Discover Euler's Method Examples

Trying to understand the proof that the numerical method of solving differential equation
$x_{i+1}=x_i+{\mathrm{\tau <!--\; \tau \; -->}}_{i}F(t_i+\frac{{\mathrm{\tau <!--\; \tau \; -->}}_i}{2},x_{i+\frac{1}{2}})$
$x_{i+\frac{1}{2}}=x_i+\frac{\mathrm{\tau <!--\; \tau \; -->}}{2}F(t_i,x_i)$
which seems to be the Cauchy-Euler's method.
I am stuck with the Taylor approximation of
$x(t+\mathrm{\tau <!--\; \tau \; -->})=x(t)+\mathrm{\tau <!--\; \tau \; -->}{x}^{\prime }(t)+\frac{1}{2}{x}^{″}(t)+O({\mathrm{\tau <!--\; \tau \; -->}}^3)$
Here it is OK for me.
I also understand, why $x^{\prime }(t)=F(t,x(t))$.
But why
$x^{″}(t)=\frac{d}{dt}F(t,x(t))={\mathrm{\partial <!--\; \partial \; -->}}_{t}F(t,x(t))+D_{x}F(t,x(t))F(t,x(t))$???
I am a little frustrated about this. Normally I am more into computer science and some mathematical concepts can be missing, so sorry for possible trivial or illposed question, thanks for patience :)

Use Euler's method to find approximate values for the solution of the initial value − problem
$\frac{dy}{dx}=x-<!--\; -\; -->y$
$y(0)=1$
on the interval [0,1] using five steps of size h=0.2.

My attempts:
I know that the recurrence relation $y_{n+1}=y_n+hf(x_n,y_n)$ however I am unable to see how the interval comes into play.An idea I had was to consider the bounds of the interval and approximate y(0) and y(1) however this does not include h so I am extremely skeptical.

I'm writing a MATLAB function to solve a system of differential equations with the Euler method. I need to estimate the local truncation error of the method but I don't know how to do.
I know that the local truncation error of a first order DE $y^{\prime }=f(y(t),t)$ is given by :
$LT{E}_i=-<!--\; -\; -->\frac{h^2}{2}{y}^{″}({\mathrm{\xi <!--\; \xi \; -->}}_i)$
where $h$ is the step size and ${\mathrm{\xi <!--\; \xi \; -->}}_i\in <!--\; \in \; -->[t_{i-<!--\; -\; -->1},t_i]$.
What is the expression for the local truncation error of the system $\stackrel{\to <!--\; \to \; -->}{y^{\prime }}=f(\stackrel{\to <!--\; \to \; -->}{y}(t),t)$?

What is the remainder when ${17}^{100000000\dots <!--\; \dots \; -->}$, with $n$ zeroes is divided by 20? I.e.:
${17}^{{10}^n}\mathrm{mod}20=?$
My approach: I can solve it by Euler's totient function ; ${17}^8=1\mathrm{mod}20$, so remainder is 1.
But I want to know if the following method will fetch the result: 17 divided by 20 gives −3 as remainder and −3 raised to any multiple of $10\mathrm{mod}20$ gives 1?
Is this logic correct? I think it works only for positive remainder.

