Discover Euler's Method Examples

Does the Modified Euler Method always overestimate the true solution values?

Here is the modified Euler Method:
$w_0=\mathrm{\alpha <!--\; \alpha \; -->}$
$w_{i+1}=w_i+\frac{h}{2}[f(t_i,w_i)+f(t_{i+1},w_i+hf(t_i,w_i))],i=0,1,\dots <!--\; \dots \; -->,N-<!--\; -\; -->1$
Because based on my calculations and data for a problem that I'm working on, I'm getting that it consistently overestimates the true solution...
Is it supposed to do that?

Show that two solutions $x_n$ and $y_n$ generated by implicit Euler satisfy the inequality $|x_0-<!--\; -\; -->y_0|$.

Can someone push me in the right direction by explaining to me how exactly implicit Euler's method works? I understand that it is some iteration method where
$u_{k+1}=u_k+hf(t_{k+1},u_{k+1}).$
And the $u_{k+1}$ is often found by some other iteration method (kind of shaky on this part here). I was wondering if there were some way to manipulate the equation to rewrite $x_n=x_{n+1}-<!--\; -\; -->hf(t_{k+1},x_{k+1})$ and $y_n$ similarly but I'm not sure if that would get anything done. We are also given that f satisfies the condition where $(x-<!--\; -\; -->y)(f(t,x)-<!--\; -\; -->f(t,y))\le <!--\; \le \; -->0$, which means $|x(t)-<!--\; -\; -->y(t)|\le <!--\; \le \; -->|x(0)-<!--\; -\; -->y(0)|$. Any help would be greatly appreciated. Thank you!

I have been given a system:
$y^{\prime }=Ay,y\in <!--\; \in \; -->{\mathbb{R}}^d,A\in <!--\; \in \; -->\mathbb{R}$
for this, I believe I should use this formula for the "upper bound" on the truncation error:
$\frac{1}{2}{h}^2{y}^{″}({\mathrm{\xi <!--\; \xi \; -->}}_j),t_j\le <!--\; \le \; -->{\mathrm{\xi <!--\; \xi \; -->}}_j\le <!--\; \le \; -->t_j+h$
However, I am having trouble wrapping my head around what exactly $y^{″}({\mathrm{\xi <!--\; \xi \; -->}}_j)$ is. Is it $\frac{d^{2}y}{d{t}^2}$, and how do I find this?

This is the question:

Find the error for Euler method of solving initial value problem
$y\prime =\frac{y}{x+1},y(0)=1$
on the interval $[0,1]$.
I'm confused by the question. Am I supposed to find something that prevents me from using the Euler method to solve this problem? If it is, how do I work through the problem to find the error? Thanks.

I have equation:
$\begin{array}{r}\frac{\mathrm{d}y}{\mathrm{d}t}=-<!--\; -\; -->0.04\sqrt{y}\end{array}$
How would I find the expression for Euler's method? I know the general expression is:
$y_n=y_{n-<!--\; -\; -->1}+h\cdot <!--\; \cdot \; -->F(t_{n-<!--\; -\; -->1},y_{n-<!--\; -\; -->1})$
but I am confused with what to use as x and y and how I could convert $-<!--\; -\; -->0.04\sqrt{y}$ into that form.
Edit: increments of 1 second, $y(0)=3$.

Consider the initial value problem
${\displaystyle \frac{dy}{dx}}=y,y(0)=1$
Approximate $y(1)$ using Euler's method with a step size of ${\displaystyle \frac{1}{n}}$, where n is an arbitrary natural number. Use this approximation to write Euler's number $e$ as a limit of an expression in $n$. How large do you have to choose $n$ in order to approximate $e$ up to an error of at most 0.1? Comment on the quality of approximate in this example.
What I did is the following:
$y(1)\approx <!--\; \approx \; -->y_1=y_0+hf(x_0,y_0)=1+hf(0,1)=1+{\displaystyle \frac{1}{n}}$
This is where I stuck, am I on the right direction? What should I do next?

I just wanted to see if I went about solving this correctly. The problem didn't provide a step so I'm using .1:
$x^{\prime }(t)=1+t\sin <!--\; \; -->(tx)$
Where $x(0)=0$ at $t=1$
So using the idea that:
$y_1=y_0+h\bullet <!--\; \bullet \; -->F(x_0,y_0)$
I did the following:
$y_1=0+.1[1+(.1)\sin <!--\; \; -->(1(0))]=.1$
$y_2=.1+.1[1+(.1)\sin <!--\; \; -->(1(.1))]=.2$
I repeated this same formula 6 more times. Did I apply Euler's method correctly?

