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Differential Equations question

2023 · 12 Apr · Shift 1 · Q25
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  5. /2023 · 12 Apr · Shift 1 · Q25

Differential Equations question

2023 · 12 Apr · Shift 1 · Q25

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y=y(x),y>0y=y(x), y \gt 0y=y(x),y>0, be a solution curve of the differential equation (1+x2)dy=y(x−y)dx\left(1+x^{2}\right) \mathrm{d} y=y(x-y) \mathrm{d} x(1+x2)dy=y(x−y)dx. If y(0)=1y(0)=1y(0)=1 and y(22)=βy(2 \sqrt{2})=\betay(22​)=β, then
  1. A
    eβ−1=e−2(3+22)e^{\beta^{-1}}=e^{-2}(3+2 \sqrt{2})eβ−1=e−2(3+22​)
  2. B
    e3β−1=e(5+2)e^{3 \beta^{-1}}=e(5+\sqrt{2})e3β−1=e(5+2​)
  3. C
    e3β−1=e(3+22)e^{3 \beta^{-1}}=e(3+2 \sqrt{2})e3β−1=e(3+22​)
  4. D
    eβ−1=e−2(5+2)e^{\beta^{-1}}=e^{-2}(5+\sqrt{2})eβ−1=e−2(5+2​)
View written solutionFree

Correct answer: C

  1. Rewrite the differential equation

Given

(1+x2) dy=y(x−y) dx.(1+x^2)\,dy = y(x-y)\,dx.(1+x2)dy=y(x−y)dx.

So,

dydx=y(x−y)1+x2.\frac{dy}{dx} = \frac{y(x-y)}{1+x^2}.dxdy​=1+x2y(x−y)​.

We are told y>0y>0y>0. To simplify, use the substitution

z=1y.z=\frac{1}{y}.z=y1​.

Then

y=1z,dydx=−1z2dzdx.y=\frac{1}{z}, \qquad \frac{dy}{dx}=-\frac{1}{z^2}\frac{dz}{dx}.y=z1​,dxdy​=−z21​dxdz​.
  1. Substitute into the differential equation

From

dydx=y(x−y)1+x2,\frac{dy}{dx}=\frac{y(x-y)}{1+x^2},dxdy​=1+x2y(x−y)​,

put y=1/zy=1/zy=1/z:

−1z2dzdx=1z(x−1z)1+x2.-\frac{1}{z^2}\frac{dz}{dx} = \frac{\frac{1}{z}\left(x-\frac{1}{z}\right)}{1+x^2}.−z21​dxdz​=1+x2z1​(x−z1​)​.

Multiply both sides by z2z^2z2:

−dzdx=xz−11+x2.-\frac{dz}{dx} = \frac{xz-1}{1+x^2}.−dxdz​=1+x2xz−1​.

Hence,

dzdx+x1+x2z=11+x2.\frac{dz}{dx} + \frac{x}{1+x^2}z = \frac{1}{1+x^2}.dxdz​+1+x2x​z=1+x21​.

This is a linear differential equation in zzz.

  1. Solve the linear equation

The integrating factor is

IF=e∫x1+x2 dx=e12ln⁡(1+x2)=1+x2.IF = e^{\int \frac{x}{1+x^2}\,dx} = e^{\frac12\ln(1+x^2)} = \sqrt{1+x^2}.IF=e∫1+x2x​dx=e21​ln(1+x2)=1+x2​.

Therefore,

ddx(z1+x2)=11+x2.\frac{d}{dx}\left(z\sqrt{1+x^2}\right)=\frac{1}{\sqrt{1+x^2}}.dxd​(z1+x2​)=1+x2​1​.

Integrate:

z1+x2=∫dx1+x2+C.z\sqrt{1+x^2} = \int \frac{dx}{\sqrt{1+x^2}} + C.z1+x2​=∫1+x2​dx​+C.

Now,

∫dx1+x2=ln⁡(x+1+x2).\int \frac{dx}{\sqrt{1+x^2}} = \ln\left(x+\sqrt{1+x^2}\right).∫1+x2​dx​=ln(x+1+x2​).

So,

z1+x2=ln⁡(x+1+x2)+C.z\sqrt{1+x^2} = \ln\left(x+\sqrt{1+x^2}\right)+C.z1+x2​=ln(x+1+x2​)+C.

Since z=1/yz=1/yz=1/y,

1+x2y=ln⁡(x+1+x2)+C.\frac{\sqrt{1+x^2}}{y} = \ln\left(x+\sqrt{1+x^2}\right)+C.y1+x2​​=ln(x+1+x2​)+C.
  1. Use the initial condition y(0)=1y(0)=1y(0)=1

At x=0x=0x=0,

1+01=ln⁡(0+1)+C.\frac{\sqrt{1+0}}{1} = \ln(0+1)+C.11+0​​=ln(0+1)+C.

Thus,

1=0+C  ⟹  C=1.1 = 0 + C \implies C=1.1=0+C⟹C=1.

Hence the solution is

1+x2y=1+ln⁡(x+1+x2).\frac{\sqrt{1+x^2}}{y} = 1+\ln\left(x+\sqrt{1+x^2}\right).y1+x2​​=1+ln(x+1+x2​).
  1. Use x=22x=2\sqrt{2}x=22​

Let

y(22)=β.y(2\sqrt2)=\beta.y(22​)=β.

Then

1+(22)2β=1+ln⁡(22+1+8).\frac{\sqrt{1+(2\sqrt2)^2}}{\beta} = 1+\ln\left(2\sqrt2+\sqrt{1+8}\right).β1+(22​)2​​=1+ln(22​+1+8​).

Compute:

9=3.\sqrt{9}=3.9​=3.

So,

3β=1+ln⁡(22+3)=1+ln⁡(3+22).\frac{3}{\beta}=1+\ln(2\sqrt2+3)=1+\ln(3+2\sqrt2).β3​=1+ln(22​+3)=1+ln(3+22​).

Therefore,

e3/β=e1+ln⁡(3+22)=e(3+22).e^{3/\beta}=e^{1+\ln(3+2\sqrt2)}=e(3+2\sqrt2).e3/β=e1+ln(3+22​)=e(3+22​).
  1. Match with the options

This is exactly:

e3β−1=e(3+22),e^{3\beta^{-1}}=e(3+2\sqrt2),e3β−1=e(3+22​),

which is Option C.

  1. Compare with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.

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