Discover Euler's Method Examples

y′=f(t,y) with Euler's Method
I'm trying to compute the approximation solution for
$y^{\prime }=-<!--\; -\; -->2y+2-<!--\; -\; -->e^{-<!--\; -\; -->4t}:0\le <!--\; \le \; -->t\le <!--\; \le \; -->5,y(0)=1$
but the answer that I get for $n=100$ intervals is −5.069 which isn't right. What's the issue with my algorithm?

valueA = 0;
valueB = 5;
double y0 = 1;
double y1 = 0;
double t = 0;
valueN = 100;
double valueM = 0;

double functionOne = 0;
double h = (valueB-valueA)/valueN;
for (double i = 0; i <= 5.00; i+=h){
valueM = ((-2*y0) + 2 - Math.exp(-4*t));

y1 = y0 + (t * valueM);
y0 = y1;
t += h;

}
return y1;

First Ten Values:
1. 1.0
2. 0.959063462346101
3. 0.9002187652733168
4. 0.8478313902772178
5. 0.8188330413428864
6. 0.8174466603785826
7. 0.8366204005777724
8. 0.8646771827937695
9. 0.8921768293608917
10. 0.9148331832363752
11. 0.9323323583816936

We have the linear first order ODE $y^{\prime }(t)=m(t)y(t)+n(t)$ where $m(t),n(t)$ are given functions. We need to show that after applying the implicit Euler method we receive a linear system of equations for the variable $y_{k+1}$.

Let $h>0$ be the step size, $t_0\in <!--\; \in \; -->\mathbb{R}$ such that the ODE suffices the initial value condition $y(t_0)=y^0$. The implicit Euler method in our case is given by $y_0=y^0$ and $y_{k+1}=y_k+h(a(t_{k+1})y_{k+1}+b(t_{k+1}))$.

This leads to $y_{k+1}=\frac{y_k+hn(t_{k+1})}{1-<!--\; -\; -->hm(t_{k+1})}$.

My questions:
1. How does this make a linear system of equations or what do we mean with that? Here we have only one equation and in what sense is this equation linear? Usually I would associate a system of linear equations with something like $Ax=b$.
2. We use the implicit Euler method. But is it correct if the result is an explicit equation? That confuses me a bit.
Any help is very appreciated.

While using Euler's method for $\frac{dy}{dx}=y$ given $y(0)=1$ I noticed that the approximate solution is $(1+1/s{)}^{sx}$ where s is the interval size. Plug in one for $x$ and you get:
$e^1=\underset{s\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(1+1/s{)}^s$
you get $e$. This has to be $e$ because that is that is the solution to the differential equation. Is this a new proof that
$e=\underset{s\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(1+1/s{)}^s$
List item

I'm new to numerical analysis and got stuck on an ODE:
$y^{″}+y^3-<!--\; -\; -->0.5=0$
with $h=0.1$, $y(0)=1$ and $y^{\prime }(0)=1$.
How should I begin?

I was just doing some practice questions for a test, but have been stumped by the following for the past couple of hours.
I'm given a system such that:
$\frac{du}{dt}=v\mathrm{\&<!--\; \&\; -->}\frac{dv}{dt}=-<!--\; -\; -->f(u)$
with Hamiltonian
$H=\frac{1}{2}{\left(\frac{du}{dt}\right)}^2+\int <!--\; \int \; -->fdu.$
I have to show that using the forward Euler method leads to a global error for $H$ that grows like $n{h}^2$ for step size $h$ and number of steps $n$.
I know that the global error can be calculated via $\mathrm{ϵ<!--\; ϵ\; -->}=|U^n-<!--\; -\; -->U(T)|$ but I'm not sure how to apply it in this case.
Thanks for any help!
EDIT: So if I understand correct, I have to calculate $|H_{n+1}-<!--\; -\; -->H_n|$

If I have the following differential equations:
$\{\begin{array}{l}\stackrel{\dot{}<!--\; \dot{}\; -->}{x_s}=\mathrm{\gamma <!--\; \gamma \; -->}{y}_s+x_s(R_c^2-<!--\; -\; -->x_s^2-<!--\; -\; -->y_s^2)\\ \stackrel{\dot{}<!--\; \dot{}\; -->}{y_s}=-<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->}{x}_s+y_s(R_c^2-<!--\; -\; -->x_s^2-<!--\; -\; -->y_s^2)\end{array}$
that represent the coordinates x and y of a robot following a limit-cycle trajectory, How to use Euler method to find the approximations of ($x_s$)?