Use Euler's method to approximate the solution for the following initial value problem. $y^{\prime }=t{e}^{3t}-<!--\; -\; -->2y,0\le <!--\; \le \; -->t\le <!--\; \le \; -->1,y(0)=0\text{ with }h=0.5$
h is the step size The initial condition is
$$\begin{gathered} y(0)=0\Rightarrow <!--\; \Rightarrow \; -->w_0=0 \\ {t}_0=0 \\ \text{thus} \\ {w}_1=w_0+h(t_0{e}^{3t_0}-<!--\; -\; -->2w_0) \\ 0+0.5[(0)e^{3(0)}-<!--\; -\; -->2(0)=0 \\ t=a+ih\text{ common distance} \\ {t}_1=.5 \end{gathered}$$
From this we can create this Table:
$\begin{array}{|cccc|}\hline i& t_i& w_i& y(t_i)\\ 0& 0& 0& 0\\ 1& .5& 0& \\ 2& 1& 1.2& \\ \hline\end{array}$
To find the the actual solutions requires one to solve the ordinary differential equation IVP problem.
A 1st order linear ODE has the form
$$\begin{gathered} y^{\prime }(x)+p(x)y=q(x) \\ y(x)=\frac{\int <!--\; \int \; -->e^{\int <!--\; \int \; -->p(x)dx}q(x)dx+C}{e^{\int <!--\; \int \; -->p(x)dt}} \\ \frac{\int <!--\; \int \; -->e^{\int <!--\; \int \; -->p(x)dx}q(x)dx+C}{e^{\int <!--\; \int \; -->p(x)dx}} \end{gathered}$$
$$\begin{gathered} \frac{\int <!--\; \int \; -->e^{\int <!--\; \int \; -->2dt}{e}^{3t}tdt+C}{e^{\int <!--\; \int \; -->2dt}} \\ \frac{\int <!--\; \int \; -->e^{5t}tdt}{e^{2t}}=\frac{t{e}^{3t}}{5}-<!--\; -\; -->\frac{e^{3t}}{25}+\frac{e^{-<!--\; -\; -->2t}C}{1} \end{gathered}$$
I was unable to reach the correct answer to the ODE which was
$\frac{1}{5}t{e}^{3t}-<!--\; -\; -->\frac{1}{25}{e}^{3t}+\frac{1}{25}{e}^{-<!--\; -\; -->2t}$
How do I correct my work and arrive to the correct conclussion?

I'm solving a system of stiff ODEs, at first I wanted to implement BDF, but it seem to be a quite complicated method, so I decided to start with Backward Euler method. Basically it says that you can solve an ODE:
$y^{\prime }=f(t,y)$, with initial guess $y(t_0)=y_0$
Using the following approximation:
$y_{k+1}=y_k+hf(t_{k+1},y_{k+1})$, where $h$ is a step size on parameter $t$
Wikipedia article says that you can solve this equation using Newton-Raphson method, which is basically a following iteration:
$x_{n+1}=x_n-<!--\; -\; -->\frac{g(x_n)}{g^{\prime }(x_n)}$
So, the question is how to correctly mix them together? What initial guess $x_0$ and function $g$ should be?
Also $f$ is quite complex in my case and I'm not sure if it possible to find another derivative of it analytically. I want to write an implementation of it by myself, so pblackefined Mathematica functions wouldn't work.

$y(0)=1$
How can we solve this equation with operator splitting methods in matlab by using forward Euler method?

I am trying to show that, for the equation
$y^{\prime }+\mathrm{\alpha <!--\; \alpha \; -->}y=0$
alternating between a forward Euler method step for $y_{2n}$ and a backward Euler step for $y_{2n+1}$ with time-step $h$ is equivalent to
$y_{n+1}=\frac{1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}h}{1+\mathrm{\alpha <!--\; \alpha \; -->}h}{y}_n$
I have that
$y_1=y_0+h(-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}{y}_0)=(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}h)y_0$
How exactly $(1+\mathrm{\alpha <!--\; \alpha \; -->}h)$ works into this I don't see. If I just press forward anyway, using the trapezoid rule for the backward Euler step next, and call ${\stackrel{^<!--\; ^\; -->}{y}}_n$ the approximation of $y_n$ using the forward Euler method, I get
$y_2=y_1+h\left(\frac{y_1+{\stackrel{^<!--\; ^\; -->}{y}}_2}{2}\right)$
I'm kind of doing this a priori but that seems right to me: the trapezoid rule is to multiply the change in the $x$-axis by the average of the bases which are $y_1$ and $y_2$. This becomes
$y_2=y_1+h\left(\frac{y_1+(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}h)y_1}{2}\right)=(1+h\left(\frac{2-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}h}{2}\right))y_1$
This doesn't seem to be shaping up to the form I'm supposed to be getting. Am I doing something wrong?

