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Summary of Lecture 12:
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Frenet-Serret Equations for a curve $\underline{r}(s)$, where $\hat{T} = d \underline{r} / d s$.
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Classification of differential equations:
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Partial vs Ordinary
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Order
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Degree
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Homogeneous vs Inhomogeneous
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Linear vs Non-Linear
Equations Soluble By Direct Integration
Given some:
$$\frac{dy}{dx} = f(x) \implies y(x) = \int^x f(x^\prime)dx\prime + c$$
We can generalise this to some repeated derivative:
$$\frac{d^ny}{dx^n} = f(x)$$
Instead of having one undetermined constant here, we now have $n$. As each round of integration picks up a factor of the integration subject, these constants will form an $n$th degree polynomial.
Example: Particle Falling
$$\frac{d^2z}{dt^2} = -g$$
$$\implies \frac{dz}{dt} = -gt + v_0$$
$$z = -\frac{1}{2}gt^2 + v_0 t + z_0$$
Here we consider our unknown constants as boundary conditions, i.e. the initial velocity and initial height.
Separable Equations
These are equations in the form:
$$\frac{dy}{dx} = \frac{f(x)}{g(x)}$$
$$g(y) dy = f(x) dx \implies \int g(y) dy = \int f(x) dx + c$$
Example I: Falling Particle with Air Resistance
$$\frac{dv}{dt}= -g-kv$$
$$\int \frac{dv}{g+kv} = - \int dt$$
$$\implies \frac{1}{k} \ln \left(g + kv\right) = -t + c$$
$$\implies k + gv = Ae^{-kt}$$
$$\implies v(t) = - \frac{g}{k} + \left(v_0 + \frac{g}{k}\right)e^{-kt}$$
Example II
$$\frac{dy}{dx} - x^2 y^2 = x^2$$
$$\frac{dy}{dx} = x^2 + x^2y^2 = x^2(y^2 + 1)$$
$$\implies \frac{1}{y^2 + 1} dy = x^2 dx$$
$$\implies \arctan y = \frac{1}{3}x^3 + c$$
$$\implies y = \tan \left(\frac{1}{3}x^3 + c\right)$$
Example III
$$\frac{dy}{dx} = -\frac{x}{y}$$
$$y dy = -x dx$$
$$\frac{1}{2}y^2 = - \frac{1}{2}x^2 + c$$
$$y^2 + x^2 = 2c$$
This is the equation for a circle.
Exact Equations
Suppose a function $y(x)$ is implicitly defined such that $f(x, y) = c$. It follows that the total differential:
$$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy = 0$$
As $f(x, y)$ is a constant for all values of $x, y$, the constant differentiates to zero.
We can set:
$$M(x, y) = \frac{\partial f}{\partial x} \qquad N(x, y) = \frac{\partial f}{\partial y}$$
To try to solve this equation:
$$M(x, y) dx + N(x, y) dy = 0$$
If we can do this (i.e. if the function can be decomposed into gradient components $M$ and $N$), the equation is called “exact”. For this to be true, we need:
$$\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = \frac{\partial^2 f}{\partial x \partial y}$$
If this relationship holds, we have an exact equation and can integrate M and N to get $f(x, y)$ to solve the equation in the form $f(x, y) = c$
Example
$$\frac{dy}{dx} = - \frac{2x+y}{x+2y}$$
$$(2x+y) dx + (x+2y) dy = 0$$
So:
$$M(x, y) = 2x+y \qquad N(x, y) = x+2y$$
And:
$$\frac{\partial M}{\partial y} = 1 \qquad \frac{\partial N}{\partial x} = 1$$
So yes, it is an exact function. It follows that:
$$\frac{\partial f}{\partial x} = 2x+y \implies f(x, y) = x^2 + xy + c(y)$$
$$\frac{\partial f}{\partial y} = x+2y \implies f(x, y) = xy + y^2 + d(x)$$
$$\implies f(x, y) = x^2 + xy + y^2 = c$$
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