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\begin{document}
\pagestyle{fancy}\thispagestyle{empty}
\null

\vfill
\begin{center}
\begin{Huge}\textbf{Differential Equations} \end{Huge}
\end{center}

\vfill
\tproof{{\Huge This version has proofs and comments for the teacher}

These come in frames like this one, and for this reason, the page numbers are not the same as on the student handout version.

\textit{The proofs given must not be understood as ``the'' proofs, but as the ones which over the years, I feel most comfortable with. When a theorem does not need anything specific to ultracalculus, the proof is omitted.}
}
\vfill

\hfill Coll\`ege Andr\'e-Chavanne

\hfill Gen\`eve


\vfill




\mypage

\section{Initial problems}
\exo{
Radioactivity is due to the decay of nuclei in the atoms. The
  following observation has been made: For a given element, it takes
  the same time for half of the nuclei to disintegrate, independently
  of how much matter there is. Which means that there is a higher
  activity if there is more matter. This leads to the conclusion that the
  activity depends on the quantity of matter. It is stated in the
  law:

``The rate of decay (with respect to time) of the nuclei of radioactive
  substances is proportional to the number of remaining nuclei.''

Let $y$ be the number of remaining nuclei and $t$ is for time.

Write the equation which corresponds to the statement of the law. 

Find $y$ as function of $t$.
}

\exo{
The velocity of a falling body is proportional to the time $t$
  during which it has already fallen.

Write the equation corresponding to this statement and show how to
find the position of this falling body, with respect to time.

What is the equation for the position of the falling body, with
respect to time?
}

\exo{
We have all observed that if something is very hot (a cake coming out
of the oven...) its temperature decreases quite quickly in terms of
degrees per minute, whereas an object hardly warmer than the
surrounding temperature will not lose as many degrees during the first
minute.

Newton's (experimental) law of cooling states (in modern units)
  that the rate at which a body at $T^\circ C$ above surrounding
  temperature cools is proportional to $T$

Find the mathematical equivalent to this law.

A body at $68^\circ C$ in a room at $16^\circ C$ is $55^\circ c$ after
5 minutes. What will its temperature be in another 5 minutes?

What will its temperature be in yet another 5 minutes?

After how long will the body reach the surrounding temperature?
Comment your answer.
}

\definition{A differential equation is an equation involving an
  independent variable $t$ and the derivatives of $y$ with respect to
  $t$, where \emph{the function} $y$ is the unknown.
  }\index{differential equation!definition}
  
\Definition{First order differential equation}{
A first order differential equation involves only the first derivative of $y$:
 \[y'=f(y,t)\]
  }

\caution The solution to a differential equation is a function or a
  whole class of functions.

\bigskip
The independent variable is traditionally $t$ rather than $x$. This is because
differential equations appeared first in physics problems related to time.

%============================================================

\section{Slope Fields}

If a graph is drawn with $t$ as horizontal coordinate and $y$ as vertical coordinate, at $\langle t,y\rangle$ the differential equation yields a specific slope $f(y,t)$. A visualisation can be provided by marking some of the slopes on a grid.

\exo{
\index{slope-field}\index{vector}

In the following drawings, each arrow represents a slope. They form a
slope-field.  (One must imagine that between two arrows the slope changes
continuously).

\begin{itemize}
\item Choose a point on the plane, follow the arrows with a pencil to
draw a curve. What does this curve represent?

\medskip
The curves that can be drawn as above are called \textbf{specific solutions}

\item Repeat this with some other points. 
\end{itemize}

\begin{enumerate}
\item \hfill

\begin{center}
\begin{tikzpicture}[scale = 1]
\draw [->](-4,0)--(4,0);
\draw [->] (0,-4)--(0,4);
\foreach \i in {-3,-2.5,...,3}
\foreach \j in {-3,-2.5,...,3}
{%
\pgfmathparse{\j}; %The differential equation
\pgfmathsetmacro\yprime{\pgfmathresult};
\pgfmathsetmacro\N{0.4}; %The length of the vector
\pgfmathparse{\N*(\yprime)/(sqrt(1+(\yprime)^2))};
\pgfmathsetmacro\Dy{\pgfmathresult/2};
\pgfmathparse{\N/(sqrt(1+(\yprime)^2))};
\pgfmathsetmacro\Dx{\pgfmathresult/2};
 \draw (\i-\Dx,\j-\Dy)--(\i ,\j )circle(0.05) -- (\i+\Dx,\j+\Dy);};
\end{tikzpicture}
\end{center}