Let
$x^{\prime }(t)=f(t,x(t)),t\in <!--\; \in \; -->(0,T)$ with $x(0)=x_0$
$f$ satifies the Lipschitz-condition $f(t,x)-<!--\; -\; -->f(t,y)\le <!--\; \le \; -->L|x-<!--\; -\; -->y|$
$h\in <!--\; \in \; -->(0,\frac{1}{L})$ is the step size and the approximation $x_k$ for $x(t_k)=hk$ is given by $x_k=x_{k-<!--\; -\; -->1}+hf(t_k,x_k)$.
Now I would be very interested how to derive the error
$|x_k-<!--\; -\; -->x(t_k)|\le <!--\; \le \; -->\frac{1}{1-<!--\; -\; -->Lh}(|x_{k-<!--\; -\; -->1}-<!--\; -\; -->x(t_{k-<!--\; -\; -->1})|+\frac{h^2}{2}\underset{s\in <!--\; \in \; -->[0,T]}{max}|x^{″}(s)|)$
I tried to look up it up in some numerical analysis books but it is always different

I know the difference between the forward (explicit) Euler method and backward (implicit) Euler method in the context of approaching solutions to ODEs.
However, I have a doubt regarding the following method:
$y_{n+1}=y_n+\frac{h}{2}(f(t_n,y_n)+f(t_{n+1},y_{n+1}))$
I am not sure whether it is an implicit or an explicit method.
I tend to say it is implicit one step methode because we have to solve an equation to calculate $y_{n+1}.$
Thank you for your help.

Consider the following problem
$y^{\prime }=\ln <!--\; \; -->\ln <!--\; \; -->(4+y^2),x\in <!--\; \in \; -->[0,1],y(0)=1$
We can formulate the problem the approximate the solution
$y_{n+1}=y_n+h\ln <!--\; \; -->\ln <!--\; \; -->(4+y_n^2),n=0,1,...,N-<!--\; -\; -->1,y_0=0$
with mesh points $x_n=nh$. I am tasked with finding the truncation error. Now considering a Taylor series we find that
$|T_n|\le <!--\; \le \; -->\frac{h}{2!}|(\ln <!--\; \; -->\ln <!--\; \; -->(4+{\mathrm{\xi <!--\; \xi \; -->}}^2){)}^{\prime }|\mathrm{\xi <!--\; \xi \; -->}\in <!--\; \in \; -->(x_n,x_{n+1})$
Now
$\ln <!--\; \; -->\ln <!--\; \; -->(4+y^2{)}^{\prime }=\frac{d}{dy}\ln <!--\; \; -->\ln <!--\; \; -->(4+y^2)=\frac{2y\ln <!--\; \; -->\ln <!--\; \; -->(4+y^2)}{\ln <!--\; \; -->(4+y^2)(4+y^2)}\le <!--\; \le \; -->\frac{2y}{4+y^2}\le <!--\; \le \; -->1/2$
$\therefore <!--\; \therefore \; -->|T_n|\le <!--\; \le \; -->\frac{h}{4}$
Is this the correct way to go about obtaining this?

I have a differntial equation $du/dx=f(x,u)$, where
$f(x,u)=\frac{d\mathrm{\psi <!--\; \psi \; -->}}{dx}+p(u-<!--\; -\; -->\mathrm{\psi <!--\; \psi \; -->}(x){)}^4,$
where $\mathrm{\psi <!--\; \psi \; -->}$ is a given function of $x$ independent of $u$.
The backward Euler Method written for step $i$ is
$u_{i+1}=u_i+hf(x_{i+1},u_{i+1}).$
So, at every step we have a 4th degree polynomial to solve for $u_{i+1}$, which can have more than one root. I am told to employ a root-finding algorithm. But I don't understand how to find the one correct root and reject all others. Bisection, Newton's and other methods seem to only work when there is one root in an interval. Would appreciate any help!

