Differential equations 2

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YouTube lecture recording from October 2020

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From Words to Maths

How do we form differential equations from words?

Example 1: a growing leaf

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1. Assume the length of the leaf is $\;y\;$ cm at time $\;t\;$ days after it was first observed.

2. Let the rate the leaf receives water be $\;k_1y\;$ where $\;k_1\;$ is a positive constant.

3. Area of leaf at time $\;t\;$ days is proportional to $\;y^2\;$ since it maintains its shape, so the leaf is losing water at a rate of $\;k_2y^2\;$ where $\;k_2\;$ is another positive constant.

$\displaystyle{{\rm d}y\over {\rm d}t}=k_1y-k_2y^2=y(k_1-k_2y)$

Example 2: solid tumour growth

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1. Denote the volume of the tumour at time $\;t\;$ by $\;V(t)\;$, and its radius by $\;R(t)$.

2. Since the tumour is a sphere, its volume $\;V\;$ is proportional to $\;R^3\;$ and its surface area to $\;R^2\;$, so the surface area is proportional to $\;V^{2/3}\;$.

3. The rate at which the tumour acquires nutrients, and hence the rate at which the volume increases, is thus proportional to $\;V^{2/3}$.

$\displaystyle\frac{{\rm d}V}{{\rm d}t} = kV^{2/3}$ > $\displaystyle\frac{{\rm d}V}{{\rm d}t} = kV^{2/3}$

$\displaystyle\int V^{-2/3}~{\rm d}V = \int k~{\rm d}t$ > $\displaystyle V = \left({kt+c\over 3}\right)^3$

Second Order Differential Equations

$\displaystyle \frac{{\rm d}^2y}{{\rm d}x^2} = \left(\frac{{\rm d}y}{{\rm d}x}\right)^2$

$\displaystyle \frac{{\rm d}z}{{\rm d}x} = z^2$

$\displaystyle \int z^{-2}~{\rm d}z = \int ~{\rm d}x \qquad\Rightarrow\qquad-{1\over z} = x + A \qquad\Rightarrow\qquad z = -{1\over x+A}$

$\displaystyle \frac{{\rm d}y}{{\rm d}x} = -{1\over x+A}$

$\displaystyle y = -\ln(x+A) + B$

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• At $\;x=0,\;$ $\;\displaystyle \frac{{\rm d}y}{{\rm d}x} = -1\;$ and $\;y=0\;$.

  We substitute the first condition into $\displaystyle \frac{{\rm d}y}{{\rm d}x} = -{1\over x+A}$ to obtain $\;A=1\;$.

  Then substitute $A$ and the second condition into the eventual solution to find $\;B=0$.

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• $\;y(0)=0\;$ and $\;y(e-1)=-1$.

  This time both conditions can be substituted into the solution:

  $\displaystyle y(0)=0 \Rightarrow 0=B-\ln(A) \Rightarrow B=\ln(A)$ > $\displaystyle y(e-1)=-1 \Rightarrow -1=\ln(A)-\ln{e-1+A} \Rightarrow A=1$

More integration tricks

$\displaystyle {{\rm d}y\over {\rm d}t}=k_1y-k_2y^2=y(k_1-k_2y)$

$\displaystyle \int {{\rm d}y\over y(k_1-k_2y)} =\int {\rm d}t.\;$

Copy
import sympy as sp
y, k1, k2 = sp.symbols('y k_1 k_2')
sp.apart(1 / (y*(k1 - k2*y)),y)

$\displaystyle - \frac{k_{2}}{k_{1} \left(- k_{1} + k_{2} y\right)} + \frac{1}{k_{1} y}$

$\displaystyle {\rm~Thus~~~~~~ }{1\over k_1}\ln y-{k_2\over k_1k_2}\ln(k_1-k_2y)=t + c$

$\displaystyle y={k_1\exp (k_1t+d)\over 1+ k_2 \exp(k_1t + d)}$

Introductory problems

Main problems

Extension problems