Sections:
- Solve first order differential equation
- Solve second order differential equation
- Form differential equation from real-life situations
- Questions
(1) Solve first order differential equation
Solve by direct integration:
Example
Find the general solution of the differential equation ${dy \over dx} = x^2 + 2$.
Answer: $ y = {1 \over 3}x^3 + 2x + C $
Solutions
\begin{align*}
y & = \int x^2 + 2 \phantom{.} dx \\
y & = {1 \over 3}x^3 + 2x + C, \text{ where } C \text{ is an arbitrary constant}
\end{align*}
Solve by variable separable method:
Example
Find the general solution of the differential equation ${dy \over dx} = y - 1$.
Answer: $ y = Ae^x + 1 $
Solutions
\begin{align*}
{dy \over dx} & = y - 1 \\
{1 \over y - 1} \left({dy \over dx}\right) & = 1 \\
\int {1 \over y - 1} \left({dy \over dx}\right) \phantom{.} dx & = \int 1 \phantom{.} dx \\
\int {1 \over y - 1} \phantom{.} dy & = \int 1 \phantom{.} dx \\
\ln |y - 1| & = x + C \\
|y - 1| & = e^{x + C} \\
y - 1 & = \pm e^{x + C} \\
y - 1 & = \pm e^C . e^x \\
y - 1 & = A e^x \\
y & = Ae^x + 1, \text{ where } A = \pm e^C \text{ and } C \text{ is an arbitrary constant}
\end{align*}
Solve by using provided substitution:
Example
Use the substitution $y = v + x^2$ to find the general solution of the differential equation $ {dy \over dx} = 1 + 2x - {y \over x^2} $.
Answer: $ y = Ae^{1 \over x} + x^2 $
Solutions
\begin{align*}
y & = v + x^2 \\
{d \over dx} (y) & = {d \over dx} (v + x^2) \\
{dy \over dx} & = {dv \over dx} + 2x \phantom{000000} [\text{Substitute this into the DE}] \\
\\
{dy \over dx} & = 1 + 2x - {y \over x^2} \\
({dv \over dx} + 2x) & = 1 + 2x - {v + x^2 \over x^2} \\
{dv \over dx} & = 1 - \left( {v \over x^2} + {x^2 \over x^2} \right) \\
{dv \over dx} & = 1 - {v \over x^2} - 1 \\
{dv \over dx} & = -{v \over x^2} \\
\\
{1 \over v} \left(dv \over dx\right) & = - {1 \over x^2} \phantom{000000000} [\text{Variable separable method}] \\
\int {1 \over v} \left(dv \over dx\right) \phantom{.} dx & = \int -{1 \over x^2} \phantom{.} dx \\
\int {1 \over v} \phantom{.} dv & = \int -{1 \over x^2} \phantom{.} dx \\
\ln |v| & = {1 \over x} + C \\
|v| & = e^{ {1 \over x} + C} \\
v & = \pm e^{ {1 \over x} + C} \\
v & = \pm e^C . e^{1 \over x} \\
y - x^2 & = A e^{1 \over x} \\
y & = Ae^{1 \over x} + x^2, \text{ where } A = \pm e^C \text{ and } C \text{ is an arbitrary constant}
\end{align*}
(2) Solve second order differential equation
Example
Find the general solution of the differential equation ${d^2 x \over dt^2} = \sin t - t $.
Answer: $ x = - \sin t - {1 \over 6}t^3 + Ct + D $
Solutions
\begin{align*}
{dx \over dt} & = \int \sin t - t \phantom{.} dt \\
& = -\cos t - {1 \over 2}t^2 + C \\
\\
x & = \int -\cos t - {1 \over 2}t^2 + C \phantom{.} dt \\
& = -\sin t - {1 \over 6}t^3 + Ct + D, \text{ where } C \text{ and } D \text{ are arbitrary constants}
\end{align*}
(3) Form differential equation from real-life situations
Example 1: Modelling decrease
A certain radioactive material is known to decay (i.e. mass reduces) at a rate inversely proportional to the amount present. The amount of material present at a given time, $t$ minutes, is $N$ grams.
Form the differential equation involving $N$ and $t$ for this situation.
