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Differential Equations question

2024 · 29 Jan · Shift 2 · Q42
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  5. /2024 · 29 Jan · Shift 2 · Q42

Differential Equations question

2024 · 29 Jan · Shift 2 · Q42

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
If sin⁡(yx)=log⁡e∣x∣+α2\sin \left(\frac{y}{x}\right)=\log _e|x|+\frac{\alpha}{2}sin(xy​)=loge​∣x∣+2α​ is the solution of the differential equation xcos⁡(yx)dydx=ycos⁡(yx)+xx \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+xxcos(xy​)dxdy​=ycos(xy​)+x and y(1)=π3y(1)=\frac{\pi}{3}y(1)=3π​, then α2\alpha^2α2 is equal to
  1. A
    12
  2. B
    9
  3. C
    4
  4. D
    3
View written solutionFree

Correct answer: D

  1. Given solution form

We are told that the solution of the differential equation is

sin⁡(yx)=log⁡e∣x∣+α2.\sin\left(\frac{y}{x}\right)=\log_e|x|+\frac{\alpha}{2}.sin(xy​)=loge​∣x∣+2α​.

We must use the condition

y(1)=π3y(1)=\frac{\pi}{3}y(1)=3π​

to find α\alphaα, and then compute α2\alpha^2α2.

  1. Apply the initial condition

At x=1x=1x=1, we have

y(1)=π3.y(1)=\frac{\pi}{3}.y(1)=3π​.

So,

yx∣x=1=π/31=π3.\frac{y}{x}\Big|_{x=1}=\frac{\pi/3}{1}=\frac{\pi}{3}.xy​​x=1​=1π/3​=3π​.

Substitute x=1x=1x=1 into the given solution:

sin⁡(yx)=log⁡e∣1∣+α2.\sin\left(\frac{y}{x}\right)=\log_e|1|+\frac{\alpha}{2}.sin(xy​)=loge​∣1∣+2α​.

Now,

sin⁡(π3)=log⁡e1+α2.\sin\left(\frac{\pi}{3}\right)=\log_e 1+\frac{\alpha}{2}.sin(3π​)=loge​1+2α​.

Since

sin⁡(π3)=32,log⁡e1=0,\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}, \qquad \log_e 1=0,sin(3π​)=23​​,loge​1=0,

we get

32=α2.\frac{\sqrt{3}}{2}=\frac{\alpha}{2}.23​​=2α​.

Hence,

α=3.\alpha=\sqrt{3}.α=3​.

Therefore,

α2=3.\alpha^2=3.α2=3.
  1. Match with the options

The value is

α2=3.\alpha^2=3.α2=3.

So the correct option is:

D: 3

  1. Comparison with stored correct answer

Stored correct answer is D, which matches our result.

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