\newpage
\item\hfill

\begin{center}
\begin{tikzpicture}[scale = 1]
\draw [->](-4,0)--(4,0);
\draw [->] (0,-4)--(0,4);
\foreach \i in {-3,-2.5,...,3}
\foreach \j in {-3,-2.5,...,3}
{%
\pgfmathparse{\i*\j}; %The differential equation
\pgfmathsetmacro\yprime{\pgfmathresult};
\pgfmathsetmacro\N{0.4}; %The length of the vector
\pgfmathparse{\N*(\yprime)/(sqrt(1+(\yprime)^2))};
\pgfmathsetmacro\Dy{\pgfmathresult/2};
\pgfmathparse{\N/(sqrt(1+(\yprime)^2))};
\pgfmathsetmacro\Dx{\pgfmathresult/2};
 \draw (\i-\Dx,\j-\Dy)--(\i ,\j )circle(0.05) -- (\i+\Dx,\j+\Dy);};
\end{tikzpicture}
\end{center}

\bigskip
\item \hfill

\begin{center}
\begin{tikzpicture}[scale = 1]
\draw [->](-4,0)--(4,0);
\draw [->] (0,-4)--(0,4);
\foreach \i in {-3,-2.5,...,3}
\foreach \j in {-3,-2.5,...,3}
{%
\pgfmathparse{\i^2-\j^2}; %The differential equation
\pgfmathsetmacro\yprime{\pgfmathresult};
\pgfmathsetmacro\N{0.4}; %The length of the vector
\pgfmathparse{\N*(\yprime)/(sqrt(1+(\yprime)^2))};
\pgfmathsetmacro\Dy{\pgfmathresult/2};
\pgfmathparse{\N/(sqrt(1+(\yprime)^2))};
\pgfmathsetmacro\Dx{\pgfmathresult/2};
 \draw (\i-\Dx,\j-\Dy)--(\i ,\j )circle(0.05) -- (\i+\Dx,\j+\Dy);};
\end{tikzpicture}
\end{center}
\end{enumerate}

}

\newpage
\exo{
The following slope-field is for
$\ds y'=y^2\cdot t$

\begin{center}
\begin{tikzpicture}[scale = 1]
\draw [->](-4,0)--(4,0);
\draw [->] (0,-4)--(0,4);
\foreach \i in {-3,-2.5,...,3}
\foreach \j in {-3,-2.5,...,3}
{%
\pgfmathparse{\j*\j*\i}; %The differential equation
\pgfmathsetmacro\yprime{\pgfmathresult};
\pgfmathsetmacro\N{0.4}; %The length of the vector
\pgfmathparse{\N*(\yprime)/(sqrt(1+(\yprime)^2))};
\pgfmathsetmacro\Dy{\pgfmathresult/2};
\pgfmathparse{\N/(sqrt(1+(\yprime)^2))};
\pgfmathsetmacro\Dx{\pgfmathresult/2};

 \draw (\i-\Dx,\j-\Dy)--(\i ,\j )circle(0.05) -- (\i+\Dx,\j+\Dy);};
\end{tikzpicture}
\end{center}

Follow the solutions for different initial values.
}

\exo{
Draw the slope-field for $$y'=2t\sqrt{1-y^2}$$ for $t\in
  [-1;1]\quad y\in[-1;1]$ (grid every 0.5)

}

\exo{
Draw the slope-field for $$y'=(1+y^2)\cdot e^t$$ for $t\in
  [-1;1]\quad y\in[-1;1]$ (grid every 0.5)
}
%============================================================
\newpage
\section{Specific solutions}
\index{graphical representation}
\tproof{This method, found in \textit{An elementary treatise on differential equations}, Bell and Sons,   London, 1939, attributed to Brodetscky and Wada, is rarely presented but has the advantage of giving graphical representations even for some unsolvable equations by methods in the syllabus. It is also a good exercise to give meaning to the first and second derivative.}

The following method enables to sketch rapidly the general form of the family of
curves which are solution of any equation of the form $y'=f(t,y)$, under the
condition that $y''$ exists.
%============================================================

\example{} Consider the equation $$y'=t(y-1)$$

Differentiating again leads to $$y''=(y-1)+t\cdot y'=(y-1)+t\cdot t(y-1)$$
$$y''=(t^2+1)(y-1)$$

We know that the first derivative gives information about the slope and the
second derivative gives information about the bending.

The first factor $t^2+1$ is positive. The second is positive if $y>1$. This means that if $y>1$ the second derivative is positive and the curve bends upwards, if $y<1$ the curve bends downwards and if $y=0$ there is no bending, hence the curve is a straight line. As for $y=0$ we have that $y'=0$ the line is horizontal.

