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Mathematics Question on Differential equations

Let the solution curve y = f(x) of the differential equation
dydx+xyx2−1=+x4+2x1−x2,x∈(−1,1)\frac{dy}{dx} + \frac{xy}{x^2 - 1} = + \frac{ x^4+2x}{\sqrt{1 - x^2}}, \quad x \in (-1, 1) pass through the origin. Then
∫−3232f(x) dx\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) \,dxis

A

π3−14\frac{π}{3}-\frac{1}{4}

B

π3−34\frac{π}{3} - \frac{\sqrt3}{4}

C

π6−34\frac{π}{6} - \frac{\sqrt3}{4}

D

π6−32\frac{π}{6} - \frac{\sqrt3}{2}

Answer

π3−34\frac{π}{3} - \frac{\sqrt3}{4}

Explanation

Solution

The correct answer is (B) : π3−34\frac{π}{3} - \frac{\sqrt3}{4}
dydx+xyx2−1=x4+2x1−x2\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^4 +2x}{\sqrt{1 - x^2}}
which is first order linear differential equation.
Integrating factor (I.F.)=e∫xx2−1dx(I.F.) = e^{∫\frac{x}{x^2−1}dx}
e12ln⁡∣x2−1∣=∣x2−1∣e^{\frac{1}{2}\ln{|x^2 - 1|}} = \sqrt{|x^2 - 1|}
=1−x2∵x∈(−1,1)=\sqrt{1−x^2} ∵x∈(−1,1)
Solution of differential equation
y1−x2=∫(x4+2x)dx=x55+x2+cy\sqrt{1−x^2}=∫(x^4+2x)dx=\frac{x^5}{5}+x^2+c
Curve is passing through origin, c = 0
y=x5+5x251−x2y=\frac{x^5+5x^2}{5\sqrt{1−x^2}}
∫−3232(x5+5x251−x2) dx=0+2∫032x21−x2 dx\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \left( \frac{x^5 +5x^2}{5\sqrt{1 - x^2}}\right) \,dx = 0 + 2\int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1 - x^2}} \,dx
put x=sin⁡θput\ x= \sin \theta
dx=cos⁡θdx = \cos \theta
I=2∫0π3sin⁡2(θ)⋅cos⁡(θ)dθcos⁡(θ)I = 2\int_{0}^{\frac{\pi}{3}} \frac{\sin^2(\theta) \cdot \cos(\theta)d\theta}{\cos(\theta)}
∫0π3(1−cos⁡2θ) dθ\int_{0}^{\frac{\pi}{3}} (1 - \cos2\theta) \,d\theta
=(θ−sin⁡⁡2θ2)∣0π3=(\theta −\frac{\sin⁡2 \theta }{2})|_{0}^{\frac{π}{3}}
=π3−34= \frac{π}{3}- \frac{\sqrt{3}}{{4}}