I have the following system:
$y^{\prime }=y+e^x$
$y(0)=0$
The problem asks for applying Euler's method and then finding an expression for the global error. Finally, supposing that
$\underset{h-<!--\; -\; -->>0}{lim}\frac{1-<!--\; -\; -->(\frac{1+h}{e^h}{)}^{\frac{x}{h}}}{\frac{x}{h}(e^h-<!--\; -\; -->(1+h))}=1$
Check that the global error tends to 0 as $h$ tends to 0 as well.
Let's first apply Euler's method.
${\mathrm{\eta <!--\; \eta \; -->}}_0=y_0=0$
$x_i=x_0+ih=ih$
${\mathrm{\eta <!--\; \eta \; -->}}_{i+1}={\mathrm{\eta <!--\; \eta \; -->}}_i+hf(x_i,{\mathrm{\eta <!--\; \eta \; -->}}_i)$
Since $x_i=ih$, we have ${\mathrm{\eta <!--\; \eta \; -->}}_{i+1}={\mathrm{\eta <!--\; \eta \; -->}}_i+h({\mathrm{\eta <!--\; \eta \; -->}}_i+e^{ih}$. Factoring by ${\mathrm{\eta <!--\; \eta \; -->}}_i$:
${\mathrm{\eta <!--\; \eta \; -->}}_{i+1}={\mathrm{\eta <!--\; \eta \; -->}}_i(1+h)+h{e}^{ih}$
And now I don't how should I proceed. Should I solve the iteration? I've done problems where I can express ${\mathrm{\eta <!--\; \eta \; -->}}_{i+1}$ in terms of ${\mathrm{\eta <!--\; \eta \; -->}}_0$ by iterations, but here I can't do so because the $h{e}^{ih}$ prevents that (or at least I haven't managed to do it).
I know that global error is defined as $e(x,h)=\mathrm{\eta <!--\; \eta \; -->}(x,h)-<!--\; -\; -->y(x)$ but I don't know how exactly should I use that formula. My guess it that I have to solve the differential equation and use the solution $y(x)$, but I don't know how to solve that system or what exactly is $\mathrm{\eta <!--\; \eta \; -->}(x,h)$.
Any help regarding to solving this problem will be highly appreciated.
Thanks!

Say we have a function $y(t)$, that satisfies the ordinary differential equation $\frac{\mathrm{d}y}{\mathrm{d}t}=f(t,y)\text{for }t\in <!--\; \in \; -->(t_0,t_{max}],$, where $t$ takes discrete values, $t_n$, with a constant step size $h=t_{n+1}-<!--\; -\; -->t_n$. We also are given the initial condition, for example, $y(0)=1$.

If we were to use the Backward Euler Method:
$y_{n+1}\approx <!--\; \approx \; -->y_n+hf(t_{n+1},y_{n+1}),$
to approximate a solution for a function, say, $f(y)=-<!--\; -\; -->y$, I am told that it is possible to use this information to show that the Backward Euler Scheme is first order, by expanding the global error as a Taylor series in powers of $h$. (The global error at fixed $t$ is the difference between the approximate solution and the exact solution, $y(t)$)

Can anyone figure out how this might be possible? I cannot find any resources online that dmeonstrate this idea.

"Whats the result of one step of Backward Euler Method with h = 0.1 applied on the IVP:
$y^{\prime }(t)=5y(t)+10$
$y(0)=1$"?
So, been trying to understand the Backward Euler Method for a while now and almost get it. However, I don't know exactly what to do and would really appreciate to see how someone who knows what they're doing tackle this problem.

Solve the first-order system that satisfies the given initial conditions using the Euler Method for $y(0.5)$ and $z(0.5)$, using a mesh size of $h=0.1$:
1. $y^{″}-<!--\; -\; -->6z^2{z}^{\prime }-<!--\; -\; -->y^{\prime }-<!--\; -\; -->x^{3}y=0;y(0)=1,y^{\prime }(0)=1.5$
2. $z^{‴}+3y^2(z^{″}{)}^2-<!--\; -\; -->5z^{\prime }-<!--\; -\; -->x^{2}z=0;z(0)=1.25,z^{\prime }(0)=1.5,z^{″}(0)=2$
Please help. I just can’t figure this problem out. We’re supposed to use $u=y$, $v=y^{\prime }$, $w=z$, $g=z^{\prime }$, $k=z^{″}$ when defining first-order system of ODE’s.