We have been told that $u^{\prime }(t)=u(t)$ and $u(0)=1$ from this information we can conclude that $u(t)=e^x$
Show that $u_k=(1+h{)}^k,k=0,1,...$ is an approximation for $u(kh)$ using Euler's method
My current thoughts are that h is the interval by which we increment, but I am not sure where to go from there.

Thus far we've taken the step size $h$ to be positive, and therefore we've developed solutions to the right of the initial point. Is Euler's method valid if we use a negative step size $h<0$ and develop a solution to the left?
I think it is not possible.How we can explain that it is possible or not?

How can we use Euler's method to approximate the solutions for the following IVP below:
$y^{\prime }=-<!--\; -\; -->y+t{y}^{1/2},\text{ with }1\le <!--\; \le \; -->t\le <!--\; \le \; -->2,y(1)=2,$
and with $h=0.5$
The main concern is the organization, i.e., set up of it for this particular example.

And, if the actual solution to the IVP above is:
$y(t)=(t-<!--\; -\; -->2+\sqrt{2}\mathrm{e}\cdot <!--\; \cdot \; -->{\mathrm{e}}^{-<!--\; -\; -->t/2}{)}^2$
then, how to compare the actual error and compare the error bound?

I'm trying to approximate the solution to this equation using the Euler's method:
$y^{\prime }(x)=3-<!--\; -\; -->\tan <!--\; \; -->(x)\cdot <!--\; \cdot \; -->y(x),y(2)=4$
When solving for the step of $0.2$, I don't know what to calculate when plugging $y(2.2)$.
What is $y(x)$? Is it the derivative of $3-<!--\; -\; -->\tan <!--\; \; -->(x)\cdot <!--\; \cdot \; -->y(x)$?

Let $y^{\prime }(s)=y^2(s),y(0)=1$ be an ode on $s\in <!--\; \in \; -->{\mathbb{R}}_{+}\setminus <!--\; \setminus \; -->\{1\}$. Let $h_n=s_{n+1}-<!--\; -\; -->s_n$. Prove the following inequality with the Euler method $\mathrm{\forall <!--\; \forall \; -->}n:s_n<1$ the following holds
$y_n\le <!--\; \le \; -->\frac{1}{1-<!--\; -\; -->s_n}$
The explicit Euler method is:
$y_{n+1}=y_n+h_{n}f(s_n,y_n)$

The problem is, I don't know if I have to use somewhere Gronwall's lemma or if it is straight calculation. The solution to the ODE is $y(s)=\frac{1}{1-<!--\; -\; -->s}$ and here is what I've tried so far
$y_{n+1}=y_n+h_{n}f(s_n,y_n)$
becomes
$y_n=y_{n+1}-<!--\; -\; -->h_{n}f(s_n,y_n)$
$y_n=y_{n+1}-<!--\; -\; -->(s_{n+1}-<!--\; -\; -->s_n)y_n^2$
But then I get stucked, since I don't see how I can remove the $y_{n+1}$ and $s_{n+1}$ terms without getting new recursive terms. Or do I have to use the geometric series?

I am approximating a solution to a first order LODE using Euler's method. I made two tables, one using a step size of .01 and another using .05 ( I had to start at x=0 and end at x=1). I am not understanding the directions for the second part of my assignment:

It states that the order of numerical methods (like Euler's) is based upon the bound for the cummulative error; i.e. for the cummulative error at, say x=2, is bounded by $C{h}^n$, where $C$ is a generally unknown constant and $n$ is the order. For Euler's method, plot the points:
$(0.1,\mathrm{\varphi <!--\; \varphi \; -->}(1)-<!--\; -\; -->y_{10}),$
$(.05,\mathrm{\varphi <!--\; \varphi \; -->}(1)-<!--\; -\; -->y_{20}),$
$(.025,\mathrm{\varphi <!--\; \varphi \; -->}(1)-<!--\; -\; -->y_{40}),$
$(.0125,\mathrm{\varphi <!--\; \varphi \; -->}(1)-<!--\; -\; -->y_{80}),$
$(.00625,\mathrm{\varphi <!--\; \varphi \; -->}(1)-<!--\; -\; -->y_{160})$
And then fit a line to the above data of the form $Ch$. I don't understand, am I supposed to plot these using a step size of .1 or .05? Or am I supposed to use another step size?
Any clarification is appreciated.
Thanks
Edit:
The LODE I am given is $y^{\prime }=x+2y,y(0)=1$ and the exact solution I found was $\mathrm{\varphi <!--\; \varphi \; -->}(x)=\frac{1}{4}(-<!--\; -\; -->2x+5e^{2x}-<!--\; -\; -->1)$.

I've been told numerical methods in solving ODEs, such as Euler's method and Runge-Kutta, are all in some way approximations to Picard's iteration, and I'm trying to understand how.

