The product substitution absorbs a linear term

Use the substitution u=xyu=xy, where uu is a function of xx, to reduce the differential equation xdydx+(1−x3)y−x2=0x\frac{dy}{dx}+(1-x^3)y-x^2=0 to dudx=(u+1)x2\frac{du}{dx}=(u+1)x^2.

Hence show that the general solution to the differential equation is of the form y=Aef(x)−1xy=\frac{Ae^{f(x)}-1}{x}, where ff is a function in xx. marks

Sketch, on separate diagrams, the particular solution curve for (i) A=2A=2 and (ii) A=−3A=-3. marks

[2012/RVHS/Prelim/II/1 (modified)]

Total marks: 7

Solution

Substitution

Product substitution

Use u=xyu=xy to reduce xdydx+(1−x3)y−x2=0x\dfrac{dy}{dx}+(1-x^3)y-x^2=0 to dudx=(u+1)x2\dfrac{du}{dx}=(u+1)x^2.

Hint

  1. Differentiate u=xyu=xy with the product rule.

  2. Substitute and simplify.

Answer · Substitution

u′=(u+1)x2u\prime=(u+1)x^2.

General solution

Separation of variables

Hence show that the general solution has the form y=Aef(x)−1xy=\dfrac{Ae^{f(x)}-1}x.

Hint

  1. Separate: 1u+1 du=x2 dx\frac1{u+1}\,du=x^2\,dx.

Answer · General solution

y=(Aex3/3−1)/xy=(Ae^{x^3/3}-1)/x; f(x)=x3/3f(x)=x^3/3.

x≠0x\ne0; includes A=0A=0.

(i)

Sketching a solution curve

Sketch the particular solution curve for A=2A=2.

Explore this idea

Hint

  1. y=2ex3/3−1xy=\frac{2e^{x^3/3}-1}x: vertical asymptote x=0x=0.

  2. Find where it crosses the xx-axis.

Particular solution A = 2
Particular solution A = 2Scroll across to see the full diagram.

Answer · (i)

A=2A=2: zero −3ln⁡23-\sqrt[3]{3\ln2}, pole 0 and left limit 0+0^+; turning branches as shown.

Explore the idea

The solution with A = 2

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 = 11.791
Read the solution at x0
Try an example
y at x = 0
—
Rate dy/dx there
—
  • Solution that starts at the purple dot
  • Closed-form answer
Key idea

Every solution has the asymptote x=0x=0; A=2A=2 gives a positive branch for x>0x>0.

Ideas to try
Slide the starting value: only some curves cross the axis on the left.
(ii)

Sketching a solution curve

Sketch the particular solution curve for A=−3A=-3.

Explore this idea

Hint

  1. y=−3ex3/3−1xy=\frac{-3e^{x^3/3}-1}x.

Particular solution A = -3
Particular solution A = -3Scroll across to see the full diagram.

Answer · (ii)

A=−3A=-3: positive branch for x<0x<0, negative branch with maximum for x>0x>0.

Explore the idea

The solution with A = −3

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 = 1-5.187
Read the solution at x0
Try an example
y at x = 0
—
Rate dy/dx there
—
  • Solution that starts at the purple dot
  • Closed-form answer
Key idea

For A=−3A=-3 the numerator is always negative, so yy has the opposite sign to xx.

Ideas to try
Compare the sign of yy on the two sides of x=0x=0.