Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Differential Equations question

2022 · 26 Jul · Shift 2 · Q28
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Differential Equations
  5. /2022 · 26 Jul · Shift 2 · Q28

Differential Equations question

2022 · 26 Jul · Shift 2 · Q28

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let the solution curve y=f(x)y=f(x)y=f(x) of the differential equation dydx+xyx2−1=x4+2x1−x2\frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}dxdy​+x2−1xy​=1−x2​x4+2x​, x∈(−1,1)x\in(-1,1)x∈(−1,1) pass through the origin. Then ∫−3232f(x)dx\int\limits_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x−23​​∫23​​​f(x)dx is equal to
  1. A
    π3−14\frac{\pi}{3}-\frac{1}{4}3π​−41​
  2. B
    π3−34\frac{\pi}{3}-\frac{\sqrt{3}}{4}3π​−43​​
  3. C
    π6−34\frac{\pi}{6}-\frac{\sqrt{3}}{4}6π​−43​​
  4. D
    π6−32\frac{\pi}{6}-\frac{\sqrt{3}}{2}6π​−23​​
View written solutionFree

Correct answer: B

  1. Given 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}}, \qquad x\in(-1,1)dxdy​+x2−1xy​=1−x2​x4+2x​,x∈(−1,1)

We must find the solution curve y=f(x)y=f(x)y=f(x) passing through the origin, i.e. f(0)=0f(0)=0f(0)=0, and then compute

I=∫−3232f(x) dx.I=\int_{-\frac{\sqrt3}{2}}^{\frac{\sqrt3}{2}} f(x)\,dx.I=∫−23​​23​​​f(x)dx.
  1. Write the equation in linear form

Since

xx2−1=−x1−x2,\frac{x}{x^2-1}=-\frac{x}{1-x^2},x2−1x​=−1−x2x​,

the equation becomes

dydx−x1−x2y=x4+2x1−x2.\frac{dy}{dx}-\frac{x}{1-x^2}y=\frac{x^4+2x}{\sqrt{1-x^2}}.dxdy​−1−x2x​y=1−x2​x4+2x​.

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),

with

P(x)=−x1−x2.P(x)=-\frac{x}{1-x^2}.P(x)=−1−x2x​.
  1. Find the integrating factor
I.F.=e∫P(x)dx=e∫−x1−x2 dx.\text{I.F.}=e^{\int P(x)dx}=e^{\int -\frac{x}{1-x^2}\,dx}.I.F.=e∫P(x)dx=e∫−1−x2x​dx.

Let u=1−x2u=1-x^2u=1−x2, then du=−2x dxdu=-2x\,dxdu=−2xdx, so

∫−x1−x2 dx=12∫duu=12ln⁡(1−x2).\int -\frac{x}{1-x^2}\,dx=\frac12\int \frac{du}{u}=\frac12\ln(1-x^2).∫−1−x2x​dx=21​∫udu​=21​ln(1−x2).

Hence,

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 differential equation by the integrating factor

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

1−x2 dydx−x1−x2y=x4+2x.\sqrt{1-x^2}\,\frac{dy}{dx}-\frac{x}{\sqrt{1-x^2}}y=x^4+2x.1−x2​dxdy​−1−x2​x​y=x4+2x.

The left side is exactly

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

So,

ddx(y1−x2)=x4+2x.\frac{d}{dx}\big(y\sqrt{1-x^2}\big)=x^4+2x.dxd​(y1−x2​)=x4+2x.

Integrating,

y1−x2=∫(x4+2x) dx=x55+x2+C.y\sqrt{1-x^2}=\int (x^4+2x)\,dx=\frac{x^5}{5}+x^2+C.y1−x2​=∫(x4+2x)dx=5x5​+x2+C.

Thus,

y=f(x)=x55+x2+C1−x2.y=f(x)=\frac{\frac{x^5}{5}+x^2+C}{\sqrt{1-x^2}}.y=f(x)=1−x2​5x5​+x2+C​.
  1. Use the condition that the curve passes through the origin

Since f(0)=0f(0)=0f(0)=0,

0=0+0+C1−0=C.0=\frac{0+0+C}{\sqrt{1-0}}=C.0=1−0​0+0+C​=C.

So,

f(x)=x55+x21−x2.f(x)=\frac{\frac{x^5}{5}+x^2}{\sqrt{1-x^2}}.f(x)=1−x2​5x5​+x2​.

Therefore,

I=∫−3232x55+x21−x2 dx.I=\int_{-\frac{\sqrt3}{2}}^{\frac{\sqrt3}{2}} \frac{\frac{x^5}{5}+x^2}{\sqrt{1-x^2}}\,dx.I=∫−23​​23​​​1−x2​5x5​+x2​dx.
  1. Use symmetry

Observe:

  • x51−x2\dfrac{x^5}{\sqrt{1-x^2}}1−x2​x5​ is an odd function, so its integral over [−a,a][-a,a][−a,a] is 000.
  • x21−x2\dfrac{x^2}{\sqrt{1-x^2}}1−x2​x2​ is even.