Suppose we have a differential equation on an interval $[x_0,x_L]$:
$\frac{dy}{dx}=f(x,y)$
with initial condition $y(x_0)=y_0$

I would like to numerically solve the equation on a set of points $\{x_0<x_1<\cdots <!--\; \cdots \; --><x_n\}$, i.e. obtain approximations $y_i$ to the true solution $y(x_i)$ for each $x_i$.
Picard's iteration works as follows:
$y_{0,0}=y_0$
$y_{0,k}(x_1)=y_0+{\int <!--\; \int \; -->}_{x_0}^{x_1}f(x,y_{0,k-<!--\; -\; -->1}(x))dx\mathrm{f}\mathrm{o}\mathrm{r}k\ge <!--\; \ge \; -->1$
suppose we stop for $k=m$, then take $y_1=y_{0,m}(x_1)$
We then repeat the process for $i\ge <!--\; \ge \; -->1$.
$y_{i,0}=y_i$
$y_{i,k}(x_{i+1})=y_i+{\int <!--\; \int \; -->}_{x_i}^{x_{i+1}}f(x,y_{i,k-<!--\; -\; -->1}(x))dx\mathrm{f}\mathrm{o}\mathrm{r}k\ge <!--\; \ge \; -->1$
$y_{i+1}=y_{i,m}(x_{i+1})$
So is the idea of a numerical method (e.g. Euler's method) to replace the integral ${\int <!--\; \int \; -->}_{x_i}^{x_{i+1}}f(x,y_{i,k-<!--\; -\; -->1}(x))dx$ with an approximation such as $(x_{i+1}-<!--\; -\; -->x_i)f(x_i,y_i)$ (Euler's method)? What I don't understand is why numerical methods only iterate once for each point $x_i$ (in other words, $m=1$) but Picard's iteration suggests you should iterate multiple (potentially many) times for each $x_i$?

We learned Euler's method today there is one hw problem totally stunned my hat off. It says: Use Euler's method to approximate ${\int <!--\; \int \; -->}_0^2{e}^{-<!--\; -\; -->u^2}du$. I know Euler's method is $y_{n+1}=y_n+hf(t_n,y_n)$, but this is for $y^{\prime }$ right? somehow I can compute an integral? This is supposed to be done using computer, I know how to use Euler's method to solve equation like $y^{\prime }=2y-<!--\; -\; -->3e^{-<!--\; -\; -->t}$.

For: $F(0)=0$ and $F^{\prime }(x)=f(x)$
Euler's method: $F(0+h)=F(0)+h{F}^{\prime }(0)=0+hf(0)$
Continuing the process, $F(10h)=hf(0)+hf(h)+hf(2h)+.....hf(9h)$
This resembles the Riemann sum: ${\mathrm{\Sigma <!--\; \Sigma \; -->}}_{i=1}^{n}f(x_i)(x_i-<!--\; -\; -->x_{i-<!--\; -\; -->1})$
Therefore my professor used Euler's method to solve integral problems.

Example: ${\int <!--\; \int \; -->}_3^{3.09}f(x)dx=\mathrm{0.81.}$ Approximate $f(3).$ $F(x)=hf(x)$
$0.81=0.09f(x)$
$f(x)=3$
My question: How did $F(x+h)=F(x)+hf(x)$ become $F(x)=hf(x)?$

The question is to use Euler's Method to approximate Y:
$Y^{″}(t)=Y^{\prime }(t)-<!--\; -\; -->2Y(t)$
$Y^{\prime }(0)=Y(0)=1$
with $t_0=0$ and $h=0.2$

So what I did:
First iteration:
$t_1=0.2$
$y(t_1)=y(t_0)+h\cdot <!--\; \cdot \; -->y^{\prime }(t_0)=1+0.2\times <!--\; \times \; -->1=1.2$
$y^{\prime }(t_1)=y^{\prime }(t_0)+h\cdot <!--\; \cdot \; -->(y^{\prime }(t_0)-<!--\; -\; -->2y(t_0))=1+0.2\times <!--\; \times \; -->(1-<!--\; -\; -->2\times <!--\; \times \; -->1)=0.8$
Second iteration:
$t_2=0.4$
$y(t_2)=y(t_1)+h\cdot <!--\; \cdot \; -->y^{\prime }(t_1)=1.2+0.2\times <!--\; \times \; -->0.8=1.36$
$y^{\prime }(t_2)=y^{\prime }(t_1)+h\cdot <!--\; \cdot \; -->(y^{\prime }(t_1)-<!--\; -\; -->2y(t_1))=0.8+0.2\times <!--\; \times \; -->(0.8-<!--\; -\; -->2\times <!--\; \times \; -->1.2)=-<!--\; -\; -->2.4$
Correct?