Methods II Tripos Questions

First Order ODEs

Example from notes

Solve: $y' + 3y = e^{-2x}$

Solution:

Integrating factor: $\mu = e^{\int 3\,dx} = e^{3x}$

$\frac{d}{dx}(e^{3x}y) = e^{3x}e^{-2x} = e^x$

$e^{3x}y = e^x + C$

$y = e^{-2x} + Ce^{-3x}$

Second Order ODEs

Example: Damped Oscillator

Solve $y'' + 4y' + 5y = 0$ with $y(0) = 1$ , $y'(0) = 0$ .

Characteristic equation: $\lambda^2 + 4\lambda + 5 = 0$

$\lambda = \frac{-4 \pm \sqrt{16 - 20}}{2} = -2 \pm i$

General solution: $y = e^{-2x}(A\cos x + B\sin x)$

Apply initial conditions:

$y(0) = A = 1$

$y'(x) = -2e^{-2x}(\cos x + B\sin x) + e^{-2x}(-\sin x + B\cos x)$

$y'(0) = -2(1) + B = 0 \Rightarrow B = 2$

Solution: $y = e^{-2x}(\cos x + 2\sin x)$

Lagrange Multipliers

Maximum of $f(x,y) = xy$ on $x^2 + 4y^2 = 1$

Set up: $\mathcal{L} = xy - \lambda(x^2 + 4y^2 - 1)$

Equations:

$y - 2\lambda x = 0$ … (i)
$x - 8\lambda y = 0$ … (ii)
$x^2 + 4y^2 = 1$ … (iii)

From (i): $\lambda = \frac{y}{2x}$

Substitute into (ii): $x - \frac{8y^2}{2x} = x - \frac{4y^2}{x} = 0$

$x^2 = 4y^2$

From (iii): $4y^2 + 4y^2 = 8y^2 = 1$

$y = \pm\frac{1}{2\sqrt{2}}$

$x = \pm\frac{1}{\sqrt{2}}$

Maximum: $f = \frac{1}{2\sqrt{2}} \cdot \frac{1}{2\sqrt{2}} = \frac{1}{8}$

(When signs match.)

Vector Calculus

Line Integral Example

Compute $\int_C \mathbf{F} \cdot d\mathbf{r}$ for $\mathbf{F} = (y, x)$ along $C$ :

Straight line from $(0,0)$ to $(1,1)$
Curve $y = x^2$ from $(0,0)$ to $(1,1)$

Check if conservative:

$\nabla \times \mathbf{F} = \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = \frac{\partial x}{\partial x} - \frac{\partial y}{\partial y} = 1 - 1 = 0$

$\mathbf{F}$ is conservative. Find potential: $\phi = xy + C$ .

Line integral = $\phi(1,1) - \phi(0,0) = 1$

Both paths give the same result: 1.

Stokes’ Theorem Verification

Verify Stokes’ theorem for $\mathbf{F} = (-y, x, 0)$ on circle $x^2 + y^2 \leq a^2$ in plane $z = 0$ .

Curl: $\nabla \times \mathbf{F} = (0, 0, 2)$

Surface integral: $\int_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \int_S 2\,dS = 2\pi a^2$

Line integral: Boundary is circle $x^2 + y^2 = a^2$ at $z = 0$ .

$\mathbf{r} = (a\cos\theta, a\sin\theta, 0)$

$d\mathbf{r} = (-a\sin\theta, a\cos\theta, 0)\,d\theta$

$\mathbf{F} = (-a\sin\theta, a\cos\theta, 0)$

$\mathbf{F} \cdot d\mathbf{r} = a^2\sin^2\theta + a^2\cos^2\theta = a^2$

$\oint \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} a^2\,d\theta = 2\pi a^2$

Both equal: $2\pi a^2$

Fourier Series

Parseval’s Theorem Application

Use Fourier series for $f(x) = x$ on $[-\pi, \pi]$ to evaluate $\sum_{n=1}^{\infty} \frac{1}{n^2}$ .

Fourier coefficients:

Since $f$ is odd: $a_n = 0$ .

$b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} x\sin(nx)\,dx = \frac{2(-1)^{n+1}}{n}$

Parseval:

$\frac{1}{2\pi}\int_{-\pi}^{\pi} x^2\,dx = \frac{1}{2}\sum_{n=1}^{\infty} b_n^2$

$\frac{\pi^2}{3} = \frac{1}{2}\sum \frac{4}{n^2}$

$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$

Fourier Series of Square Wave

$f(x) = \begin{cases} -1 & -\pi < x < 0 \\ 1 & 0 < x < \pi \end{cases}$

Even though $f$ is odd, this gives a series with only sine terms:

$f(x) = \frac{4}{\pi}\left(\sin x + \frac{1}{3}\sin 3x + \frac{1}{5}\sin 5x + \cdots\right)$

$= \frac{4}{\pi}\sum_{k=0}^{\infty} \frac{\sin((2k+1)x)}{2k+1}$

Note: This can also be obtained by differentiating the triangle wave series.