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Differential Equations question

2023 · 24 Jan · Shift 2 · Q30
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  5. /2023 · 24 Jan · Shift 2 · Q30

Differential Equations question

2023 · 24 Jan · Shift 2 · Q30

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (x2−3y2)dx+3xy dy=0,y(1)=1(x^2-3y^2)dx+3xy~dy=0,y(1)=1(x2−3y2)dx+3xy dy=0,y(1)=1. Then 6y2(e)6y^2(e)6y2(e) is equal to
  1. A
    32e2\frac{3}{2}\mathrm{e}^223​e2
  2. B
    3e23\mathrm{e}^23e2
  3. C
    e2\mathrm{e}^2e2
  4. D
    2e22\mathrm{e}^22e2
View written solutionFree

Correct answer: D

  1. Given differential equation
(x2−3y2) dx+3xy dy=0(x^2-3y^2)\,dx+3xy\,dy=0(x2−3y2)dx+3xydy=0

with initial condition

y(1)=1.y(1)=1.y(1)=1.

We need to find 6y2(e)6y^2(e)6y2(e).

  1. Rewrite in differential form

Divide by dxdxdx:

(x2−3y2)+3xydydx=0(x^2-3y^2)+3xy\frac{dy}{dx}=0(x2−3y2)+3xydxdy​=0

so

3xydydx=3y2−x2.3xy\frac{dy}{dx}=3y^2-x^2.3xydxdy​=3y2−x2.

Thus,

dydx=3y2−x23xy.\frac{dy}{dx}=\frac{3y^2-x^2}{3xy}.dxdy​=3xy3y2−x2​.
  1. Substitute v=y2v=y^2v=y2

Since y dy/dxy\,dy/dxydy/dx appears naturally, let

v=y2.v=y^2.v=y2.

Then

dvdx=2ydydx.\frac{dv}{dx}=2y\frac{dy}{dx}.dxdv​=2ydxdy​.

From

3xydydx=3y2−x2,3xy\frac{dy}{dx}=3y^2-x^2,3xydxdy​=3y2−x2,

multiply by 222:

6xydydx=2(3y2−x2).6xy\frac{dy}{dx}=2(3y^2-x^2).6xydxdy​=2(3y2−x2).

Using 2y dy/dx=dv/dx2y\,dy/dx=dv/dx2ydy/dx=dv/dx,

3xdvdx=6v−2x2.3x\frac{dv}{dx}=6v-2x^2.3xdxdv​=6v−2x2.

So,

dvdx−2xv=−23x.\frac{dv}{dx}-\frac{2}{x}v=-\frac{2}{3}x.dxdv​−x2​v=−32​x.

This is a linear differential equation in vvv.

  1. Find integrating factor
I.F.=e∫−2xdx=e−2ln⁡x=x−2.\text{I.F.}=e^{\int -\frac{2}{x}dx}=e^{-2\ln x}=x^{-2}.I.F.=e∫−x2​dx=e−2lnx=x−2.

Multiply the equation by x−2x^{-2}x−2:

x−2dvdx−2x3v=−23x.x^{-2}\frac{dv}{dx}-\frac{2}{x^3}v=-\frac{2}{3x}.x−2dxdv​−x32​v=−3x2​.

Left side becomes:

ddx(vx−2)=−23x.\frac{d}{dx}(vx^{-2})=-\frac{2}{3x}.dxd​(vx−2)=−3x2​.

Integrate:

vx−2=∫−23x dx=−23ln⁡x+C.vx^{-2}=\int -\frac{2}{3x}\,dx=-\frac{2}{3}\ln x + C.vx−2=∫−3x2​dx=−32​lnx+C.

Hence,

v=x2(C−23ln⁡x).v=x^2\left(C-\frac{2}{3}\ln x\right).v=x2(C−32​lnx).

Since v=y2v=y^2v=y2,

y2=x2(C−23ln⁡x).y^2=x^2\left(C-\frac{2}{3}\ln x\right).y2=x2(C−32​lnx).
  1. Use initial condition

Given y(1)=1y(1)=1y(1)=1:

12=12(C−23ln⁡1).1^2=1^2\left(C-\frac{2}{3}\ln 1\right).12=12(C−32​ln1).

Since ln⁡1=0\ln 1=0ln1=0, we get

C=1.C=1.C=1.

Therefore,

y2=x2(1−23ln⁡x).y^2=x^2\left(1-\frac{2}{3}\ln x\right).y2=x2(1−32​lnx).
  1. Evaluate at x=ex=ex=e
y2(e)=e2(1−23ln⁡e)=e2(1−23)=e23.y^2(e)=e^2\left(1-\frac{2}{3}\ln e\right) =e^2\left(1-\frac{2}{3}\right) =\frac{e^2}{3}.y2(e)=e2(1−32​lne)=e2(1−32​)=3e2​.

Therefore,

6y2(e)=6⋅e23=2e2.6y^2(e)=6\cdot \frac{e^2}{3}=2e^2.6y2(e)=6⋅3e2​=2e2.
  1. Compare with options
2e22e^22e2

matches Option D.

  1. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

So they agree.

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