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Differential Equations question

2024 · 4 Apr · Shift 2 · Q57
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  5. /2024 · 4 Apr · Shift 2 · Q57

Differential Equations question

2024 · 4 Apr · Shift 2 · Q57

JEE MainMathematicsDifferential EquationsNumerical+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (x+y+2)2dx=dy,y(0)=−2(x+y+2)^2 d x=d y, y(0)=-2(x+y+2)2dx=dy,y(0)=−2. Let the maximum and minimum values of the function y=y(x)y=y(x)y=y(x) in [0,π3]\left[0, \frac{\pi}{3}\right][0,3π​] be α\alphaα and β\betaβ, respectively. If (3α+π)2+β2=γ+δ3,γ,δ∈Z(3 \alpha+\pi)^2+\beta^2=\gamma+\delta \sqrt{3}, \gamma, \delta \in \mathbb{Z}(3α+π)2+β2=γ+δ3​,γ,δ∈Z, then γ+δ\gamma+\deltaγ+δ equals ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 31

  1. Given differential equation

We have

(x+y+2)2 dx=dy(x+y+2)^2\,dx = dy(x+y+2)2dx=dy

so

dydx=(x+y+2)2.\frac{dy}{dx}=(x+y+2)^2.dxdy​=(x+y+2)2.

Also, the initial condition is

y(0)=−2.y(0)=-2.y(0)=−2.
  1. Simplify using substitution

Let

z=x+y+2.z=x+y+2.z=x+y+2.

Then

dzdx=1+dydx=1+z2.\frac{dz}{dx}=1+\frac{dy}{dx}=1+z^2.dxdz​=1+dxdy​=1+z2.

So the differential equation becomes

dzdx=1+z2.\frac{dz}{dx}=1+z^2.dxdz​=1+z2.

This is separable:

dz1+z2=dx.\frac{dz}{1+z^2}=dx.1+z2dz​=dx.

Integrating,

tan⁡−1z=x+C.\tan^{-1} z = x + C.tan−1z=x+C.

Hence

z=tan⁡(x+C).z=\tan(x+C).z=tan(x+C).

Now use the initial condition. Since

z=x+y+2,z=x+y+2,z=x+y+2,

at x=0x=0x=0, y=−2y=-2y=−2, so

z(0)=0+(−2)+2=0.z(0)=0+(-2)+2=0.z(0)=0+(−2)+2=0.

Thus

0=tan⁡C  ⟹  C=00=\tan C \implies C=00=tanC⟹C=0

(choosing the branch consistent near x=0x=0x=0).

Therefore,

z=tan⁡x.z=\tan x.z=tanx.

So

x+y+2=tan⁡xx+y+2=\tan xx+y+2=tanx

and hence

y=tan⁡x−x−2.y=\tan x - x -2.y=tanx−x−2.
  1. Find maximum and minimum on [0,π3]\left[0,\frac{\pi}{3}\right][0,3π​]

We have

y(x)=tan⁡x−x−2.y(x)=\tan x - x -2.y(x)=tanx−x−2.

Differentiate:

y′(x)=sec⁡2x−1=tan⁡2x≥0.y'(x)=\sec^2 x -1 = \tan^2 x \ge 0.y′(x)=sec2x−1=tan2x≥0.

So yyy is increasing on [0,π3]\left[0,\frac{\pi}{3}\right][0,3π​].

Hence:

  • minimum value occurs at x=0x=0x=0
  • maximum value occurs at x=π3x=\frac{\pi}{3}x=3π​

Thus

β=y(0)=tan⁡0−0−2=−2,\beta=y(0)=\tan 0 - 0 -2 = -2,β=y(0)=tan0−0−2=−2, α=y(π3)=tan⁡π3−π3−2=3−π3−2.\alpha=y\left(\frac{\pi}{3}\right)=\tan\frac{\pi}{3}-\frac{\pi}{3}-2=\sqrt3-\frac{\pi}{3}-2.α=y(3π​)=tan3π​−3π​−2=3​−3π​−2.
  1. Compute (3α+π)2+β2(3\alpha+\pi)^2+\beta^2(3α+π)2+β2

First,

3α=3(3−π3−2)=33−π−6.3\alpha = 3\left(\sqrt3-\frac{\pi}{3}-2\right)=3\sqrt3-\pi-6.3α=3(3​−3π​−2)=33​−π−6.

So

3α+π=33−6.3\alpha+\pi = 3\sqrt3-6.3α+π=33​−6.

Then

(3α+π)2=(33−6)2=27+36−363=63−363.(3\alpha+\pi)^2=(3\sqrt3-6)^2 = 27+36-36\sqrt3=63-36\sqrt3.(3α+π)2=(33​−6)2=27+36−363​=63−363​.

Also,

β2=(−2)2=4.\beta^2=(-2)^2=4.β2=(−2)2=4.

Therefore,

(3α+π)2+β2=(63−363)+4=67−363.(3\alpha+\pi)^2+\beta^2 = (63-36\sqrt3)+4 = 67-36\sqrt3.(3α+π)2+β2=(63−363​)+4=67−363​.

So in the form

γ+δ3,\gamma+\delta\sqrt3,γ+δ3​,

we get

γ=67,δ=−36.\gamma=67,\qquad \delta=-36.γ=67,δ=−36.

Hence

γ+δ=67−36=31.\gamma+\delta=67-36=31.γ+δ=67−36=31.
  1. Final answer
31\boxed{31}31​

The derived answer matches the stored correct answer.

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