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Differential Equations question

2025 · 22 Jan · Shift 2 · Q47
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  5. /2025 · 22 Jan · Shift 2 · Q47

Differential Equations question

2025 · 22 Jan · Shift 2 · Q47

JEE MainMathematicsDifferential EquationsNumerical+4 / −1
Let y=f(x)y=f(x)y=f(x) be the solution of the differential equation dy dx+xyx2−1=x6+4x1−x2,−1<x<1\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{x y}{x^2-1}=\frac{x^6+4 x}{\sqrt{1-x^2}},-1\lt x\lt 1 dxdy​+x2−1xy​=1−x2​x6+4x​,−1<x<1 such that f(0)=0f(0)=0f(0)=0. If 6∫−1/21/2f(x)dx=2π−α6 \int_{-1 / 2}^{1 / 2} f(x) \mathrm{d} x=2 \pi-\alpha6∫−1/21/2​f(x)dx=2π−α then α2\alpha^2α2 is equal to ‾\underline{\hspace{2cm}}​ .
Numerical answer
View written solutionFree

Correct answer: 27

  1. Given differential equation

We have

dydx+xyx2−1=x6+4x1−x2,−1<x<1\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^6+4x}{\sqrt{1-x^2}},\qquad -1<x<1dxdy​+x2−1xy​=1−x2​x6+4x​,−1<x<1

with initial condition

f(0)=0.f(0)=0.f(0)=0.

We must find

6∫−1/21/2f(x) dx=2π−α6\int_{-1/2}^{1/2} f(x)\,dx = 2\pi-\alpha6∫−1/21/2​f(x)dx=2π−α

and then compute α2\alpha^2α2.

  1. Put the equation in linear form

This is a linear differential equation:

dydx+P(x)y=Q(x)\frac{dy}{dx}+P(x)y=Q(x)dxdy​+P(x)y=Q(x)

where

P(x)=xx2−1,Q(x)=x6+4x1−x2.P(x)=\frac{x}{x^2-1},\qquad Q(x)=\frac{x^6+4x}{\sqrt{1-x^2}}.P(x)=x2−1x​,Q(x)=1−x2​x6+4x​.

The integrating factor is

I.F.=e∫P(x)dx=e∫xx2−1dx.\text{I.F.}=e^{\int P(x)dx}=e^{\int \frac{x}{x^2-1}dx}.I.F.=e∫P(x)dx=e∫x2−1x​dx.

Let u=x2−1u=x^2-1u=x2−1, then du=2x dxdu=2x\,dxdu=2xdx, so

∫xx2−1dx=12∫duu=12ln⁡∣x2−1∣.\int \frac{x}{x^2-1}dx=\frac12\int \frac{du}{u}=\frac12\ln|x^2-1|.∫x2−1x​dx=21​∫udu​=21​ln∣x2−1∣.

Since −1<x<1-1<x<1−1<x<1, we have x2−1<0x^2-1<0x2−1<0, so ∣x2−1∣=1−x2|x^2-1|=1-x^2∣x2−1∣=1−x2. Thus

I.F.=e12ln⁡(1−x2)=1−x2.\text{I.F.}=e^{\frac12\ln(1-x^2)}=\sqrt{1-x^2}.I.F.=e21​ln(1−x2)=1−x2​.
  1. Multiply the equation by the integrating factor

Multiplying throughout by 1−x2\sqrt{1-x^2}1−x2​,

1−x2 dydx+xx2−11−x2 y=x6+4x.\sqrt{1-x^2}\,\frac{dy}{dx}+\frac{x}{x^2-1}\sqrt{1-x^2}\,y=x^6+4x.1−x2​dxdy​+x2−1x​1−x2​y=x6+4x.

Now observe that

ddx(y1−x2)=1−x2 dydx+yddx(1−x2).\frac{d}{dx}\big(y\sqrt{1-x^2}\big)=\sqrt{1-x^2}\,\frac{dy}{dx}+y\frac{d}{dx}(\sqrt{1-x^2}).dxd​(y1−x2​)=1−x2​dxdy​+ydxd​(1−x2​).

Also,

ddx(1−x2)=−x1−x2.\frac{d}{dx}(\sqrt{1-x^2})=\frac{-x}{\sqrt{1-x^2}}.dxd​(1−x2​)=1−x2​−x​.

So

ddx(y1−x2)=1−x2 dydx−x1−x2y.\frac{d}{dx}\big(y\sqrt{1-x^2}\big)=\sqrt{1-x^2}\,\frac{dy}{dx}-\frac{x}{\sqrt{1-x^2}}y.dxd​(y1−x2​)=1−x2​dxdy​−1−x2​x​y.

But since

xx2−11−x2=−x1−x2,\frac{x}{x^2-1}\sqrt{1-x^2}=-\frac{x}{\sqrt{1-x^2}},x2−1x​1−x2​=−1−x2​x​,

we indeed get

ddx(y1−x2)=x6+4x.\frac{d}{dx}\big(y\sqrt{1-x^2}\big)=x^6+4x.dxd​(y1−x2​)=x6+4x.

