AA Topic 5.18 · Analysis and Approaches HL

Four methods, and the thirty seconds that picks one

Separable, homogeneous, integrating factor and Euler, and how to recognise which equation is which.

Higher Level

Nearly all the difficulty here is choosing. Pick an equation and the method, the reason for it and the working appear together.

separable

How to recognise each one

If it looks likeUse
dy/dx = (something in x) × (something in y)separable
dy/dx = a function of y⁄x onlyhomogeneous, substitute y = vx
dy/dx + P(x)y = Q(x)integrating factor
none of the above, or a numerical answer is wantedEuler

Test for homogeneous quickly: replace x with λx and y with λy. If every λ cancels, it is homogeneous and y = vx will work. That test takes ten seconds and saves you from attempting a separation that cannot happen.

The integrating factor

For dydx + P(x)y = Q(x), multiply everything by e∫P dx. The left side then becomes the derivative of a product, by design, and you integrate both sides.

y′ + yx= x, so P = 1x IF= e∫(1/x)dx = eln x = x xy′ + y= x², and the left side is exactly ddx(xy) that is the whole trick xy= x³3 + C, so y = x²3 + Cx

Euler's method

When nothing solves, step along the gradient: yn+1 = yn + h f(xn, yn), with xn+1 = xn + h.

nxyf = x + ynext y
00111.1
10.11.11.21.22
20.21.221.421.362
30.31.3621.6621.5282
40.41.52821.92821.72102

The exact solution here is y = 2ex − x − 1, giving 1.7974 at x = 0.5, so Euler is low by about 0.076. Keep every digit in the table and round only at the end.

Your turn

1. Which method does dydx + 2yx = x³ need?

2. For dydx = xy with y(0) = 1, find y when x = 2, to 3 decimal places.

3. Using Euler with h = 0.1 on dy/dx = x + y, y(0) = 1, what is y₂?

Where the marks go

Identifying the method and saying why. A wrong choice costs the whole question, and the right one is often worth a mark on its own.

On an integrating factor, showing e∫P dx simplified. On a homogeneous equation, stating y = vx and dy/dx = v + x dv/dx before substituting.

Using the initial condition at the end. A general solution with an unfound constant is incomplete whenever a condition was given.

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