Year 1 Sem 2/LC Mathematics for Physicists 1B/Lecture 14

Lecture 14:
Thu 19 Feb 2026 09:00

Recap of last lecture:

Soluble by direct integration:
$$\frac{dy}{dx} = f(x) \implies y(x) = \int f(x) dx + c$$
Separable equations:
$$\frac{dy}{dx} = \frac{f(x)}{g(y)} \implies \int f(x) dx = \int g(y) dy$$
Exact equations: Given in the form (or re-arrangeable to the form):
$$M(x, y) dx + N(x, y) dy = 0$$
These are soluble if:
$$M(x, y) = \frac{df}{dx} \qquad N(x, y) = \frac{df}{dy}$$
With $f(x, y) = c$. The condition for an equation being exact is:
$$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$

“Homogeneous” Equations

Here, the word homogenous has a different meaning to classification of differentiating equations. If we have some function that satisfied this:

$$g(\lambda x, \lambda y) = \lambda^p g(x, y)$$

We call it a homogeneous function of order $p$.

“Homogeneous” differential equations in this case are functions of the form:

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

Here, we make the substitution $v(x) = y(x)/x$. We can write:

$$y(x) = v(x) x$$

And substitute back in. We use the product rule where:

$$\frac{dy}{dx} = x \frac{dv}{dx} + v = f(v)$$

$$x \frac{dv}{dx} = f(v) - v$$

This is now separable, so:

$$\int \frac{dv}{f(v) - v} = \int \frac{dx}{x}$$

$$2xy dy - (x^2 + y^2) dx = 0$$

This is not exact, so we put it into a standard form and substitute:

$$2xy dy = (x^2 + y^2) dx$$

$$\frac{dy}{dx} = \frac{x^2 + y^2}{2xy} = \frac{(y/x)^2 + 1}{2 (y/x)}$$

Let $u(x) = y/x$: