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

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<b>Lecture 14: </b><br>
Thu 19 Feb 2026 09:00
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Recap of last lecture:

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Soluble by direct integration: <div class="pn-import-equation"><nowiki>$$\frac{dy}{dx} = f(x) \implies y(x) = \int f(x) dx + c$$</nowiki></div>

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Separable equations: <div class="pn-import-equation"><nowiki>$$\frac{dy}{dx} = \frac{f(x)}{g(y)} \implies \int f(x) dx = \int g(y) dy$$</nowiki></div>

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Exact equations:

Given in the form (or re-arrangeable to the form): <div class="pn-import-equation"><nowiki>$$M(x, y) dx + N(x, y) dy = 0$$</nowiki></div>

These are soluble if: <div class="pn-import-equation"><nowiki>$$M(x, y) = \frac{df}{dx} \qquad N(x, y) = \frac{df}{dy}$$</nowiki></div> With <nowiki>$f(x, y) = c$</nowiki>.

The condition for an equation being exact is: <div class="pn-import-equation"><nowiki>$$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$</nowiki></div>

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== “Homogeneous” Equations ==

Here, the word homogenous has a different meaning to classification of differentiating equations. If we have some function that satisfied this: <div class="pn-import-equation"><nowiki>$$g(\lambda x, \lambda y) = \lambda^p g(x, y)$$</nowiki></div>

We call it a homogeneous function of order <nowiki>$p$</nowiki>.

“Homogeneous” differential equations in this case are functions of the form: <div class="pn-import-equation"><nowiki>$$\frac{dy}{dx} = f \left(\frac{y}{x}\right)$$</nowiki></div>

Here, we make the substitution <nowiki>$v(x) = y(x)/x$</nowiki>. We can write: <div class="pn-import-equation"><nowiki>$$y(x) = v(x) x$$</nowiki></div>

And substitute back in. We use the product rule where: <div class="pn-import-equation"><nowiki>$$\frac{dy}{dx} = x \frac{dv}{dx} + v = f(v)$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$x \frac{dv}{dx} = f(v) - v$$</nowiki></div>

This is now separable, so: <div class="pn-import-equation"><nowiki>$$\int \frac{dv}{f(v) - v} = \int \frac{dx}{x}$$</nowiki></div>

=== Example ===

<div class="pn-import-equation"><nowiki>$$2xy dy - (x^2 + y^2) dx = 0$$</nowiki></div> This is not exact, so we put it into a standard form and substitute: <div class="pn-import-equation"><nowiki>$$2xy dy = (x^2 + y^2) dx$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\frac{dy}{dx} = \frac{x^2 + y^2}{2xy} = \frac{(y/x)^2 + 1}{2 (y/x)}$$</nowiki></div>

Let <nowiki>$u(x) = y/x$</nowiki>:

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