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Differential Equations question

2022 · 25 Jul · Shift 2 · Q41
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  5. /2022 · 25 Jul · Shift 2 · Q41

Differential Equations question

2022 · 25 Jul · Shift 2 · Q41

JEE MainMathematicsDifferential EquationsNumerical+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation dydx=4y3+2yx23xy2+x3,y(1)=1\frac{d y}{d x}=\frac{4 y^{3}+2 y x^{2}}{3 x y^{2}+x^{3}}, y(1)=1dxdy​=3xy2+x34y3+2yx2​,y(1)=1. If for some n∈N,y(2)∈[n−1,n)n \in \mathbb{N}, y(2) \in[n-1, n)n∈N,y(2)∈[n−1,n), then nnn is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 3

  1. Identify the type of differential equation

Given

dydx=4y3+2yx23xy2+x3,y(1)=1.\frac{dy}{dx}=\frac{4y^3+2yx^2}{3xy^2+x^3}, \qquad y(1)=1.dxdy​=3xy2+x34y3+2yx2​,y(1)=1.

Factor numerator and denominator:

dydx=2y(2y2+x2)x(3y2+x2).\frac{dy}{dx}=\frac{2y(2y^2+x^2)}{x(3y^2+x^2)}.dxdy​=x(3y2+x2)2y(2y2+x2)​.

This is a homogeneous differential equation because the RHS is a function of y/xy/xy/x.

  1. Substitute y=vxy=vxy=vx

Let

y=vx⇒dydx=v+xdvdx.y=vx \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}.y=vx⇒dxdy​=v+xdxdv​.

Substitute into the differential equation:

v+xdvdx=2(vx)(2v2x2+x2)x(3v2x2+x2).v+x\frac{dv}{dx} = \frac{2(vx)\big(2v^2x^2+x^2\big)}{x\big(3v^2x^2+x^2\big)}.v+xdxdv​=x(3v2x2+x2)2(vx)(2v2x2+x2)​.

Simplify:

v+xdvdx=2v(2v2+1)3v2+1.v+x\frac{dv}{dx} = \frac{2v(2v^2+1)}{3v^2+1}.v+xdxdv​=3v2+12v(2v2+1)​.

Hence,

xdvdx=2v(2v2+1)3v2+1−v.x\frac{dv}{dx} = \frac{2v(2v^2+1)}{3v^2+1}-v.xdxdv​=3v2+12v(2v2+1)​−v.

Take vvv common:

xdvdx=v(2(2v2+1)3v2+1−1).x\frac{dv}{dx} = v\left(\frac{2(2v^2+1)}{3v^2+1}-1\right).xdxdv​=v(3v2+12(2v2+1)​−1).

Now simplify the bracket:

4v2+2−(3v2+1)3v2+1=v2+13v2+1.\frac{4v^2+2-(3v^2+1)}{3v^2+1}=\frac{v^2+1}{3v^2+1}.3v2+14v2+2−(3v2+1)​=3v2+1v2+1​.

So,

xdvdx=v(v2+1)3v2+1.x\frac{dv}{dx}=\frac{v(v^2+1)}{3v^2+1}.xdxdv​=3v2+1v(v2+1)​.
  1. Separate variables
3v2+1v(v2+1) dv=dxx.\frac{3v^2+1}{v(v^2+1)}\,dv = \frac{dx}{x}.v(v2+1)3v2+1​dv=xdx​.

Now decompose:

3v2+1v(v2+1)=Av+Bv+Cv2+1.\frac{3v^2+1}{v(v^2+1)} = \frac{A}{v}+\frac{Bv+C}{v^2+1}.v(v2+1)3v2+1​=vA​+v2+1Bv+C​.

So,

3v2+1=A(v2+1)+v(Bv+C)=(A+B)v2+Cv+A.3v^2+1=A(v^2+1)+v(Bv+C)=(A+B)v^2+Cv+A.3v2+1=A(v2+1)+v(Bv+C)=(A+B)v2+Cv+A.

Comparing coefficients:

  • constant term: A=1A=1A=1
  • vvv term: C=0C=0C=0
  • v2v^2v2 term: A+B=3⇒B=2A+B=3 \Rightarrow B=2A+B=3⇒B=2

Thus,

3v2+1v(v2+1)=1v+2vv2+1.\frac{3v^2+1}{v(v^2+1)}=\frac{1}{v}+\frac{2v}{v^2+1}.v(v2+1)3v2+1​=v1​+v2+12v​.

Therefore,

∫(1v+2vv2+1)dv=∫dxx.\int \left(\frac{1}{v}+\frac{2v}{v^2+1}\right)dv = \int \frac{dx}{x}.∫(v1​+v2+12v​)dv=∫xdx​.

Integrate:

ln⁡∣v∣+ln⁡(v2+1)=ln⁡∣x∣+C.\ln|v|+\ln(v^2+1)=\ln|x|+C.ln∣v∣+ln(v2+1)=ln∣x∣+C.

So,

ln⁡∣v(v2+1)∣=ln⁡∣x∣+C.\ln|v(v^2+1)|=\ln|x|+C.ln∣v(v2+1)∣=ln∣x∣+C.

Hence,

v(v2+1)=Cx.v(v^2+1)=Cx.v(v2+1)=Cx.
  1. Return to yyy and use initial condition

Since

v=yx,v=\frac{y}{x},v=xy​,

we get

yx(y2x2+1)=Cx.\frac{y}{x}\left(\frac{y^2}{x^2}+1\right)=Cx.xy​(x2y2​+1)=Cx.

Multiply through:

y(y2+x2)x3=Cx.\frac{y(y^2+x^2)}{x^3}=Cx.x3y(y2+x2)​=Cx.

So,

y(y2+x2)=Cx4.y(y^2+x^2)=Cx^4.y(y2+x2)=Cx4.

Use y(1)=1y(1)=1y(1)=1:

1(12+12)=C⋅14⇒2=C.1(1^2+1^2)=C\cdot 1^4 \Rightarrow 2=C.1(12+12)=C⋅14⇒2=C.

Therefore the solution satisfies

y(y2+x2)=2x4.y(y^2+x^2)=2x^4.y(y2+x2)=2x4.
  1. Find y(2)y(2)y(2)

At x=2x=2x=2,

y(y2+4)=2⋅24=32.y(y^2+4)=2\cdot 2^4=32.y(y2+4)=2⋅24=32.

So,

y3+4y−32=0.y^3+4y-32=0.y3+4y−32=0.

Check integer root y=2y=2y=2:

23+4(2)−32=8+8−32=−16≠0.2^3+4(2)-32=8+8-32=-16 \neq 0.23+4(2)−32=8+8−32=−16=0.

Check y=3y=3y=3:

33+4(3)−32=27+12−32=7>0.3^3+4(3)-32=27+12-32=7>0.33+4(3)−32=27+12−32=7>0.

Since the cubic is continuous, the positive root lies in

(2,3).(2,3).(2,3).

Thus,

y(2)∈[2,3).y(2)\in [2,3).y(2)∈[2,3).

Given y(2)∈[n−1,n)y(2)\in [n-1,n)y(2)∈[n−1,n), we get

n=3.n=3.n=3.
  1. Comparison with stored answer

Derived answer: 333.

Stored correct answer: 333.

They agree.

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