Homogeneous differential equations (HL)

Separable differential equations, part II.

Contents

Overloaded definition

Homogeneous here refers to

 dy dx=f(yx)\frac{\d y}{\d x} = f\left(\frac yx\right)

Outside of IB, “homogeneous” may also refer to differential equations where every term contains either a derivative or yy, and no term only depends on xx or constants. See the Wikipedia article on homogeneous differential equations .

Signals

The question does not always remind you to use this method to solve a differential equation, though they do for Euler’s method and Maclaurin series. Then write the ODE with  dy dx=\displaystyle \frac{\d y}{\d x} = form see if it is repeated integration, separation of variables, or integrating factors. If it is not any of those forms, then it’s probably a homogeneous differential equation.

A y=vxy = vx substitution is very likely to be successful when both xx and yy

  1. have the same highest degree and
  2. exist in similar places in the expression.

Process

For a homogeneous differential equation, we first make the substitution y=vxy = vx then solve using separations.

Also

 dy dx= d(vx) dx=v+x dv dx\begin{align*}\frac{\d y}{\d x} &= \frac{\d (vx)}{\d x} \\ &= v+x\frac{\d v}{\d x}\end{align*}

Example: (May 2018 HL Paper 3 Calculus #5) Consider the differential equation

x dy dx−y=xp+1x\frac{\d y}{\d x} - y = x^p + 1

where x∈R,x≠0x\in\mathbb R, x\neq0 and pp is a positive integer, p>1p>1. Given y=−1y = -1 when x=1x = 1, solve for yy as a function of xx.

x(v+x dv dx)−vx=xp+1xv+x2 dv dx−vx=xp+1 dv dx=xp−2+1x2\begin{align*}x\left(v+x\frac{\d v}{\d x}\right)-vx &= x^p + 1 \\ xv + x^2\frac{\d v}{\d x} - vx &= x^p + 1\\ \frac{\d v}{\d x} &= x^{p-2} + \frac{1}{x^2}\end{align*}

Here we can use integration to solve. Though usually you end up with a separable differential equation with a non-constant function of yy.

v=1p−1xp−1−1x+Cv = \frac{1}{p-1}x^{p-1} - \frac{1}{x} + C

Then substitute back to solve for y=f(x)y = f(x)

y=vx=1p−1xp−1+Cxy = vx = \frac{1}{p-1}x^{p} - 1 + Cx

Then substitute in (1,−1)(1, -1) to solve for CC

−1=1p−1(1)p−1+C(1)−1+1=1p−1+C−1p−1=C\begin{align*}-1 &= \frac{1}{p-1}(1)^{p} - 1 + C(1) \\ -1 + 1 &= \frac{1}{p-1} + C \\ -\frac{1}{p-1} &= C \end{align*}

Putting everything together

y=xpp−1−1−xp−1■y = \frac{x^p}{p-1} - 1 - \frac{x}{p - 1} \qed

If you are OCD about your ODE you can also simplify further

y=xp−xp−1−1y = \frac{x^p - x}{p-1} - 1
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