${dN \over dt} = $ $ -{k \over N}, \text{ where } k \text{ is a positive constant} $
Example 2: Modelling increase and decrease
The growth of a type of insects is studied. The birth rate of the insects is proportional to the number in thousands, $x$, of insects, at any time $t$ days after the start of the study. The insects die at a constant rate of $500$ insects daily.
Form the differential equation involving $x$ and $t$ for this situation.
${dx \over dt} = $ $ kx - {500 \over 1000}, \text{ where } k \text{ is a positive constant} $
(4) Questions
1. A student is investigating the rate of change of temperature ($\theta$ degree celsius) of a cup of hot coffee at time $t$ minutes. The student suggests that $\theta$ and $t$ are related by the differential equation:
$ {d \theta \over dt } = - k \theta + b$, where $k$ and $b$ are positive constants
(a) Given that $ \theta = 80 $ when $ t = 0 $, find $ \theta $ in terms of $b$, $k$ and $t$.
Answer: $ \theta = {b \over k} - \left({b \over k} - 80\right) e^{-kt} $
Solutions
\begin{align*}
{d \theta \over dt} & = -k \theta + b \\
{1 \over -k \theta + b} \left(d \theta \over dt\right) & = 1 \\
\int {1 \over -k \theta + b} \left(d \theta \over dt\right) \phantom{.} dt & = \int 1 \phantom{.} dt \\
\int {1 \over -k \theta + b} \phantom{.} d\theta & = \int 1 \phantom{.} dt \\
-{1 \over k} \int \underbrace{{-k \over -k \theta + b}}_{\int {f'(x) \over f(x)} \phantom{.} dx } \phantom{.} d\theta & = \int 1 \phantom{.} dt \\
\int {-k \over -k \theta + b} \phantom{.} d\theta & = -k \int 1 \phantom{.} dt \\
\ln | -k\theta + b| & = -kt + c \\
|-k \theta + b| & = e^{-kt + c} = e^{-kt} \times e^c \\
-k \theta + b & = \pm e^c (e^{-kt}) \\
-k\theta + b & = A e^{-kt}, \text{ where } A = \pm e^c \\
-k\theta & = A e^{-kt} - b \\
k\theta & = b - A e^{-kt} \\
\theta & = {b \over k} - {A \over k} e^{-kt} \\
\\
\text{When } & t = 0 \text{ and } \theta = 80, \\
80 & = {b \over k} - {A \over k} e^0 \\
80 & = {b \over k} - {A \over k} \\
80k & = b - A \\
A & = b - 80k \\
\\
\theta & = {b \over k} - {b - 80k \over k} e^{-kt} \\
\theta & = {b \over k} - \left({b \over k} - {80k \over k}\right) e^{-kt} \\
\theta & = {b \over k} - \left({b \over k} - 80\right) e^{-kt}
\end{align*}
(b) Explain what will eventually happen to the temperature of the cup of coffee using the student's model.
Solutions
\begin{align*}
& \theta = {b \over k} - \left({b \over k} - 80\right) e^{-kt} \\
\\
& \text{As } t \rightarrow \infty, e^{-kt} \rightarrow 0 \text{ and } \theta \rightarrow {b \over k} \\
\\
& \text{Temperature of the cup of coffee will approach } {b \over k} \text{ degree celsius}
\end{align*}
2. With the launch of a new mobile application, analysts model the spread of the application among a fixed group of users. At any time $t$ hours after launch, users are classified into two groups: those who have installed the application and those who have not.
Let
- $x$ denote the number of users, in thousands, who installed the application at time $t$,
- $M$ denote the total numbers of users in the group, in thousands, where $M$ is a constant.
It is assumed that the rate at which new users install the application is proportional to the product of the number of users who have already installed it and the number of users who have not yet installed it.
(a) Write down a differential equation for ${dx \over dt}$ in terms of $x$, $M$ and a constant of proportionality $k$.
Answer: $ {dx \over dt} = kx (M - x) $
It is given that when the application is first launched ($t = 0$), $x = 2$ and ${dx \over dt} = {M - 2 \over 8}$.