Hence,

I=∫−aax21−x2dx,a=32.I=\int_{-a}^{a}\frac{x^2}{\sqrt{1-x^2}}dx, \qquad a=\frac{\sqrt3}{2}.I=∫−aa​1−x2​x2​dx,a=23​​.

So,

I=2∫032x21−x2dx.I=2\int_0^{\frac{\sqrt3}{2}}\frac{x^2}{\sqrt{1-x^2}}dx.I=2∫023​​​1−x2​x2​dx.
  1. Evaluate the integral

Put

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

with limits:

x=0⇒θ=0,x=32⇒θ=π3.x=0 \Rightarrow \theta=0, \qquad x=\frac{\sqrt3}{2} \Rightarrow \theta=\frac{\pi}{3}.x=0⇒θ=0,x=23​​⇒θ=3π​.

Then

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

so

x21−x2dxna=sin⁡2θcos⁡θ(cos⁡θ dθ)=sin⁡2θ dθ.\frac{x^2}{\sqrt{1-x^2}}dx na=\frac{\sin^2\theta}{\cos\theta}(\cos\theta\,d\theta)=\sin^2\theta\,d\theta.1−x2​x2​dxna=cosθsin2θ​(cosθdθ)=sin2θdθ.

Thus,

I=2∫0π/3sin⁡2θ dθ.I=2\int_0^{\pi/3}\sin^2\theta\,d\theta.I=2∫0π/3​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{\theta}{2}-\frac{\sin2\theta}{4}.∫sin2θdθ=2θ​−4sin2θ​.

Therefore,

I=2[θ2−sin⁡2θ4]0π/3=2(π6−sin⁡(2π/3)4).I=2\left[\frac{\theta}{2}-\frac{\sin2\theta}{4}\right]_0^{\pi/3} =2\left(\frac{\pi}{6}-\frac{\sin(2\pi/3)}{4}\right).I=2[2θ​−4sin2θ​]0π/3​=2(6π​−4sin(2π/3)​).

Since

sin⁡(2π3)=32,\sin\left(\frac{2\pi}{3}\right)=\frac{\sqrt3}{2},sin(32π​)=23​​,

we have

I=2(π6−38)=π3−34.I=2\left(\frac{\pi}{6}-\frac{\sqrt3}{8}\right) =\frac{\pi}{3}-\frac{\sqrt3}{4}.I=2(6π​−83​​)=3π​−43​​.
  1. Match with the options
I=π3−34\boxed{I=\frac{\pi}{3}-\frac{\sqrt3}{4}}I=3π​−43​​​

So the correct option is:

B\boxed{\text{B}}B​
  1. Comparison with stored correct answer

Stored correct answer: B

Our derived answer is also B, so they agree.

PreviousNext

More from Differential Equations

  • Suppose y=y(x) be the solution curve to the differential equation dxdy​−y=2−e−x such that x→∞lim​y(x) is finite. If a and b are respectively the x- and y-intercepts of the tangent to…2022 · Numerical
  • Let the solution curve y = y(x) of the differential equation (4+x2)dy−2x(x2+3y+4)dx=0 pass through the origin. Then y(2) is equal to ​.2022 · Numerical
  • Let S=(0,2π)−{2π​,43π​,23π​,47π​}. Let y=y(x), x ∈ S, be the solution curve of the differential equation dxdy​=1+sin2x1​,y(4π​)=21​…2022 · Numerical
  • If y=y(x) is the solution of the differential equation xdxdy​+2y=xex, y(1)=0 then the local maximum value of the function z(x)=x2y(x)−ex,x∈R is :2022 · MCQ
  • If the solution of the differential equation dxdy​+ex(x2−2)y=(x2−2x)(x2−2)e2x satisfies y(0)=0, then the value of y(2) is…2022 · MCQ
  • Let y=y1​(x) and y=y2​(x) be two distinct solutions of the differential equation dxdy​=x+y, with y1​(0)=0 and y2​(0)=1 respectively. Then, the number of points of intersection of y=y1​(x) and y=y2​(x) is2022 · MCQ
  • Let y=y(x) be the solution curve of the differential equation sin(2x2)loge​(tanx2)dy+(4xy−42​xsin(x2−4π​))dx=0, 0<x<2π​​…2022 · Numerical
  • If dxdy​+2x−12x−y(2y−1)​=0, x, y > 0, y(1) = 1, then y(2) is equal to :2022 · MCQ