Discover Euler's Method Examples

I want to analytically solve
$\frac{\mathrm{\partial <!--\; \partial \; -->}y}{\mathrm{\partial <!--\; \partial \; -->}t}=-\alpha y,$
where initial conditions are $y=1$ at $t=0$

and show the forward Euler method gets a smaller answer than the backward Euler method $\mathrm{\forall <!--\; \forall \; -->}$ $t>0$, provided that 0 $<$${\mathrm{\alpha <!--\; \alpha \; -->}}^2$ $\mathrm{\Delta <!--\; \Delta \; -->}{t}^2$ $<$ 1

Any help whatsoever will be appreciated!

How do you prove Euler's Theorem
$du=\left(\frac{\mathrm{\partial <!--\; \partial \; -->}u}{\mathrm{\partial <!--\; \partial \; -->}x}\right)dx+\left(\frac{\mathrm{\partial <!--\; \partial \; -->}u}{\mathrm{\partial <!--\; \partial \; -->}y}\right)dy$
if $u=f(x,y)$.
I also heard that Ramanujan developed another method can somebody know that?

Finding solution of differential equation using euler numerical method ${\displaystyle \frac{dy}{dx}=0.3y-<!--\; -\; -->10}$ and $0\le <!--\; \le \; -->x\le <!--\; \le \; -->3$ and $x=0,y=40.$
What i have tried
${\displaystyle y_n=y_{n-<!--\; -\; -->1}+h\cdot <!--\; \cdot \; -->f(x_{n-<!--\; -\; -->1},y_{n-<!--\; -\; -->1})}$
Where $f(x,y)=0.3y-<!--\; -\; -->10.$
I have a doubt that what should i take width interval $h$ in above question.(which is not mention here)

Can anyone please explain me what i take value of $h$. Thanks

I am struggling to find advantages and disadvantages of the following: Forward Euler Method, Trapezoidal Method, and Modified Euler Mathod (pblackictor-corrector).
Any help or books I can use to get these?

I need to solve the equation below with Euler's method:
$y^{″}+\mathrm{\pi <!--\; \pi \; -->}y{e}^{x/3}(2y^{\prime }\sin <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}x)+\mathrm{\pi <!--\; \pi \; -->}y\cos <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}x))=\frac{y}{9}$
for the initial conditions $y(0)=1$, $y^{\prime }(0)=-<!--\; -\; -->1/3$
So I know I need to turn the problem into a system of two first order differential equations.
Therefore $u_1=y^{\prime }$ and $u_2=y^{″}$ I can now write the system as:
$$\begin{gathered} u_1=y^{\prime } \\ {u}_2={\displaystyle \frac{y}{9}}-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}y{e}^{x/3}(2u_1\sin <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}x)-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}y\cos <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}x)) \end{gathered}$$
How do I proceed from here?

Suppose I am given a Cauchy-Euler form second order differential equation
$x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+y=f(x).$
The usual textbook method for solving the Cauchy Euler equation is to blackuce it to a linear differential equation with constant coefficients by the transformation $x=e^t$. But I have a fundamental doubt here, we know that $e^t>0$ $\mathrm{\forall <!--\; \forall \; -->}t\in <!--\; \in \; -->\mathbb{R}$. But when we are using the above transformation we are subconsciously assuming $x>0$. How does this make sense?

Substituting $t=\ln <!--\; \; -->(x)$ also makes no difference as ln is defined on ${\mathbb{R}}_{>0}$. So I now doubt the validity of the method followed to solve the Cauchy-Euler equation.

Can someone give me a proper explanation of what exactly going on here and why the process is valid?

I've just discoveblack that I can use Euler's method to estimating the slope of $y=(x+c{)}^2$ with perfect accuracy, regardless of what $\mathrm{\Delta <!--\; \Delta \; -->}x$ is used. What causes this?

Does it happen with other functions, what must they look like for it to happen?

I am using this method to estimate the slope, $m$
$m\approx <!--\; \approx \; -->\frac{(x+\mathrm{\Delta <!--\; \Delta \; -->}x+c{)}^2-<!--\; -\; -->(x-<!--\; -\; -->\mathrm{\Delta <!--\; \Delta \; -->}x+c{)}^2}{2\mathrm{\Delta <!--\; \Delta \; -->}x}$
Doesn't matter what $\mathrm{\Delta <!--\; \Delta \; -->}x$ is, the slope is always correct ...I would expect this only to work for tiny $\mathrm{\Delta <!--\; \Delta \; -->}x$, but works with large ones too.

