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Differential Equations question

2023 · 25 Jan · Shift 1 · Q37
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  5. /2023 · 25 Jan · Shift 1 · Q37

Differential Equations question

2023 · 25 Jan · Shift 1 · Q37

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y=y(x)y = y(x)y=y(x) be the solution curve of the differential equation dydx=yx(1+xy2(1+log⁡ex)),x>0,y(1)=3{{dy} \over {dx}} = {y \over x}\left( {1 + x{y^2}(1 + {{\log }_e}x)} \right),x \gt 0,y(1) = 3dxdy​=xy​(1+xy2(1+loge​x)),x>0,y(1)=3. Then y2(x)9{{{y^2}(x)} \over 9}9y2(x)​ is equal to :
  1. A
    x25−2x3(2+log⁡ex3){{{x^2}} \over {5 - 2{x^3}(2 + {{\log }_e}{x^3})}}5−2x3(2+loge​x3)x2​
  2. B
    x23x3(1+log⁡ex2)−2{{{x^2}} \over {3{x^3}(1 + {{\log }_e}{x^2}) - 2}}3x3(1+loge​x2)−2x2​
  3. C
    x27−3x3(2+log⁡ex2){{{x^2}} \over {7 - 3{x^3}(2 + {{\log }_e}{x^2})}}7−3x3(2+loge​x2)x2​
  4. D
    x22x3(2+log⁡ex3)−3{{{x^2}} \over {2{x^3}(2 + {{\log }_e}{x^3}) - 3}}2x3(2+loge​x3)−3x2​
View written solutionFree

Correct answer: A

  1. Given differential equation

We have

dydx=yx(1+xy2(1+ln⁡x)),x>0,\frac{dy}{dx}=\frac{y}{x}\left(1+x y^2(1+\ln x)\right), \quad x>0,dxdy​=xy​(1+xy2(1+lnx)),x>0,

with initial condition

y(1)=3.y(1)=3.y(1)=3.

We need to find y2(x)9\dfrac{y^2(x)}{9}9y2(x)​.

  1. Rewrite the equation

Expand the right-hand side:

dydx=yx+y3(1+ln⁡x).\frac{dy}{dx}=\frac{y}{x}+y^3(1+\ln x).dxdy​=xy​+y3(1+lnx).

This suggests converting in terms of y−2y^{-2}y−2.

  1. Substitute v=1y2v=\dfrac{1}{y^2}v=y21​

Let

v=y−2.v=y^{-2}.v=y−2.

Then

dvdx=−2y−3dydx.\frac{dv}{dx}=-2y^{-3}\frac{dy}{dx}.dxdv​=−2y−3dxdy​.

Using

dydx=yx+y3(1+ln⁡x),\frac{dy}{dx}=\frac{y}{x}+y^3(1+\ln x),dxdy​=xy​+y3(1+lnx),

we get

dvdx=−2y−3(yx+y3(1+ln⁡x)).\frac{dv}{dx}=-2y^{-3}\left(\frac{y}{x}+y^3(1+\ln x)\right).dxdv​=−2y−3(xy​+y3(1+lnx)).

Simplify:

dvdx=−2xy−2−2(1+ln⁡x).\frac{dv}{dx}=-\frac{2}{x}y^{-2}-2(1+\ln x).dxdv​=−x2​y−2−2(1+lnx).

Since v=y−2v=y^{-2}v=y−2,

dvdx+2xv=−2(1+ln⁡x).\frac{dv}{dx}+\frac{2}{x}v=-2(1+\ln x).dxdv​+x2​v=−2(1+lnx).

This is a linear differential equation in vvv.

  1. Solve the linear equation

The integrating factor is

I.F.=e∫2xdx=e2ln⁡x=x2.\text{I.F.}=e^{\int \frac{2}{x}dx}=e^{2\ln x}=x^2.I.F.=e∫x2​dx=e2lnx=x2.

Multiplying throughout by x2x^2x2:

x2dvdx+2xv=−2x2(1+ln⁡x).x^2\frac{dv}{dx}+2xv=-2x^2(1+\ln x).x2dxdv​+2xv=−2x2(1+lnx).

So,

ddx(x2v)=−2x2(1+ln⁡x).\frac{d}{dx}(x^2v)=-2x^2(1+\ln x).dxd​(x2v)=−2x2(1+lnx).

Integrate:

x2v=∫−2x2(1+ln⁡x) dx+C.x^2v=\int -2x^2(1+\ln x)\,dx + C.x2v=∫−2x2(1+lnx)dx+C.

Now compute

I=∫x2(1+ln⁡x) dx.I=\int x^2(1+\ln x)\,dx.I=∫x2(1+lnx)dx.