The original equation also yields $y'=0$ for $t=0$

It is also possible to see that the closer to $y=1$ the closer to zero $y'$ will be, hence the curve will be flatter. 

\exo{Draw the $y'=0$ lines. Draw the $y''=0$ lines. Sketch some of the specific solutions.}

\exo{
$$y'=y+e^t$$

leads to $y''=y'+e^t=y+2e^t$

The locus of inflexions ($y''=0$) is the curve $y=-2e^t$. 


\begin{itemize}
\item What are the values of $y'$ when $t$ takes large positive values? When $t$
takes large negative values?
\item Draw the fixed points curve and inflexion curve.
\item Draw some of the specific solutions.
\end{itemize}

}

\newpage
\exo{
Sketch some of the specific solutions of the following:

\begin{enumerate}
\item $y'=y(1-t)$
\item $y'=t^2y$
\item $y'=y+t^2$
\item $y'=t+e^y$
\end{enumerate}
}
%-------------------------------------------------------------------------------

\section{Solving}

\textbf{Reminder:} The differential $dy$ is the first order approximation (along the tangent line). If $y$ is a function of $t$, then
\[dy=y'\cdot dt\]
and \[y=\frac{dy}{dt}\]

\tproof{Since $dy$ and $dt$ are values (and not limits of values) no specific differential forms are needed.

The chain rule is hidden here:

If $u$ and $v$ are functions of $t$ and  $du=dv$, then we have $u'\cdot dt=v'\cdot dt$ hence $u'=v'$ and $u=v+k$.
}

\Definition{First Order}{
If only the first derivative appears in the equation, then it is a First Order
Differential Equation 
$$y'=\frac{dy}{dt}=f(t,y)$$
}\index{differential equation!first order}

\Definition{Separable Variables}{
If the differential equation is of the form
$$\frac{dy}{dt}=f(t)\cdot g(y)$$
then it is a differential equation with separable variables}
\index{differential equation!separable variables}

\bigskip
When solving differential equations we consider $df(x)$ and $dx$ as "separable". The easy way to understand this is to imagine that $dx$ is an infinitely small quantity and that $df(x)$ is the corresponding variation of the function along the tangent line to $f$ at point $<x;f(x)>$ 

\Definition{Differential}{
The derivative of $f(x)$ with respect to $x$ is noted either $f'(x)$ or
$$\frac{df(x)}{dx}$$.

The differential is $$df(x)=f'(x)\cdot dx$$
}\index{differential}



\definition{An initial value is a given value for $t$ and the
  corresponding value for $y$ generally in the form 
$$y(t_0)=y_0$$
 }\index{initial value}

%============================================================




%============================================================
\exo{
Find the general solutions of the following:

\begin{multicols}{2}
\begin{enumerate}
\item $\ds y'=y\cdot(y+1)$
\item $\ds y'=t\cdot\sin(t^2)$
\item $\ds y'=e^{-y}$
\item $\ds y'=y\cdot\tan(t)$
\item $\ds (t-1)\cdot y'-2y=0$
\item $\ds y'=t\cdot y\cdot\ln(t)$
\end{enumerate}
\end{multicols}
}

\exo{
Solve the initial value problems:

\begin{enumerate}
\item $\ds y'=y^2\cdot t^2$\qquad initial value: $y(1)=2 $
\item $\ds y'=t\cdot\sqrt{y}$\qquad  initial value $ y(0)=3$
\item $\ds y'=\frac{\ln(t)}{y}$\qquad  initial value $ y(1)=-2$
\item $\ds y'=t\cdot y-y+2t-2$\qquad  initial value $y(0)=0$
\item $\ds y'=(y^2-3y+2)\cdot\sqrt{t}$\qquad  initial value $y(1)=2$
\end{enumerate}
}

\exo{
Let $\Gamma_k$ be a given differentiable curve such that the
  tangent through each point $<x;y>$ of $\Gamma_k$ intercepts the
  vertical axis at $<0;k\cdot y>$. 

Which are the curves that satisfy this property?
(What can be said of $\mbox{k=0, k=1}$)

(You will need to write the general formula for the equation of the
straight line tangent to $\Gamma_k$ at $<x_0;y_0>$ using $y$ as
dependent variable of the curve $\Gamma_k$)
}

%============================================================
\index{homogeneous}\index{linear}\index{first order}
\index{differential equation!homogeneous linear}

\Definition{First Order Homogeneous Linear DE}{
An equation of the form $$y'+p(t)\cdot y=0$$ is called a First Order
Homogeneous Linear Differential Equation.