$Y^{\prime }=AY$
$A=\left[\begin{array}{ccc}1& 2& -<!--\; -\; -->1\\ 1& 0& 1\\ 4& -<!--\; -\; -->4& 5\end{array}\right]$
$Y(0)=\left(\begin{array}{c}-<!--\; -\; -->1\\ 0\\ 0\end{array}\right)$
I have to solve this system using Eulers method, but what's Eulers method, like I know how to approach this using the eigenvalue method, but the problem explicitly states that it has to be solved using that method, yet my textbook doesn't provide an algorithm for such a method.

This question was asked in CSIR. please help me to find out correct choice

Let $y(t)$ satisfy the differential equation
$y^{\prime }=\mathrm{\lambda <!--\; \lambda \; -->}y;y(0)=1$
. Then the backward Euler method for $n>1$ and $h>0$
$\frac{y_n-<!--\; -\; -->y_{n-<!--\; -\; -->1}}{h}=\mathrm{\lambda <!--\; \lambda \; -->}{y}_n;y(0)=1$
yields
1. a first order approximation to $e^{\mathrm{\lambda <!--\; \lambda \; -->}nh}$
2. a polynomial approximation to $e^{\mathrm{\lambda <!--\; \lambda \; -->}nh}$
3. a rational function approximation to $e^{\mathrm{\lambda <!--\; \lambda \; -->}nh}$
4. a Chebyshev polynomial approximation to $e^{\mathrm{\lambda <!--\; \lambda \; -->}nh}$

I don't quite understand the discretisation error of Euler's method
$L(t,h)=\frac{x(t+h)-<!--\; -\; -->x(t)}{h}-<!--\; -\; -->f(t,x(t))$
What I don't understand is: $\frac{x(t+h)-<!--\; -\; -->x(t)}{h}$ is the "gradient" and $f(t,x(t)$ is the value of the function. How can we get the error by subtracting the former by the latter?

The linear backward Euler's Method is given by
$y^{j+1}=y^j+\mathrm{\tau <!--\; \tau \; -->}{k}_1$
$(I-<!--\; -\; -->\mathrm{\tau <!--\; \tau \; -->}J)k_1=f(y^j),J:=f^{\prime }(y^j)$
To show, is that $y^{j+1}$ is equal to $\stackrel{~<!--\; ~\; -->}{y^{j+1}}$ the Output of the Backward Euler's Method, if you proccess one Newton-Step with
$g-<!--\; -\; -->(y^j+\mathrm{\tau <!--\; \tau \; -->}f(g))=0$
for $\stackrel{~<!--\; ~\; -->}{y^{j+1}}=g$
$g^0=y^j$
This is what I started with :
$g^1=g^0+\mathrm{\tau <!--\; \tau \; -->}\frac{1}{I-<!--\; -\; -->\mathrm{\tau <!--\; \tau \; -->}{f}^{\prime }(g^0)}f(g^0)$
$⟺<!--\; ⟺\; -->y^{j+1}=y^j+\mathrm{\tau <!--\; \tau \; -->}\frac{1}{I-<!--\; -\; -->\mathrm{\tau <!--\; \tau \; -->}{f}^{\prime }(y^j)}f(y^j)$
$⟺<!--\; ⟺\; -->y^{j+1}-<!--\; -\; -->y^j=\mathrm{\tau <!--\; \tau \; -->}\frac{1}{I-<!--\; -\; -->\mathrm{\tau <!--\; \tau \; -->}{f}^{\prime }(y^j)}f(y^j)$
$⟺<!--\; ⟺\; -->\mathrm{\tau <!--\; \tau \; -->}{k}_1=\mathrm{\tau <!--\; \tau \; -->}\frac{1}{I-<!--\; -\; -->\mathrm{\tau <!--\; \tau \; -->}{f}^{\prime }(y^j)}f(y^j)$
$⟺<!--\; ⟺\; -->k_1(I-<!--\; -\; -->\mathrm{\tau <!--\; \tau \; -->}{f}^{\prime }(y^j))=f(y^j)$
Did I miss something or is there anything wrong? Any help is very appreciated

We know that the solution of
$y^{\prime }=y$
with $y(0)=1$ is
$y=e^x$
As far as I see the Euler's method is explicitly used only to find the numerical approximation of e.g. $y(3)$. Can we use the Euler's method to solve this differential equation and find the exact solution?