Integrating,

y1−x2=∫(x6+4x)dx=x77+2x2+C.y\sqrt{1-x^2}=\int (x^6+4x)dx=\frac{x^7}{7}+2x^2+C.y1−x2​=∫(x6+4x)dx=7x7​+2x2+C.

Hence

y=f(x)=x77+2x2+C1−x2.y=f(x)=\frac{\frac{x^7}{7}+2x^2+C}{\sqrt{1-x^2}}.y=f(x)=1−x2​7x7​+2x2+C​.
  1. Use the initial condition

Given f(0)=0f(0)=0f(0)=0,

0=0+0+C1  ⟹  C=0.0=\frac{0+0+C}{1}\implies C=0.0=10+0+C​⟹C=0.

Thus

f(x)=x77+2x21−x2.f(x)=\frac{\frac{x^7}{7}+2x^2}{\sqrt{1-x^2}}.f(x)=1−x2​7x7​+2x2​.
  1. Set up the integral

We need

I=∫−1/21/2f(x) dx=∫−1/21/2x77+2x21−x2 dx.I=\int_{-1/2}^{1/2} f(x)\,dx=\int_{-1/2}^{1/2}\frac{\frac{x^7}{7}+2x^2}{\sqrt{1-x^2}}\,dx.I=∫−1/21/2​f(x)dx=∫−1/21/2​1−x2​7x7​+2x2​dx.

Now,

  • x71−x2\dfrac{x^7}{\sqrt{1-x^2}}1−x2​x7​ is an odd function,
  • 2x21−x2\dfrac{2x^2}{\sqrt{1-x^2}}1−x2​2x2​ is an even function.

Therefore the odd part integrates to 000 over [−1/2,1/2][-1/2,1/2][−1/2,1/2], so

I=∫−1/21/22x21−x2 dx=4∫01/2x21−x2 dx.I=\int_{-1/2}^{1/2}\frac{2x^2}{\sqrt{1-x^2}}\,dx =4\int_0^{1/2}\frac{x^2}{\sqrt{1-x^2}}\,dx.I=∫−1/21/2​1−x2​2x2​dx=4∫01/2​1−x2​x2​dx.

Thus

6I=24∫01/2x21−x2 dx.6I=24\int_0^{1/2}\frac{x^2}{\sqrt{1-x^2}}\,dx.6I=24∫01/2​1−x2​x2​dx.
  1. Evaluate the integral using x=sin⁡θx=\sin\thetax=sinθ

Let

x=sin⁡θ,dx=cos⁡θ dθ,x=\sin\theta,\qquad dx=\cos\theta\,d\theta,x=sinθ,dx=cosθdθ,

with

x=0⇒θ=0,x=12⇒θ=π6.x=0\Rightarrow \theta=0,\qquad x=\frac12\Rightarrow \theta=\frac\pi6.x=0⇒θ=0,x=21​⇒θ=6π​.

Also,

1−x2=cos⁡θ.\sqrt{1-x^2}=\cos\theta.1−x2​=cosθ.

So

∫01/2x21−x2dx=∫0π/6sin⁡2θcos⁡θcos⁡θ dθ=∫0π/6sin⁡2θ dθ.\int_0^{1/2}\frac{x^2}{\sqrt{1-x^2}}dx =\int_0^{\pi/6}\frac{\sin^2\theta}{\cos\theta}\cos\theta\,d\theta =\int_0^{\pi/6}\sin^2\theta\,d\theta.∫01/2​1−x2​x2​dx=∫0π/6​cosθsin2θ​cosθdθ=∫0π/6​sin2θdθ.

Using

sin⁡2θ=1−cos⁡2θ2,\sin^2\theta=\frac{1-\cos2\theta}{2},sin2θ=21−cos2θ​,

we get

∫sin⁡2θ dθ=θ2−sin⁡2θ4.\int \sin^2\theta\,d\theta=\frac\theta2-\frac{\sin2\theta}{4}.∫sin2θdθ=2θ​−4sin2θ​.

Hence

∫0π/6sin⁡2θ dθ=[θ2−sin⁡2θ4]0π/6=π12−sin⁡(π/3)4=π12−38.\int_0^{\pi/6}\sin^2\theta\,d\theta =\left[\frac\theta2-\frac{\sin2\theta}{4}\right]_0^{\pi/6} =\frac{\pi}{12}-\frac{\sin(\pi/3)}{4} =\frac{\pi}{12}-\frac{\sqrt3}{8}.∫0π/6​sin2θdθ=[2θ​−4sin2θ​]0π/6​=12π​−4sin(π/3)​=12π​−83​​.

Therefore

6I=24(π12−38)=2π−33.6I=24\left(\frac{\pi}{12}-\frac{\sqrt3}{8}\right)=2\pi-3\sqrt3.6I=24(12π​−83​​)=2π−33​.

Comparing with

6∫−1/21/2f(x) dx=2π−α,6\int_{-1/2}^{1/2}f(x)\,dx=2\pi-\alpha,6∫−1/21/2​f(x)dx=2π−α,

we get

α=33.\alpha=3\sqrt3.α=33​.

So

α2=(33)2=27.\alpha^2=(3\sqrt3)^2=27.α2=(33​)2=27.
  1. Comparison with stored answer

Derived answer:

27\boxed{27}27​

This matches the stored correct answer.

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