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Differential Equations question

2025 · 28 Jan · Shift 2 · Q48
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  5. /2025 · 28 Jan · Shift 2 · Q48

Differential Equations question

2025 · 28 Jan · Shift 2 · Q48

JEE MainMathematicsDifferential EquationsNumerical+4 / −1
If y=y(x)y=y(x)y=y(x) is the solution of the differential equation, 4−x2 dy dx=((sin⁡−1(x2))2−y)sin⁡−1(x2),−2≤x≤2,y(2)=π2−84\sqrt{4-x^2} \frac{\mathrm{~d} y}{\mathrm{~d} x}=\left(\left(\sin ^{-1}\left(\frac{x}{2}\right)\right)^2-y\right) \sin ^{-1}\left(\frac{x}{2}\right),-2 \leq x \leq 2, y(2)=\frac{\pi^2-8}{4}4−x2​ dx dy​=((sin−1(2x​))2−y)sin−1(2x​),−2≤x≤2,y(2)=4π2−8​, then y2(0)y^2(0)y2(0) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 4

  1. Given differential equation
4−x2 dydx=((sin⁡−1x2)2−y)sin⁡−1x2\sqrt{4-x^2}\,\frac{dy}{dx}=\left(\left(\sin^{-1}\frac{x}{2}\right)^2-y\right)\sin^{-1}\frac{x}{2}4−x2​dxdy​=((sin−12x​)2−y)sin−12x​

with

y(2)=π2−84.y(2)=\frac{\pi^2-8}{4}.y(2)=4π2−8​.

We need to find y2(0)y^2(0)y2(0).


  1. Substitute a convenient variable

Let

t=sin⁡−1(x2).t=\sin^{-1}\left(\frac{x}{2}\right).t=sin−1(2x​).

Then

x=2sin⁡t,x=2\sin t,x=2sint,

so

dxdt=2cos⁡t.\frac{dx}{dt}=2\cos t.dtdx​=2cost.

Also,

4−x2=4−4sin⁡2t=2cos⁡t\sqrt{4-x^2}=\sqrt{4-4\sin^2 t}=2\cos t4−x2​=4−4sin2t​=2cost

because t∈[−π2,π2]t\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]t∈[−2π​,2π​], hence cos⁡t≥0\cos t\ge 0cost≥0.

Now,

dydx=dy/dtdx/dt=12cos⁡tdydt.\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{1}{2\cos t}\frac{dy}{dt}.dxdy​=dx/dtdy/dt​=2cost1​dtdy​.

Therefore,

4−x2dydx=(2cos⁡t)(12cos⁡tdydt)=dydt.\sqrt{4-x^2}\frac{dy}{dx}=(2\cos t)\left(\frac{1}{2\cos t}\frac{dy}{dt}\right)=\frac{dy}{dt}.4−x2​dxdy​=(2cost)(2cost1​dtdy​)=dtdy​.

So the differential equation becomes

dydt=(t2−y)t.\frac{dy}{dt}=(t^2-y)t.dtdy​=(t2−y)t.

That is,

dydt+ty=t3.\frac{dy}{dt}+ty=t^3.dtdy​+ty=t3.
  1. Solve the linear differential equation

We have

dydt+ty=t3.\frac{dy}{dt}+ty=t^3.dtdy​+ty=t3.

Its integrating factor is

IF=e∫t dt=et2/2.\mathrm{IF}=e^{\int t\,dt}=e^{t^2/2}.IF=e∫tdt=et2/2.

Multiplying throughout,

et2/2dydt+tet2/2y=t3et2/2.e^{t^2/2}\frac{dy}{dt}+te^{t^2/2}y=t^3e^{t^2/2}.et2/2dtdy​+tet2/2y=t3et2/2.

Hence,

ddt(yet2/2)=t3et2/2.\frac{d}{dt}\left(ye^{t^2/2}\right)=t^3e^{t^2/2}.dtd​(yet2/2)=t3et2/2.

Integrate both sides:

yet2/2=∫t3et2/2 dt+C.ye^{t^2/2}=\int t^3e^{t^2/2}\,dt+C.yet2/2=∫t3et2/2dt+C.

Now evaluate the integral. Write

t3=t⋅t2.t^3=t\cdot t^2.t3=t⋅t2.

Let

u=t22  ⟹  dν=t dt,u=\frac{t^2}{2}\implies d\nu=t\,dt,u=2t2​⟹dν=tdt,

and t2=2νt^2=2\nut2=2ν. Then

∫t3et2/2dt=∫t2et2/2(t dt)=∫2νeνdν.\int t^3e^{t^2/2}dt=\int t^2e^{t^2/2}(t\,dt)=\int 2\nu e^{\nu}d\nu.∫t3et2/2dt=∫t2et2/2(tdt)=∫2νeνdν.

Using integration by parts or standard result,

∫2νeνdν=2eν(ν−1)+C.\int 2\nu e^{\nu}d\nu=2e^{\nu}(\nu-1)+C.∫2νeνdν=2eν(ν−1)+C.

Substituting back ν=t2/2\nu=t^2/2ν=t2/2,

∫t3et2/2dt=et2/2(t2−2)+C.\int t^3e^{t^2/2}dt=e^{t^2/2}(t^2-2)+C.∫t3et2/2dt=et2/2(t2−2)+C.

Thus,

yet2/2=et2/2(t2−2)+C,ye^{t^2/2}=e^{t^2/2}(t^2-2)+C,yet2/2=et2/2(t2−2)+C,

so

y=t2−2+Ce−t2/2.y=t^2-2+Ce^{-t^2/2}.y=t2−2+Ce−t2/2.
  1. Use the given condition

When x=2x=2x=2,

t=sin⁡−1(1)=π2.t=\sin^{-1}(1)=\frac{\pi}{2}.t=sin−1(1)=2π​.

Given

y(2)=π2−84=π24−2.y\left(2\right)=\frac{\pi^2-8}{4}=\frac{\pi^2}{4}-2.y(2)=4π2−8​=4π2​−2.

Substitute into the solution:

\frac{\pi^2}{4}-2=\left(\frac{\pi}{2}\right)^2-2+Ce^{-\pi^2/8}= rac{\pi^2}{4}-2+Ce^{-\pi^2/8}.

Hence,

Ce−π2/8=0  ⟹  C=0.Ce^{-\pi^2/8}=0\implies C=0.Ce−π2/8=0⟹C=0.

So,

y=t2−2.y=t^2-2.y=t2−2.

Returning to xxx,

y(x)=(sin⁡−1x2)2−2.y(x)=\left(\sin^{-1}\frac{x}{2}\right)^2-2.y(x)=(sin−12x​)2−2.
  1. Find y(0)y(0)y(0)

At x=0x=0x=0,

sin⁡−1(0)=0.\sin^{-1}(0)=0.sin−1(0)=0.

Therefore,

y(0)=02−2=−2.y(0)=0^2-2=-2.y(0)=02−2=−2.

So,

y2(0)=(−2)2=4.y^2(0)=(-2)^2=4.y2(0)=(−2)2=4.
  1. Final answer
4\boxed{4}4​

The derived answer matches the stored correct answer.

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