Linear, because $y'$ and $y$ only appear linearly.

Homogeneous, because the right hand side of the equation is zero.
}


\exo{
Prove the following theorem:

\theorem{
Let $P(t)$ be an antiderivative of $p(t)$,
  then $$y'+p(t)\cdot y=0$$ has the following solution:

$$y(t)=C\cdot e^{-P(t)}$$}
}

\exo{
A bacterial culture grows at a rate proportional to its
  population. If it has a population of one million at time $t=0$
  and $1.5$ million at $t=1h$, find its population as a function
  of $t$
}

\exo{
A radioactive element decays with a half life of 6
  years. Starting with 10kg at time $t=0$, find the amount of initial matter as a
  function of $t$
}

\exo{
Solve the initial condition problem:

$$y'+y\cdot\cos(t)=0$$

\begin{enumerate}
\item with $y(0)=\frac{1}{2}$
\item with $y(\pi/4)=\frac{1}{2}$
\end{enumerate}

Make a rough sketch of the two curves.
}

\exo{
Solve $$\ds y'+3t^{-1}y=0$$ with $y(1)=2$

Check what is happening in the neighbourhood of $t=0$

Make a rough sketch of the solution.
}

%============================================================

\exo{
Note that a homogeneous DE always has $y=0$ (the constant
function 0) as a solution. Thus in the following, non trivial
solutions are required.

\bigskip

Solve:

\begin{multicols}{2}
\begin{enumerate}
\item $\ds y'+5y=0$
\item $\ds y'+\frac{y}{1+t^2}=0$
\item $\ds y'-2y=0$
\item $\ds y'+t^2y=0$
\end{enumerate}
\end{multicols}
}

\exo{
Solve the initial condition equations:

\begin{enumerate}
\item $\ds y'+y=0$\qquad initial value: $y(0)=4$
\item $\ds y'+y\sin(t)=0$\qquad initial value: $y(\pi)=1$
\item $\ds y'+y\sqrt{1+t^2}=0$\qquad initial value: $y(0)=0$
\item $\ds y'+y\cos(e^t)=0$\qquad initial value: $y(0)=0$
\item $\ds ty'-2y=0$\qquad initial value: $y(1)=4\quad t>0$
\item $\ds t^2y'+y=0$\qquad initial value: $y(1)=-2\quad t>0$
\item $\ds t^3y'=2y$\qquad initial value: $y(1)=1\quad t>0$
\item $\ds t^3y'=2y$\qquad initial value: $y(1)=0\quad t>0$
\item $\ds y'-3y=0$\qquad initial value: $y(1)=-2$
\item $\ds y'+ye^t=0$\qquad initial value: $y(0)=e$
\end{enumerate}
}

%============================================================
\index{non homogeneous}
\index{differential equation!non homogeneous}

\Definition{First Order Linear Differential Equation}{An equation of the form $$y'+p(t)\cdot y=f(t)$$ is called
a First Order Linear Differential Equation.}

Example: A population has a net birthrate $b(t)$. Hence its rate of
variation will depend on the size of the population and the birthrate:
$b(t)\cdot y$. 

There is also a net immigration rate $f(t)$ which only depends on
time. 

Therefore the variation of population will be 
$$y'=b(t)\cdot y+f(t)$$

\bigskip
The idea is to make the left hand side look like the derivative of a product. Multiply both sides by a function $w$ (function of $t$: in fact $w(t)$) - called an \textbf{integrating factor:}

\[w\cdot y'+w\cdot p\cdot y=w\cdot f\]

\exo{What should $w$ be so that
$w(t)\cdot y'(t)+w(t)\cdot p(t)\cdot y(t)=(w(t)\cdot y(t))'$?

(Do not forget the chain rule! )
}

Once $w$ is determined, we have

\[wy=\int w(t)f(t) dt+C\]

dividing by $w$ gives the solution. 

\tproof{I find this method -- with integrating factor $w$ much more elegant than the variation of constants method. In addition, the idea of varying constants is very upsetting for students and sometimes leads to major misunderstandings. (is the integration constant really constant?). 

It is given as a method, hence there is no theorem for this solving method.
}

\exo{
Use this method to solve $$y'+3t^{-1}\cdot y=t^2$$
for $t>0$ and $y(1)=\frac{1}{2}$
}


\exo{
A population has a net birthrate of 2\% per year and a net
  immigration rate of $100 000\sin(t\cdot 2\pi)$ (The sine function
  reproduces the seasonal variation). At $t=0$ the population is
  $10^6$.

