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Differential Equations question

2024 · 1 Feb · Shift 1 · Q51
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Differential Equations question

2024 · 1 Feb · Shift 1 · Q51

JEE MainMathematicsDifferential EquationsNumerical+4 / −1
If x=x(t)x=x(t)x=x(t) is the solution of the differential equation (t+1)dx=(2x+(t+1)4)dt,x(0)=2(t+1) \mathrm{d} x=\left(2 x+(t+1)^4\right) \mathrm{dt}, x(0)=2(t+1)dx=(2x+(t+1)4)dt,x(0)=2, then, x(1)x(1)x(1) equals ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 14

  1. Rewrite the differential equation

Given

(t+1) dx=(2x+(t+1)4)dt(t+1)\,dx = \left(2x + (t+1)^4\right)dt(t+1)dx=(2x+(t+1)4)dt

with initial condition x(0)=2.x(0)=2.x(0)=2.

Divide both sides by dtdtdt and then by (t+1)(t+1)(t+1):

dxdt=2x+(t+1)4t+1=2xt+1+(t+1)3.\frac{dx}{dt} = \frac{2x+(t+1)^4}{t+1} = \frac{2x}{t+1} + (t+1)^3.dtdx​=t+12x+(t+1)4​=t+12x​+(t+1)3.

So the equation is

dxdt−2t+1x=(t+1)3.\frac{dx}{dt} - \frac{2}{t+1}x = (t+1)^3.dtdx​−t+12​x=(t+1)3.

This is a linear differential equation of the form

dxdt+P(t)x=Q(t),\frac{dx}{dt} + P(t)x = Q(t),dtdx​+P(t)x=Q(t),

where

P(t)=−2t+1,Q(t)=(t+1)3.P(t)=-\frac{2}{t+1}, \qquad Q(t)=(t+1)^3.P(t)=−t+12​,Q(t)=(t+1)3.
  1. Find the integrating factor

The integrating factor is

I.F.=e∫P(t) dt=e∫−2t+1 dt=e−2ln⁡(t+1)=(t+1)−2.I.F. = e^{\int P(t)\,dt} = e^{\int -\frac{2}{t+1}\,dt} = e^{-2\ln(t+1)} = (t+1)^{-2}.I.F.=e∫P(t)dt=e∫−t+12​dt=e−2ln(t+1)=(t+1)−2.
  1. Multiply the equation by the integrating factor

Multiplying throughout by (t+1)−2(t+1)^{-2}(t+1)−2:

(t+1)−2dxdt−2t+1(t+1)−2x=(t+1)3(t+1)−2.(t+1)^{-2}\frac{dx}{dt} - \frac{2}{t+1}(t+1)^{-2}x = (t+1)^3(t+1)^{-2}.(t+1)−2dtdx​−t+12​(t+1)−2x=(t+1)3(t+1)−2.

That gives

(t+1)−2dxdt−2(t+1)−3x=t+1.(t+1)^{-2}\frac{dx}{dt} - 2(t+1)^{-3}x = t+1.(t+1)−2dtdx​−2(t+1)−3x=t+1.

The left-hand side is

ddt(x(t+1)−2).\frac{d}{dt}\left(x(t+1)^{-2}\right).dtd​(x(t+1)−2).

So,

ddt(x(t+1)2)=t+1.\frac{d}{dt}\left(\frac{x}{(t+1)^2}\right)=t+1.dtd​((t+1)2x​)=t+1.
  1. Integrate both sides

Integrating,

x(t+1)2=∫(t+1) dt=(t+1)22+C.\frac{x}{(t+1)^2} = \int (t+1)\,dt = \frac{(t+1)^2}{2} + C.(t+1)2x​=∫(t+1)dt=2(t+1)2​+C.

Hence,

x=(t+1)2((t+1)22+C)=(t+1)42+C(t+1)2.x = (t+1)^2\left(\frac{(t+1)^2}{2}+C\right) = \frac{(t+1)^4}{2} + C(t+1)^2.x=(t+1)2(2(t+1)2​+C)=2(t+1)4​+C(t+1)2.
  1. Use the initial condition

Given x(0)=2x(0)=2x(0)=2:

2=(1)42+C(1)2=12+C.2 = \frac{(1)^4}{2} + C(1)^2 = \frac12 + C.2=2(1)4​+C(1)2=21​+C.

Therefore,

C=2−12=32.C = 2-\frac12 = \frac32.C=2−21​=23​.

So the solution is

x(t)=(t+1)42+32(t+1)2.x(t)=\frac{(t+1)^4}{2}+\frac32(t+1)^2.x(t)=2(t+1)4​+23​(t+1)2.
  1. Find x(1)x(1)x(1)

Substitute t=1t=1t=1:

x(1)=242+32⋅22=162+32⋅4=8+6=14.x(1)=\frac{2^4}{2}+\frac32\cdot 2^2 = \frac{16}{2} + \frac32\cdot 4 = 8+6=14.x(1)=224​+23​⋅22=216​+23​⋅4=8+6=14.

Therefore, 14\boxed{14}14​

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