Edit:
I thought of the following:
$y(t_1+\mathrm{\delta <!--\; \delta \; -->}t)=y(t_0)+y^{\prime }(t_1)\mathrm{\delta <!--\; \delta \; -->}t$
$y(t_1+2\mathrm{\delta <!--\; \delta \; -->}t)=y(t_1+\mathrm{\delta <!--\; \delta \; -->}t)+y^{\prime }(t_1+\mathrm{\delta <!--\; \delta \; -->}t)\mathrm{\delta <!--\; \delta \; -->}t$
$...$
and from that we could find a recursive solution for $y(t_0)$. Then by taking the limit $\underset{\mathrm{\delta <!--\; \delta \; -->}t\to <!--\; \to \; -->0}{lim}$ of it, it could be possible to find the general solution of $y$ for every $t_0$ in the domain.
Is it impossible?

I tried to do euler's method with this problem:
$\frac{dw}{dt}=(3-<!--\; -\; -->w)(w+1),w(0)=0,\text{and }t\in <!--\; \in \; -->[0,5].\mathrm{\Delta <!--\; \Delta \; -->}t=\mathrm{0.5.}$
But am getting weird results like for $y_2=5.25$, $y_3=-<!--\; -\; -->15.84375$; but checked against the book but this is not correct. Can you please work out some iterations?
$y_2=1.5+(3-<!--\; -\; -->1.5)(1.5+1)(1)=5.25$, for example

Consider the IVP
$\begin{array}{r}{y}^{″}=-<!--\; -\; -->y\end{array}$
for $t\ge <!--\; \ge \; -->0$, and $y(0)=1$, $y^{\prime }(0)=2$.

I have rewritten this differential equation as a system of first-order ODE's such that
$\begin{array}{r}{u}^{\prime }=v\\ v^{\prime }=-<!--\; -\; -->u\end{array}$
with $u(0)=1,v(0)=2$.

The solution is $y=2\sin <!--\; \; -->(t)+\cos <!--\; \; -->(t)$, $y^{\prime }=2\cos <!--\; \; -->(t)-<!--\; -\; -->\sin <!--\; \; -->(t)$

I am asked to perform one step of Euler's method with $h=0.5$, and determine if Euler's method is stable for this problem.

For the first part, I find that one step of Euler's method yields
$\begin{array}{r}{y}_1=y_0+hf(t_0,y_0)=1+(0.5)(2\cos <!--\; \; -->(0)-<!--\; -\; -->\sin <!--\; \; -->(0))=2.\end{array}$
But how do I determine if Euler's method is stable? I know that for the equation $y^{\prime }=\mathrm{\lambda <!--\; \lambda \; -->}y$, Euler's method is stable for $|1+h\mathrm{\lambda <!--\; \lambda \; -->}|\le <!--\; \le \; -->1$, but since this problem is in a different form I'm not sure what I need to do here.

I am wondering if it is possible to derive Eulers method without doing it graphically.

I am trying to give a detailed derivation of Eulers method, and how the formula is found
$y_i=y_{i-<!--\; -\; -->1}+hf(x_{i-<!--\; -\; -->1},y_{i-<!--\; -\; -->1})$
If anyone could derive this formula and show how its found would be great, or even a link to a website/textbook. Thanks!

Is there a rule that characterizes when Euler's method over-estimates or under-estimates?
For example, "if $f(x)$ is increasing, then Euler's method underestimates," or something similar?

How do you work out if Euler's method overestimates the actual solution, for the ODE:
$\frac{dy}{dx}=24\tan <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}x)$
With steps of 0.25 from $1\le <!--\; \le \; -->x\le <!--\; \le \; -->2$?

Use Euler's method with step size 0.2 to estimate y(0.6) where y(x) is the solution of the initial value problem y' = cos(x+y), y(0) = 0
So on a high level, we're using Euler's method to approximate a function that we don't know but only know the derivative equation for and the initial. We're going to use Euler's to get approximate values of the unknown function using approximations.
So is this right:
original $x=0$, $y=0$
$y_1=0+0.2(cos(0))=0.2$
so new $x=0.2$ and new $y=0.2$
$y_2=0.2+0.2(cos(0.2+0.2))\approx <!--\; \approx \; -->0.384$
$y_3=0.384+0.2(cos(0.4+0.384))=0.384+0.141\approx <!--\; \approx \; -->0.525$
Is that right? The numbers are just going to be hard to work with right?

