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Differential Equations question

2025 · 4 Apr · Shift 2 · Q32
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  5. /2025 · 4 Apr · Shift 2 · Q32

Differential Equations question

2025 · 4 Apr · Shift 2 · Q32

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
If a curve y=y(x)y=y(x)y=y(x) passes through the point (1,π2)\left(1, \frac{\pi}{2}\right)(1,2π​) and satisfies the differential equation (7x4cot⁡y−excosec⁡y)dx dy=x5,x≥1\left(7 x^4 \cot y-\mathrm{e}^x \operatorname{cosec} y\right) \frac{\mathrm{d} x}{\mathrm{~d} y}=x^5, x \geq 1(7x4coty−excosecy) dydx​=x5,x≥1, then at x=2x=2x=2, the value of cos⁡y\cos ycosy is :
  1. A
    2e2+e64\frac{2 \mathrm{e}^2+\mathrm{e}}{64}642e2+e​
  2. B
    2e2−e64\frac{2 \mathrm{e}^2-\mathrm{e}}{64}642e2−e​
  3. C
    2e2−e128\frac{2 \mathrm{e}^2-\mathrm{e}}{128}1282e2−e​
  4. D
    2e2+e128\frac{2 \mathrm{e}^2+\mathrm{e}}{128}1282e2+e​
View written solutionFree

Correct answer: C

  1. Given differential equation

We have

(7x4cot⁡y−excsc⁡y)dxdy=x5,\left(7x^4\cot y-e^x\csc y\right)\frac{dx}{dy}=x^5,(7x4coty−excscy)dydx​=x5,

with the point (1,π2)\left(1,\frac{\pi}{2}\right)(1,2π​) on the curve.

We need cos⁡y\cos ycosy at x=2x=2x=2.

  1. Rewrite the equation in a more usable form

Since

dxdy=1dy/dx,\frac{dx}{dy} = \frac{1}{dy/dx},dydx​=dy/dx1​,

it is easier to write the equation in differential form directly.

Multiply both sides by dydydy:

(7x4cot⁡y−excsc⁡y)dx=x5dy.\left(7x^4\cot y-e^x\csc y\right)dx=x^5dy.(7x4coty−excscy)dx=x5dy.

Now divide by dxdxdx:

x5dydx=7x4cot⁡y−excsc⁡y.x^5\frac{dy}{dx}=7x^4\cot y-e^x\csc y.x5dxdy​=7x4coty−excscy.

So

dydx=7cot⁡yx−excsc⁡yx5.\frac{dy}{dx}=\frac{7\cot y}{x}-\frac{e^x\csc y}{x^5}.dxdy​=x7coty​−x5excscy​.
  1. Convert into an equation involving cos⁡y\cos ycosy

Multiply both sides by sin⁡y\sin ysiny:

sin⁡y dydx=7cos⁡yx−exx5.\sin y\,\frac{dy}{dx}=\frac{7\cos y}{x}-\frac{e^x}{x^5}.sinydxdy​=x7cosy​−x5ex​.

Now use

ddx(cos⁡y)=−sin⁡y dydx.\frac{d}{dx}(\cos y)=-\sin y\,\frac{dy}{dx}.dxd​(cosy)=−sinydxdy​.

Hence,

−ddx(cos⁡y)=7cos⁡yx−exx5.-\frac{d}{dx}(\cos y)=\frac{7\cos y}{x}-\frac{e^x}{x^5}.−dxd​(cosy)=x7cosy​−x5ex​.

So,

ddx(cos⁡y)+7xcos⁡y=exx5.\frac{d}{dx}(\cos y)+\frac{7}{x}\cos y=\frac{e^x}{x^5}.dxd​(cosy)+x7​cosy=x5ex​.

Let

z=cos⁡y.z=\cos y.z=cosy.

Then we get the linear differential equation

dzdx+7xz=exx5.\frac{dz}{dx}+\frac{7}{x}z=\frac{e^x}{x^5}.dxdz​+x7​z=x5ex​.
  1. Solve the linear differential equation

The integrating factor is

I.F.=e∫7xdx=e7ln⁡x=x7.\text{I.F.}=e^{\int \frac{7}{x}dx}=e^{7\ln x}=x^7.I.F.=e∫x7​dx=e7lnx=x7.

Multiplying the equation by x7x^7x7:

x7dzdx+7x6z=x2ex.x^7\frac{dz}{dx}+7x^6z=x^2e^x.x7dxdz​+7x6z=x2ex.

Thus,

ddx(x7z)=x2ex.\frac{d}{dx}(x^7 z)=x^2e^x.dxd​(x7z)=x2ex.

Integrate:

x7z=∫x2ex dx+C.x^7 z=\int x^2e^x\,dx + C.x7z=∫x2exdx+C.

Now,

∫x2ex dx=ex(x2−2x+2).\int x^2e^x\,dx=e^x(x^2-2x+2).∫x2exdx=ex(x2−2x+2).

Therefore,

x7z=ex(x2−2x+2)+C.x^7 z=e^x(x^2-2x+2)+C.x7z=ex(x2−2x+2)+C.

So,

z=cos⁡y=ex(x2−2x+2)+Cx7.z=\cos y=\frac{e^x(x^2-2x+2)+C}{x^7}.z=cosy=x7ex(x2−2x+2)+C​.
  1. Use the initial condition

At (1,π2)\left(1,\frac{\pi}{2}\right)(1,2π​),

cos⁡(π2)=0.\cos\left(\frac{\pi}{2}\right)=0.cos(2π​)=0.

So at x=1x=1x=1,

0=e1(1−2+2)+C17=e+C.0=\frac{e^1(1-2+2)+C}{1^7}=e+C.0=17e1(1−2+2)+C​=e+C.

Hence,

C=−e.C=-e.C=−e.

Thus,

cos⁡y=ex(x2−2x+2)−ex7.\cos y=\frac{e^x(x^2-2x+2)-e}{x^7}.cosy=x7ex(x2−2x+2)−e​.
  1. Find cos⁡y\cos ycosy at x=2x=2x=2

Substitute x=2x=2x=2:

cos⁡y=e2(4−4+2)−e27=2e2−e128.\cos y=\frac{e^2(4-4+2)-e}{2^7} =\frac{2e^2-e}{128}.cosy=27e2(4−4+2)−e​=1282e2−e​.
  1. Match with the options
cos⁡y=2e2−e128\boxed{\cos y=\frac{2e^2-e}{128}}cosy=1282e2−e​​

This is Option C.

  1. Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

They match.

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