Write it as

I=∫x2dx+∫x2ln⁡x dx.I=\int x^2dx+\int x^2\ln x\,dx.I=∫x2dx+∫x2lnxdx.

We know

∫x2dx=x33.\int x^2dx=\frac{x^3}{3}.∫x2dx=3x3​.

For ∫x2ln⁡x dx\int x^2\ln x\,dx∫x2lnxdx, use integration by parts:

  • u=ln⁡xu=\ln xu=lnx, dv=x2dxdv=x^2dxdv=x2dx
  • du=1xdxdu=\frac{1}{x}dxdu=x1​dx, v=x33v=\frac{x^3}{3}v=3x3​

Thus,

∫x2ln⁡x dx=x33ln⁡x−13∫x2dx=x33ln⁡x−x39.\int x^2\ln x\,dx=\frac{x^3}{3}\ln x-\frac{1}{3}\int x^2dx =\frac{x^3}{3}\ln x-\frac{x^3}{9}.∫x2lnxdx=3x3​lnx−31​∫x2dx=3x3​lnx−9x3​.

Hence,

I=x33+x33ln⁡x−x39=2x39+x33ln⁡x.I=\frac{x^3}{3}+\frac{x^3}{3}\ln x-\frac{x^3}{9} =\frac{2x^3}{9}+\frac{x^3}{3}\ln x.I=3x3​+3x3​lnx−9x3​=92x3​+3x3​lnx.

Therefore,

∫−2x2(1+ln⁡x)dx=−2I=−4x39−2x33ln⁡x.\int -2x^2(1+\ln x)dx =-2I =-\frac{4x^3}{9}-\frac{2x^3}{3}\ln x.∫−2x2(1+lnx)dx=−2I=−94x3​−32x3​lnx.

So,

x2v=−4x39−2x33ln⁡x+C.x^2v=-\frac{4x^3}{9}-\frac{2x^3}{3}\ln x+C.x2v=−94x3​−32x3​lnx+C.

Since v=1y2v=\dfrac{1}{y^2}v=y21​,

x2y2=C−4x39−2x33ln⁡x.\frac{x^2}{y^2}=C-\frac{4x^3}{9}-\frac{2x^3}{3}\ln x.y2x2​=C−94x3​−32x3​lnx.
  1. Use the initial condition y(1)=3y(1)=3y(1)=3

At x=1x=1x=1, y=3y=3y=3, so

1232=19=C−49−23(1)ln⁡1.\frac{1^2}{3^2}=\frac{1}{9}=C-\frac{4}{9}-\frac{2}{3}(1)\ln 1.3212​=91​=C−94​−32​(1)ln1.

Since ln⁡1=0\ln 1=0ln1=0,

19=C−49.\frac{1}{9}=C-\frac{4}{9}.91​=C−94​.

Thus,

C=59.C=\frac{5}{9}.C=95​.

Hence,

x2y2=59−4x39−2x33ln⁡x=19(5−4x3−6x3ln⁡x).\frac{x^2}{y^2}=\frac{5}{9}-\frac{4x^3}{9}-\frac{2x^3}{3}\ln x =\frac{1}{9}\left(5-4x^3-6x^3\ln x\right).y2x2​=95​−94x3​−32x3​lnx=91​(5−4x3−6x3lnx).

So,

y2=9x25−4x3−6x3ln⁡x.y^2=\frac{9x^2}{5-4x^3-6x^3\ln x}.y2=5−4x3−6x3lnx9x2​.

Therefore,

y29=x25−4x3−6x3ln⁡x.\frac{y^2}{9}=\frac{x^2}{5-4x^3-6x^3\ln x}.9y2​=5−4x3−6x3lnxx2​.

Now simplify the denominator using log identities:

4x3+6x3ln⁡x=2x3(2+3ln⁡x).4x^3+6x^3\ln x=2x^3(2+3\ln x).4x3+6x3lnx=2x3(2+3lnx).

And since

ln⁡(x3)=3ln⁡x,\ln(x^3)=3\ln x,ln(x3)=3lnx,

we get

2+3ln⁡x=2+ln⁡(x3).2+3\ln x=2+\ln(x^3).2+3lnx=2+ln(x3).

So,

y29=x25−2x3(2+ln⁡(x3)).\frac{y^2}{9}=\frac{x^2}{5-2x^3(2+\ln(x^3))}.9y2​=5−2x3(2+ln(x3))x2​.
  1. Compare with options

This matches:

A: x25−2x3(2+ln⁡(x3))\boxed{\text{A: } \frac{x^2}{5-2x^3(2+\ln(x^3))}}A: 5−2x3(2+ln(x3))x2​​
  1. Comparison with stored answer

Stored correct answer: A

Our derived answer is also A. So they agree.

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