Euler's Method for series associates with a given series $\underset{j=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}(-<!--\; -\; -->1{)}^j{a}_j$ the transformed series $\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{{\mathrm{\Delta <!--\; \Delta \; -->}}^n{a}_0}{2^{n+1}}$ where ${\mathrm{\Delta <!--\; \Delta \; -->}}^0{a}_j=a_j$, ${\mathrm{\Delta <!--\; \Delta \; -->}}^n{a}_j={\mathrm{\Delta <!--\; \Delta \; -->}}^{n-<!--\; -\; -->1}{a}_j-<!--\; -\; -->{\mathrm{\Delta <!--\; \Delta \; -->}}^{n-<!--\; -\; -->1}{a}_{j+1}$, $j=1,2,\cdots <!--\; \cdots \; -->.$.
I am given that $\ln <!--\; \; -->2=1-<!--\; -\; -->\frac{1}{2}+\frac{1}{3}-<!--\; -\; -->\frac{1}{4}+-<!--\; -\; -->\cdots <!--\; \cdots \; -->=\frac{1}{1\cdot <!--\; \cdot \; -->2}+\frac{1}{2\cdot <!--\; \cdot \; -->2^2}+\frac{1}{3\cdot <!--\; \cdot \; -->2^3}+\frac{1}{4\cdot <!--\; \cdot \; -->2^4}\cdots <!--\; \cdots \; -->$.

I need to prove that $\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{{\mathrm{\Delta <!--\; \Delta \; -->}}^n{a}_0}{2^{n+1}}=\frac{1}{2}+\frac{1}{2\cdot <!--\; \cdot \; -->2^2}+\frac{1}{3\cdot <!--\; \cdot \; -->2^3}+\frac{1}{4\cdot <!--\; \cdot \; -->2^4}\cdots <!--\; \cdots \; -->$, but I can not figure out how. I attempt to by induction and the base case is simple enough. However, when trying to use the induction hypothesis I get stuck.

If I assume that $\underset{n=0}{\overset{k}{\sum <!--\; \sum \; -->}}\frac{{\mathrm{\Delta <!--\; \Delta \; -->}}^n{a}_0}{2^{n+1}}=\frac{1}{2}+\frac{1}{2\cdot <!--\; \cdot \; -->2^2}+\frac{1}{3\cdot <!--\; \cdot \; -->2^3}+\frac{1}{4\cdot <!--\; \cdot \; -->2^4}\cdots <!--\; \cdots \; -->\frac{1}{(k+1)\cdot <!--\; \cdot \; -->2^{k+1}}$ and I add $\frac{1}{(k+2)\cdot <!--\; \cdot \; -->2^{k+2}}$ to both sides. Then I just need to show that ${\mathrm{\Delta <!--\; \Delta \; -->}}^{k+1}{a}_0=\frac{1}{k+2}$. However I don't see how I can.

So I know that ${\mathrm{\Delta <!--\; \Delta \; -->}}^{k+1}{a}_0={\mathrm{\Delta <!--\; \Delta \; -->}}^k{a}_0-<!--\; -\; -->{\mathrm{\Delta <!--\; \Delta \; -->}}^k{a}_1$. From the induction hypothesis I know that ${\mathrm{\Delta <!--\; \Delta \; -->}}^k{a}_0=\frac{1}{k+1}$. But I would have to work with ${\mathrm{\Delta <!--\; \Delta \; -->}}^k{a}_1$ recursively. So I am not sure how to prove this.

I am attempting to use Euler's method to express the following system of ODE's...
$\frac{d{y}_i}{dx}=c_i(y_{i-<!--\; -\; -->1}^3-<!--\; -\; -->y_i^3)+f_i(\text{for }i=1,\dots <!--\; \dots \; -->,n)$
using the discretised equations, which are presented below in matrix notation:
$\mathit{y}(t_{k+1})={\mathit{C}}_1\mathit{y}(t_k)+{\mathit{C}}_2{\mathit{y}}^3(t_k)+\mathit{f}$
Note $f_i=1$ when $i=1$, but otherwise equals 0 (this considers the initial condition). Whilst I am familiar with Euler's method, I am unsure of the steps to get to the first equation to the second. Can anyone assist me with this? Understanding this is necessary for a project I'm working on, so I would really appreciate any help.