Consider the IVP
$\frac{dy}{dt}(t)=f(t,y(t)),t\in <!--\; \in \; -->[a,b],y(a)=\mathrm{\alpha <!--\; \alpha \; -->}\in <!--\; \in \; -->\mathbb{R}.$
One can show that Euler's method, i.e., the scheme
$y_{i+1}=y_i+hf(t_i.y_i),y_0=\mathrm{\alpha <!--\; \alpha \; -->},i=0,1,2,\dots <!--\; \dots \; -->$
has local truncation error $\mathcal{O}(h^2)$ and global error $\mathcal{O}(h).$
For higher order ODEs one can rewrite the ODE as a first order system and then apply Euler as before.
Should the local truncation error still be $\mathcal{O}(h^2)$?
I tried this for a second order ODE
$y^{″}(t)=f(t,y(t),y^{\prime }(t)),t\in <!--\; \in \; -->[a,b],y(a)=\mathrm{\alpha <!--\; \alpha \; -->},y^{\prime }(a)=\mathrm{\beta <!--\; \beta \; -->}$
and arrived at
$y_{i+2}=2y_{i+1}-<!--\; -\; -->y_i+h^{2}f(t_i,y_i,f(t_i,y_i,\frac{y_{i+1}-<!--\; -\; -->y_i}{h})).$
Then the local truncation error is
$y(t_{i+1})-<!--\; -\; -->y_{i+1}=\mathcal{O}(h^3).$
This of course contradicts my expectation that the error would still be $\mathcal{O}(h^2)$.

$x^{″}-<!--\; -\; -->t{x}^{\prime }+x^2=t$
$x(0)=1$
$x^{\prime }(0)=1$
a) Restate the problem solving a system of first-order ODEs.
$x_1^{\prime }=x_2$
$x_2^{\prime }=t-<!--\; -\; -->x_1^2+t{x}_2$
$x_1(0)=1$
$x_2(0)=1$
b) Use part a) and Euler's method with h = 0.1 to find x(0.2).
$x(0.1)=1+(0.1)(0-<!--\; -\; -->1+0)=0.9$
Is my work up to this point correct? I'm unsure whether this should actually be two equations.

Using Euler's method (the method of eigenvalues and eigenvectors, the case of different complex eigenvalues) solve the sytem of differential equations $\stackrel{\to <!--\; \to \; -->}{y^{\prime }}$ = A$\stackrel{\to <!--\; \to \; -->}{y}$ or solve the initial value problem $\stackrel{\to <!--\; \to \; -->}{y^{\prime }}$= A$\stackrel{\to <!--\; \to \; -->}{y}$, $\stackrel{\to <!--\; \to \; -->}{y}$(0) = ${\stackrel{\to <!--\; \to \; -->}{y}}_0$, if
$\left[\begin{array}{cc}-<!--\; -\; -->7& 1\\ -<!--\; -\; -->2& -5\end{array}\right]$
My attempt: (but unfortunately it is not Euler's method, i don't understand this method):
So i have started from forming a new matrix by subtracting $\lambda$ from the diagonal entries of the given matrix:
$\left[\begin{array}{cc}-\lambda -<!--\; -\; -->7& 1\\ -<!--\; -\; -->2& -\lambda -5\end{array}\right]$
The determinant of the obtained matrix is ${\lambda }_2+12\lambda +37$
Solve the equation ${\lambda }^2+12\lambda +37=0.$
The roots are $\lambda 1=-6-i,\lambda 2=-6+i$, these are the eigenvalues.
Next, i have found the eigenvectors.
1. $\lambda =-6-i$
$\left[\begin{array}{cc}-\lambda -<!--\; -\; -->7& 1\\ -<!--\; -\; -->2& -\lambda -5\end{array}\right]=\left[\begin{array}{cc}-1+i& 1\\ -<!--\; -\; -->2& i+1\end{array}\right]$
The null space of this matrix is $\left[\begin{array}{c}1/2+i/2\\ 1\end{array}\right]$
This is the eigenvector.
2. $\lambda =-6+i$
$\left[\begin{array}{cc}-\lambda -<!--\; -\; -->7& 1\\ -<!--\; -\; -->2& -\lambda -5\end{array}\right]=\left[\begin{array}{cc}-1+i& 1\\ -<!--\; -\; -->2& i-<!--\; -\; -->1\end{array}\right]$
The null space of this matrix is $\left[\begin{array}{c}1/2-<!--\; -\; -->i/2\\ 1\end{array}\right]$
and this is the eigenvector.
And i dont know what i should do next to calculate the fundamental equations (formula) or any other solution is of the following form

