/(4)] _ 0 ^ (π)/(3) = 2((π)/(6)) - 2( 3 8 ) = (π)/(3) - 3 4

JEE Advanced 2014Paper 2

Exam: JEE Advanced
Year: 2014
Paper: Paper 2
Subject: Mathematics
Chapter: Differential Equations

JEE Advanced 2014 · Paper 2 · Mathematics · Differential Equations

The function y = f(x) is the solution of the differential equation...

The function

y = f\left(x\right)

is the solution of the differential equation

\dfrac{dy}{dx} + \dfrac{xy}{{x}^{2} - 1} = \dfrac{{x}^{4} + 2x}{\sqrt{1 - {x}^{2}}}

in (- 1 , 1) satisfying

f\left(0\right) = 0 .

Then

\mathop{∫}_{- \dfrac{\sqrt{3}}{2}}^{\dfrac{\sqrt{3}}{2}}f\left(x\right)dx

Options

A $\dfrac{\pi}{3} - \dfrac{\sqrt{3}}{2}$
B $\dfrac{\pi}{3} - \dfrac{\sqrt{3}}{4}$
C $\dfrac{\pi}{6} - \dfrac{\sqrt{3}}{4}$
D $\dfrac{\pi}{6} - \dfrac{\sqrt{3}}{2}$

Correct answer

B. $\dfrac{\pi}{3} - \dfrac{\sqrt{3}}{4}$

Step-by-step solution

\dfrac{dy}{dx} + \dfrac{x}{{x}^{2} - 1}y = \dfrac{{x}^{4} + 2x}{\sqrt{1 - {x}^{2}}}

This is a linear differential equation
I.F.

= {e}^{∫ \dfrac{x}{{x}^{2} - 1}dx} = {e}^{\dfrac{1}{2}In{x}^{2} - 1} = \sqrt{1 - {x}^{2}}

⇒ solutionis

y\sqrt{1 - {x}^{2}} = ∫ \dfrac{x\left({x}^{3} + 2\right)}{\sqrt{1 - {x}^{2}}} . \sqrt{1 - {x}^{2}}dx

y\sqrt{1 - {x}^{2}} = ∫ \left({x}^{4} + 2x\right)dx = \dfrac{{x}^{5}}{5} + {x}^{2} + c

f\left(0\right) = 0 ⇒ c = 0

⇒ f\left(x\right)\sqrt{1 - {x}^{2}} = \dfrac{{x}^{5}}{5} + {x}^{2}

Now,

\mathop{∫}_{- \dfrac{\sqrt{3}}{2}}^{\dfrac{\sqrt{3}}{2}}f\left(x\right)dx =\mathop{∫}_{- \dfrac{\sqrt{3}}{2}}^{\dfrac{\sqrt{3}}{2}}\dfrac{{x}^{2}}{\sqrt{1 - {x}^{2}}}dx

(Using property)

= 2\mathop{∫}_{0}^{\dfrac{\sqrt{3}}{2}}\dfrac{{x}^{2}}{\sqrt{1 - {x}^{2}}}dx = 2\mathop{∫}_{0}^{\dfrac{\pi}{3}}\dfrac{\sin^{2} \theta}{cos \theta}cos \thetad\theta

(Taking

x = sin \theta

)

= 2\mathop{∫}_{0}^{\dfrac{\pi}{3}}\sin^{2} \thetad\theta = 2{{\left[\dfrac{\theta}{2} - \dfrac{\sin 2 \theta}{4}\right]}_{0}}^{\dfrac{\pi}{3}} = 2\left(\dfrac{\pi}{6}\right) - 2\left(\dfrac{\sqrt{3}}{8}\right) = \dfrac{\pi}{3} - \dfrac{\sqrt{3}}{4}

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