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Differential Equations question

2024 · 30 Jan · Shift 1 · Q45
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  5. /2024 · 30 Jan · Shift 1 · Q45

Differential Equations question

2024 · 30 Jan · Shift 1 · Q45

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation sec⁡x dy+{2(1−x)tan⁡x+x(2−x)}dx=0\sec x \mathrm{~d} y+\{2(1-x) \tan x+x(2-x)\} \mathrm{d} x=0secx dy+{2(1−x)tanx+x(2−x)}dx=0 such that y(0)=2y(0)=2y(0)=2. Then y(2)y(2)y(2) is equal to:
  1. A
    2{sin⁡(2)+1}2\{\sin (2)+1\}2{sin(2)+1}
  2. B
    2
  3. C
    1
  4. D
    2{1−sin⁡(2)}2\{1-\sin (2)\}2{1−sin(2)}
View written solutionFree

Correct answer: B

  1. Write the differential equation in standard form

Given

sec⁡x dy+{2(1−x)tan⁡x+x(2−x)} dx=0.\sec x\,dy+\{2(1-x)\tan x+x(2-x)\}\,dx=0.secxdy+{2(1−x)tanx+x(2−x)}dx=0.

So,

sec⁡x dydx+2(1−x)tan⁡x+x(2−x)=0.\sec x\,\frac{dy}{dx}+2(1-x)\tan x+x(2-x)=0.secxdxdy​+2(1−x)tanx+x(2−x)=0.

Multiply by cos⁡x\cos xcosx:

dydx+2(1−x)sin⁡x+x(2−x)cos⁡x=0.\frac{dy}{dx}+2(1-x)\sin x+x(2-x)\cos x=0.dxdy​+2(1−x)sinx+x(2−x)cosx=0.

Hence,

dydx=−2(1−x)sin⁡x−x(2−x)cos⁡x.\frac{dy}{dx}=-2(1-x)\sin x-x(2-x)\cos x.dxdy​=−2(1−x)sinx−x(2−x)cosx.
  1. Look for a convenient antiderivative

We notice that

ddx[x(2−x)sin⁡x]=(2−2x)sin⁡x+x(2−x)cos⁡x.\frac{d}{dx}\big[x(2-x)\sin x\big]=(2-2x)\sin x+x(2-x)\cos x.dxd​[x(2−x)sinx]=(2−2x)sinx+x(2−x)cosx.

Since 2−2x=2(1−x)2-2x=2(1-x)2−2x=2(1−x), this becomes

ddx[x(2−x)sin⁡x]=2(1−x)sin⁡x+x(2−x)cos⁡x.\frac{d}{dx}\big[x(2-x)\sin x\big]=2(1-x)\sin x+x(2-x)\cos x.dxd​[x(2−x)sinx]=2(1−x)sinx+x(2−x)cosx.

Therefore,

dydx=−ddx[x(2−x)sin⁡x].\frac{dy}{dx}=-\frac{d}{dx}\big[x(2-x)\sin x\big].dxdy​=−dxd​[x(2−x)sinx].

So,

y=−x(2−x)sin⁡x+C.y=-x(2-x)\sin x+C.y=−x(2−x)sinx+C.
  1. Use the initial condition

Given y(0)=2y(0)=2y(0)=2:

2=−0⋅(2−0)sin⁡0+C=C.2=-0\cdot(2-0)\sin 0+C=C.2=−0⋅(2−0)sin0+C=C.

Thus,

C=2.C=2.C=2.

Hence the solution is

y=2−x(2−x)sin⁡x.y=2-x(2-x)\sin x.y=2−x(2−x)sinx.
  1. Find y(2)y(2)y(2)

Substitute x=2x=2x=2:

y(2)=2−2(2−2)sin⁡2=2−0=2.y(2)=2-2(2-2)\sin 2=2-0=2.y(2)=2−2(2−2)sin2=2−0=2.
  1. Compare with the options

Thus,

y(2)=2.\boxed{y(2)=2}.y(2)=2​.

So the correct option is B.

  1. Compare with stored correct answer

Stored correct answer: B

Our derived answer is also B, so they agree.

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