I want to understand the explicit and the implicit Euler method better. Assume I have the initial value problem $y^{″}+y=0$, $y(0)=0$, $y^{\prime }(0)=1$, which is of course solved by $y=\sin <!--\; \; -->(x)$, and I convert this into a system of first-degree linear equations
$\left(\begin{array}{c}{y}^{\prime }\\ z^{\prime }\end{array}\right)=\left(\begin{array}{cc}0& 1\\ -<!--\; -\; -->1& 0\end{array}\right)\left(\begin{array}{c}y\\ z\end{array}\right)\text{.}$
When I apply the explicit and the implicit Euler methods to the differential equation $y^{″}+y=0$, with stepwidth $h>0$, I get approximation points that satisfy:
$\begin{array}{rcll}{y}_k& =& \frac{ih}{2}((1-<!--\; -\; -->ih{)}^k-<!--\; -\; -->(1+ih{)}^k)& \text{for the explicit Euler method}\\ {\stackrel{~<!--\; ~\; -->}{y}}_k& =& \frac{ih}{2(1+h^2{)}^k}((1-<!--\; -\; -->ih{)}^k-<!--\; -\; -->(1+ih{)}^k)& \text{for the implicit Euler method}\end{array}$
I am interested in seeing how fast these sequences grow with k. My guess is that the first sequence is unbounded, whereas the second converges to zero. At least that is what I get for , but I have no idea how to prove this. Any suggestions?

I have the following system:
$\begin{array}{rl}\stackrel{\dot{}<!--\; \dot{}\; -->}{x}& =-<!--\; -\; -->y\\ \stackrel{\dot{}<!--\; \dot{}\; -->}{y}& =x\end{array}$
Given that $(x_i^n,y_i^n)$ are the points obtained for $i=1,2\dots <!--\; \dots \; -->n^2$ using a time-step $h=1/n$ starting at the initial point $(x_0,y_0)=(1,0)$.
I am struggling to find the following limit:
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(x_n^n,y_n^n)$
I can intutively think of the above limit ending up somewhere on the unit circle but I am unable to obtain the value of the exact limit. Pardon if its too simple, but I am surely missing out something. Please help out, thanks in advance.

Consider this differential equation,
$dy/dx=x+\sin <!--\; \; -->(y)$
with initial condition $y=0.5$ when $x=1.2$:
1. Write down the recurrence relation for Euler's numerical method applied above.
2. With step size h = 0.1, calculate the approximations to y(1.3) and y(1.4).

1. This is my answer:
$Y_i+1=Y_i+0.1(X_i+\sin <!--\; \; -->(Y_i))$
2. I have problem solving this one...
From my txtbk reference, I sub $i=1$ into the eqn thus giving me
$Y_1=Y_0+0.1(1.2+\sin <!--\; \; -->(0.5))=0.1679$
Then, $i=2$
$Y_2=Y_1+0.1(1.3+\sin <!--\; \; -->(0.6)$
Im sure my method in part 2 is wrong and I don't really understand how to solve it via my txtbk.

