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Differential Equations question

2025 · 23 Jan · Shift 2 · Q45
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  5. /2025 · 23 Jan · Shift 2 · Q45

Differential Equations question

2025 · 23 Jan · Shift 2 · Q45

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let x=x(y)x=x(y)x=x(y) be the solution of the differential equation y=(x−y dx dy)sin⁡(xy),y>0y=\left(x-y \frac{\mathrm{~d} x}{\mathrm{~d} y}\right) \sin \left(\frac{x}{y}\right), y\gt 0y=(x−y dy dx​)sin(yx​),y>0 and x(1)=π2x(1)=\frac{\pi}{2}x(1)=2π​. Then cos⁡(x(2))\cos (x(2))cos(x(2)) is equal to :
  1. A
    2(log⁡e2)−12\left(\log _e 2\right)-12(loge​2)−1
  2. B
    1−2(log⁡e2)21-2\left(\log _e 2\right)^21−2(loge​2)2
  3. C
    1−2(log⁡e2)1-2\left(\log _{\mathrm{e}} 2\right)1−2(loge​2)
  4. D
    2(log⁡e2)2−12\left(\log _e 2\right)^2-12(loge​2)2−1
View written solutionFree

Correct answer: D

  1. Given differential equation

We have

y=(x−ydxdy)sin⁡(xy),y>0y=\left(x-y\frac{dx}{dy}\right)\sin\left(\frac{x}{y}\right), \qquad y>0y=(x−ydydx​)sin(yx​),y>0

with initial condition

x(1)=π2.x(1)=\frac{\pi}{2}.x(1)=2π​.

We need to find cos⁡(x(2))\cos(x(2))cos(x(2)).

  1. Use the substitution

Since the equation contains xy\frac{x}{y}yx​, let

v=xy⇒x=vy.v=\frac{x}{y} \quad \Rightarrow \quad x=vy.v=yx​⇒x=vy.

Then

dxdy=v+ydvdy.\frac{dx}{dy}=v+y\frac{dv}{dy}.dydx​=v+ydydv​.

So,

x−ydxdy=vy−y(v+ydvdy)=vy−vy−y2dvdy=−y2dvdy.x-y\frac{dx}{dy}=vy-y\left(v+y\frac{dv}{dy}\right)=vy-vy-y^2\frac{dv}{dy}=-y^2\frac{dv}{dy}.x−ydydx​=vy−y(v+ydydv​)=vy−vy−y2dydv​=−y2dydv​.

Substitute into the differential equation:

y=(−y2dvdy)sin⁡v.y=\left(-y^2\frac{dv}{dy}\right)\sin v.y=(−y2dydv​)sinv.

Since y>0y>0y>0, divide by yyy:

1=−ydvdysin⁡v.1=-y\frac{dv}{dy}\sin v.1=−ydydv​sinv.

Hence,

dvdy=−1ysin⁡v.\frac{dv}{dy}=-\frac{1}{y\sin v}.dydv​=−ysinv1​.
  1. Separate variables

Rearrange:

sin⁡v dv=−dyy.\sin v\, dv=-\frac{dy}{y}.sinvdv=−ydy​.

Integrate:

∫sin⁡v dv=∫−dyy.\int \sin v\, dv=\int -\frac{dy}{y}.∫sinvdv=∫−ydy​.

So,

−cos⁡v=−ln⁡y+C.-\cos v=-\ln y + C.−cosv=−lny+C.

Therefore,

cos⁡v=ln⁡y+C1.\cos v=\ln y + C_1.cosv=lny+C1​.

Since v=xyv=\frac{x}{y}v=yx​,

cos⁡(xy)=ln⁡y+C1.\cos\left(\frac{x}{y}\right)=\ln y + C_1.cos(yx​)=lny+C1​.
  1. Apply the initial condition

Given x(1)=π2x(1)=\frac{\pi}{2}x(1)=2π​, when y=1y=1y=1,

v=xy=π/21=π2.v=\frac{x}{y}=\frac{\pi/2}{1}=\frac{\pi}{2}.v=yx​=1π/2​=2π​.

Thus,

cos⁡(π2)=ln⁡1+C1.\cos\left(\frac{\pi}{2}\right)=\ln 1 + C_1.cos(2π​)=ln1+C1​.

That is,

0=0+C1⇒C1=0.0=0+C_1 \Rightarrow C_1=0.0=0+C1​⇒C1​=0.

Hence,

cos⁡(xy)=ln⁡y.\cos\left(\frac{x}{y}\right)=\ln y.cos(yx​)=lny.
  1. Find x(2)x(2)x(2)

At y=2y=2y=2,

cos⁡(x(2)2)=ln⁡2.\cos\left(\frac{x(2)}{2}\right)=\ln 2.cos(2x(2)​)=ln2.

Let

θ=x(2)2.\theta=\frac{x(2)}{2}.θ=2x(2)​.

Then

cos⁡θ=ln⁡2.\cos\theta=\ln 2.cosθ=ln2.

We need

cos⁡(x(2))=cos⁡(2θ).\cos(x(2))=\cos(2\theta).cos(x(2))=cos(2θ).

Using

cos⁡(2θ)=2cos⁡2θ−1,\cos(2\theta)=2\cos^2\theta-1,cos(2θ)=2cos2θ−1,

we get

cos⁡(x(2))=2(ln⁡2)2−1.\cos(x(2))=2(\ln 2)^2-1.cos(x(2))=2(ln2)2−1.
  1. Match with options

This is

2(log⁡e2)2−1,2(\log_e 2)^2-1,2(loge​2)2−1,

which is Option D.

  1. Comparison with stored answer

Stored correct answer: D
Derived answer: D

So the derived answer agrees with the stored answer.

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