(b) Show that the solution of the differential equation in part (i) can be written in the form
$ x = { BM \over B - (2 - M)e^{-{1 \over 16}Mt}}$, where $B$ is a positive constant
Solutions
\begin{align*}
{dx \over dt} & = kx (M - x) \\
\\
\text{When } t = 0, & \phantom{0} x = 2 \text{ and } {dx \over dt} = {M - 2 \over 8}, \\
{M - 2 \over 8} & = k(2) (M - 2) \\
{1 \over 8} & = 2k \\
{1 \over 16} & = k \\
\\
{dx \over dt} & = {1 \over 16} x (M - x) \\
{1 \over x(M - x)} \left(dx \over dt\right) & = {1 \over 16} \\
\int {1 \over x(M - x)} \left(dx \over dt\right) dt & = \int {1 \over 16} \phantom{.} dt \\
\int {1 \over - x^2 + Mx} \phantom{.} dx & = \int {1 \over 16} \phantom{.} dt \\
- \int {1 \over x^2 - Mx} \phantom{.} dx & = \int {1 \over 16} \phantom{.} dt \\
- \int {1 \over \left(x - {M \over 2}\right)^2 - \left(M \over 2\right)^2} \phantom{.} dx & = \int {1 \over 16} \phantom{.} dt \\
- {1 \over 2 \left(M \over 2\right)} \ln \left| {x - {M \over 2} - {M \over 2} \over x - {M \over 2} + {M \over 2} } \right| & = \int {1 \over 16} \phantom{.} dt \\
- {1 \over M} \ln \left| { x - M \over x } \right| & = \int {1 \over 16} \phantom{.} dt \\
\ln \left| { x - M \over x } \right| & = -{1 \over 16}Mt + c \\
\left| { x - M \over x } \right| & = e^{-{1 \over 16}Mt + c} \\
{ x - M \over x } & = \pm e^{-{1 \over 16}Mt} \times e^c \\
{ x - M \over x } & = A e^{-{1 \over 16}Mt}, \text{ where } A = \pm e^c \\
\\
\text{When } & t = 0 \text{ and } x = 2, \\
{2 - M \over 2} & = A (1) \\
{2 - M \over 2} & = A \\
\\
{x - M \over x} & = A e^{-{1 \over 16}Mt} \\
x - M & = A e^{-{1 \over 16}Mt} x \\
x - A e^{-{1 \over 16}Mt} x & = M \\
x (1 - A e^{-{1 \over 16}Mt} ) & = M \\
x & = {M \over 1 - A e^{-{1 \over 16}Mt}} \\
x & = {M \over 1 - {2 - M \over 2} e^{-{1 \over 16}Mt}} \times {2 \over 2} \\
x & = {2M \over 2 - (2 - M) e^{-{1 \over 16}Mt}} \text{ (Shown)}
\end{align*}
(c) Given that the total number of users is $800 \phantom{.} 000$, find the particular solution for $x$ in terms of $t$. Hence, sketch the graph of $x$ against $t$, stating clearly any asymptotes and intercepts.
Solutions
\begin{align*}
x & = {2M \over 2 - (2 - M) e^{-{1 \over 16}Mt}} \\
\\
\text{When } & M = 800, \\
x & = {2(800) \over 2 - (2 - 800) e^{-{1 \over 16}(800)t}} \\
x & = {1600 \over 2 + 798 e^{-50t}} \\
x & = {800 \over 1 + 399 e^{-50t}} \phantom{000000} [\text{Particular solution}]
\end{align*}
(d) Find the time taken, to the nearest minute, for $95\%$ of the users to have installed the application.
Answer: $ 11 \text{ minutes} $
Solutions
\begin{align*}
800 \times {95 \over 100} & = 760 \\
\\
760 & = {800 \over 1 + 399 e^{-50t}} \\
1 + 399 e^{-50t} & = {800 \over 760} \\
399e^{-50t} & = {20 \over 19} - 1 \\
399 e^{-50t} & = {1 \over 19} \\
e^{-50t} & = {1 \over 19} \div 399 \\
-50t & = \ln {1 \over 7581} \\
t & = { \ln {1 \over 7581} \over -50} \\
t & = 0.17867 \text{ hours} \\
t & \approx 11 \text{ minutes (to nearest minute)}
\end{align*}