Find the population as a function of $t$.
}


\exo{

Find the general solution of the differential equation:
\begin{multicols}{2}[]
\begin{enumerate}
\item $y'+4y=8$ 
\item $y'-2y=6$ 
\item $y'+ty=5t$ 
\item $y'+e^ty=-2e^t$ 
\item $y'-y=t^2$ 
\item $2y'+y=t$ 
\item $ty'+2y=1/t\quad t>0$ 
\item $ty'+y=\sqrt{t}\quad t>0$
\end{enumerate}
\end{multicols}
}

\exo{
Writing equations:

\begin{enumerate}

\item Consider the following problem proposed by Claude Perrault, from
  Paris in 1674. It describes the path of a silver pocket watch which
  is pulled by its chain. The curve is called the ``tractrix.''\index{tractrix}

Take a chain-watch and pull it by the chain, the end of the chain following the straight edge of the table. If the starting position of the watch is when the chain is perpendicualr to the table, at the exact distance if the length of the chain, what will be the equation of the path of the watch?

$a$ is the length of the watch chain (constant). $d$ can be calculated
with respect to $y$ and $x$ and $a$

The chain is tangent to the curve of the path, because the watch is
pulled in the direction of the chain. 

\begin{enumerate}
\item Write the (differential) equation for $y$. In fact, it will be
  $x$ with respect to $y$.
\item This is not easy to solve, but check that
$$x=-\sqrt{a^2-y^2}-a\cdot\ln\left(\frac{a-\sqrt{a^2-y^2}}{y}\right)$$ is
  the solution.
\end{enumerate}

\end{enumerate}
}

%============================================================
\section{Numeric approximation: the Euler method}
\index{numeric approximation!Euler method}\index{Euler approximation}
Very often, an explicit solution cannot be found.

Let $y'=f(t;y)$ with $y(a)=y_0$

then $Y(a+\Delta t)=y_0+f(a;y_0)\cdot \Delta t$ where $Y(x)$ is an
approximation of the exact solution $y(s)$

\tproof{Describe the approximation along a path with steps of, say, $0.5$. Then with steps of $h$. Set $h$ positive ultrasmall and you have the expression for the integral.}

\exo{

\begin{enumerate}
\item Write $Y(a+2\Delta t)$
\item Write $Y(a+k\Delta t)$ using the $\sum$ sign.
\item Show that if $a+k\Delta t=s$ then
$$Y(s)=y_0+\sum_{t=a}^sf(t;y(t))\cdot \Delta t$$
\end{enumerate}
}

\exo{
The equation $y'=t-y^2$ is not linear, not separable.

Compute the Euler method for $y(0)=0$, $\Delta t=0.2$ and $t\in[0;1]$

Sketch the curve.
}

\exo{
Same equation and initial value. $\Delta t=0.1$ Compute and
  sketch the curve for the same interval.
}

\exo{
Compute the Euler approximation for $\ds y'=\frac{t\sin(t)}{y}$
  \quad $y(2)=5$ for $t\in[2;8]$

(Use of a spreadsheet may help.) 
}



\end{document}


\exo{
In relativity, the mass is $m=m_0/\sqrt{1-(\frac{v}{c})^2}$ at velocity
$v$. ($c$ is the velocity of light). Using former results, show that
if $v$ is small compared to $c$, then $m=m_0+?$ 
}


\exo{
Imagine an empty universe except for the earth and one object
  infinitely far away. This object is attracted to the earth (and vice
  versa, but being earth-people we will consider us to be the centre
  of interest...)

At first, the gravitational pull is very weak and the object starts to
move slowly towards the earth. Gradually it builds up velocity \dots\ until it
hits the ground. The final velocity is (up to its direction) the one
that is needed as initial velocity to go back to infinity. This velocity
is the escape velocity -- the one that it needed to escape from
earth's attraction, the one that a rocket must reach to leave the
orbit of the earth.

Consider energy: the total work is $F\cdot d$ -- force times distance,
where as above $F=G\dfrac{m_1\cdot m_2}{d^2}$ depends on the distance
$d$. Work is energy.

The kinetic energy is $\frac{1}{2}m\cdot v^2$ and by the law of
conservation of energy, these two quantities must be equal.

$G=6.7259\cdot 10^{-11}[N\cdot m^2\cdot kg^{-2}]$, the mass of the
earth is $5.9742\cdot 10^{24}[kg]$ and the radius of the earth is
$6.36\cdot 10^6[m]$

Find the total work of the object and hence find the escape velocity.
}

\end{document}