Solve the initial value problem related to $H(x,y)=x^2+y^2-<!--\; -\; -->1=0$ and $(x_0,y_0)=(1,0)$, using the Euler method.
The problems I've seen on the internet always give the $y^{\prime }$ in terms of $y$ and $x$ so what I did first was tro try to find this $y^{\prime }$ since I don't know it yet.
By implicit differentiation I have that $y^{\prime }=-<!--\; -\; -->x/y$
Now, If I set my $h=0.1$ for example, you can see on the first iteration I'll have an undefined value:
$x_0=1,y_0=0$
$x_1=x_0+h=1.01,y_1=y_0+y^{\prime }(x_0,y_0)=0+-<!--\; -\; -->1/0\to <!--\; \to \; -->UNDEFINED$
But I don't think this makes any sense, perhaps I differentiated incorrectly or this is just not the right way to go. I'd like someone to help me understand the very first step which is obtaining an ODE for a given function.

use euler's method with step size $h=0.15$ to compute the approximate value of $y(0.6)$,correct upto five decimal places from initial value problem
$y^{\prime }=x(y+x)-<!--\; -\; -->2,y(0)=2$
Actually, I have found the answer for $y(0.6)$ with $h=0.15$ but i doesn't know wether the answer is correct upto 5 decimal places or not. How can i check the error and get the answer correct upto 5 decimal places?

How can I show that the Euler method fails to approximate the exact solution
$y(x)=(2x/3{)}^{3/2}$
to the IVP
$y^{\prime }=y^{\frac{1}{3}}$
$y(0)=0$
Here we have $f(t,y)=y^{\frac{1}{3}}$, $y_0=0$ and so $f(t_0,y_0)=f(0,0)=0$ and
$y_{n+1}=y_n+hf(t_n,y_n)$
Thus
$$\begin{gathered} y_1=0 \\ {y}_2=0 \\ ⋮<!--\; ⋮\; --> \\ {y}_n=0 \end{gathered}$$
So, I can't understand why it fails. Could you help me?

$y^{″}=11-<!--\; -\; -->y$
$y(2)=1;y^{\prime }(2)=-<!--\; -\; -->4$
and asked to use Euler's method to find $y(2.2)$ for $h=0.1$
To find $y^{\prime }$ I simply took the integral of $y^{″}$ to get:
$y^{\prime }=11x-<!--\; -\; -->yx$
However, this does not satisfy the condition given above, that $y^{\prime }(2)=-<!--\; -\; -->4$. Is this not the correct way of obtaining the first order derivative?

Use the two second-order multi-step methods
${\omega }_{i+1}={\omega }_i+\frac{h}{2}(3f_i-f_{i-1})$
and
${\omega }_{i+1}={\omega }_i+\frac{h}{2}(f_{i+1}+f_i)$
as a pblackictor-corrector method to compute an approximation to $y(0.3)$, with stepsize $h=0.1$, for the IVP;
$y^{\prime }(t)=3ty,y(0)=-1.$
Use Euler’s method to start.

I do not understand how to use these methods to approximate $y(0.3)$. Moreover I am not sure how Euler's method fits into this question. Could someone clarify this question please?

If we want to apply the forward Euler method to $x^{″}=x$ with $x(0)=0,x^{\prime }(0)=1$, we can introduce a new function
$u:=\left[\begin{array}{c}{x}^{\prime }\\ x\end{array}\right]$
then
$u^{\prime }=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]u\text{and}u(0)=\left[\begin{array}{c}1\\ 0\end{array}\right].$
So then first step of the forward Euler would be
$\begin{array}{rl}{u}_1& =u(0)+h{u}^{\prime }(0)\\ & =\left[\begin{array}{c}1\\ 0\end{array}\right]+h\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ & =\left[\begin{array}{c}1\\ h\end{array}\right].\end{array}$
And if we translate it back we get $x_1=h$.
How can we do a similar thing with the equation $x^{″}=x^2$ (with $x(0)=0,x^{\prime }(0)=1$)? It doesn't look like it can be written as a system, so there should be some other trick.

I am attempting to compute an approximation of the solution with the forward Euler method in $[0,1]$ with step lengths $h_1=0.2$, $h_2=0.1$ given the initial value problem below
$\frac{dy}{dz}=\frac{1}{1+z}-<!--\; -\; -->y(z)y(0)=1$
I am not sure what to do when I am given two step sizes instead of one. I know how to compute it if it was given with a step size. Am I supposed to find out the approximation for two different step sizes? Or is there anything I am missing?

How does the Velocity Verlet method differ from the standard Euler method? Why do we need to add Acceleration / 2 to calculate position?