I am asked to solve the following problem. Use Euler's modified method to solve the initial value problem $\frac{dx}{dt}=\frac{1+x^2}{t},1\le <!--\; \le \; -->t\le <!--\; \le \; -->4,x(1)=0$
The step size is not given here. Now, I can solve this using Euler's method but I am not able to find the formula for modified one in the mentioned textbook. So, I need the method here. I need to use the following as textbook: Brain Bradie, A friendly Introduction to Numerical Analysis (Pearson)

I have a question from my book and it says essentially, consider the IVP $x^{\bullet <!--\; \bullet \; -->}=-<!--\; -\; -->x$ with $x(0)=1$, what is the exact value of $x(1)$, then using Eulers method with step size1 , estimate $x(1)$ call this $x^{\ast <!--\; \ast \; -->}(1)$ , then repeat for step sizes of ${10}^{-<!--\; -\; -->n}$ for $n=1,2,3,4$ then finally plot $E=|x^{\ast <!--\; \ast \; -->}(1)-<!--\; -\; -->x(1)|$ as a function of step size and then as $lnE$ vs $lnt$.
Now I am having some issues and ill explain,
for the first part I get that $x(1)=e^{-<!--\; -\; -->1}$
We have $f(x)=-<!--\; -\; -->x$ and $x_0=1$
so I have that by Euler method
$x_1=x_0+f(x_0)t$
which would imply that for $t=1,x_1=0$
and then for the second part it would imply $x_1=0.9$, then 0.99, then 0.999 and finally 0.9999.
But this doesn't seem to make any sense to me. Seeing as none of these are close to $e^{-<!--\; -\; -->1}$
So I am confused in regard to where I am making mistakes, or where everything is going wrong. For the plotting part, I am also stuck because of this. I am looking for any help and advice.

Now I consider the harmonic oscillator problem.
The ordinal differential equation is
$\begin{array}{rl}\stackrel{\dot{}<!--\; \dot{}\; -->}{q}& =p\\ \stackrel{\dot{}<!--\; \dot{}\; -->}{p}& =-<!--\; -\; -->q\end{array}$
In symplectic Euler method, where
$\begin{array}{rl}\frac{p^{(m+1)}-<!--\; -\; -->p^{(m)}}{\mathrm{\Delta <!--\; \Delta \; -->}t}& =-<!--\; -\; -->q^{(m)}\\ \frac{q^{(m+1)}-<!--\; -\; -->q^{(m)}}{\mathrm{\Delta <!--\; \Delta \; -->}t}& =p^{(m+1)}\end{array}$
the flow operator ${\mathrm{\psi <!--\; \psi \; -->}}_{\mathrm{d},\mathrm{\Delta <!--\; \Delta \; -->}}=((1-<!--\; -\; -->\mathrm{\Delta <!--\; \Delta \; -->}{t}^2)q+\mathrm{\Delta <!--\; \Delta \; -->}t\cdot <!--\; \cdot \; -->p,-<!--\; -\; -->\mathrm{\Delta <!--\; \Delta \; -->}t\cdot <!--\; \cdot \; -->q+p)$ is symplectic.
Here, the textbook suggests that this flow operator stricktly integrates the modified Hamiltonian and this Hamiltonian is invariant. This modified hamiltonian is
$\begin{array}{r}\stackrel{~<!--\; ~\; -->}{H}=\frac{q^2+p^2}{2}-<!--\; -\; -->\frac{qp}{2}\mathrm{\Delta <!--\; \Delta \; -->}t\end{array}$
I cannot understand how to derive $\stackrel{~<!--\; ~\; -->}{H}$.

Having trouble wrapping my brain around this question.. don't know where to start.
A. Consider the differential equation $f^{\prime }(x)=(x+1)f(x)$ with $f(0)=1$. What is the exact formula for $f(x)$?
I tried solving for $f(x)$ but that just gives me $(x+1)f^{\prime }(x)$. I feel like that's not sufficient here.
B. Solve for $f(.2)$ using Eulers approximation method with increment $h=.01$ for $x\in <!--\; \in \; -->[0,0.2]$.
Edit- my work for part A.
integral f'(x) = integral(x+1) *f(x)+(x+1) * integral (f(x)
so f(x) = ((x^2+x)/2)*f(x) + (x+1) * (f(x)^2)/2.
I don't believe this is the right answer.