A removable extension of the differential equation

(i) Find the general solution of the differential equation (x2+1x)dydx=2\left(\frac{x^2+1}{x}\right)\frac{dy}{dx}=2. marks

(ii) Sketch the particular solution of the differential equation in (i) for which y=0y=0 and x=0x=0. marks

(iii) Describe the gradient of every solution curve as x→+∞x\to+\infty. mark

[2010/MI/Prelim/I/11(b) (modified)]

Total marks: 5

Solution

(i)

Direct integration

Find the general solution of (x2+1x)dydx=2\left(\dfrac{x^2+1}x\right)\dfrac{dy}{dx}=2.

Hint

  1. Make dydx\frac{dy}{dx} the subject.

  2. Integrate with respect to xx.

Answer · (i)

y=ln⁡(1+x2)+Cy=\ln(1+x^2)+C

Original DE requires x≠0x\ne0 on each connected interval.

(ii)

Particular solution

Sketch the particular solution for which y=0y=0 when x=0x=0.

Explore this idea

Hint

  1. Use the point to find CC.

  2. Sketch y=ln⁡(x2+1)y=\ln(x^2+1).

Particular solution y = ln(x² + 1)
Particular solution y = ln(x² + 1)Scroll across to see the full diagram.

Answer · (ii)

y=ln⁡(1+x2)y=\ln(1+x^2): even, minimum (0,0), inflections at ±1\pm1.

Uses the intended smooth extension through 0.

Explore the idea

The family y = ln(x² + 1) + C

Each dash shows the slope the differential equation gives at that point. Change the starting value and the solution follows the dashes.

Starting value of y at x = 00
Read the solution at x2
Try an example
y at x = 2
1.609
Rate dy/dx there
0.8
  • Solution that starts at the purple dot
  • Closed-form answer
Key idea

All solutions are copies of ln⁡(1+x2)\ln(1+x^2) shifted vertically. The particular one through the origin has constant C=0C=0.

Ideas to try
  1. Move the starting value up and down: the curves are vertical shifts of each other.
  2. Read the gradient at x=1x=1 and at x=4x=4.
(iii)

Long-term gradient

Describe the gradient of every solution curve as x→+∞x\to+\infty.

Hint

  1. Every solution has the same gradient function 2xx2+1\frac{2x}{x^2+1}.

Answer · (iii)

y′→0y\prime\to0 as x→+∞x\to+\infty.