Help to Integral Calculus Solutions & Examples

$y^{″}=11-<!--\; -\; -->y$
$y(2)=1;y^{\prime }(2)=-<!--\; -\; -->4$
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?

Is it proper to say "increases/decreases from no bound"?
We commonly use the expression increases without bound to describe certain divergent behaviour of functions (e.g. the function $f(x)=x^2$ increases without bound on $[0,\mathrm{\infty <!--\; \infty \; -->})$). What would be the proper way of describing the behaviour of the curve of f(x) if I want to move towards the right on $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0]$?
In the curve sketching unit in my calculus course, I tend to describe all key elements (critical numbers intervals of increase/decrease, points of inflection, intercepts, etc.) from left to right on the x-axis, and so to increase consistency, I want to describe the increase/decrease of a function from left to right. So far, I have been saying "the function increases/decreases without bound towards the left", but this has been directly opposite to what that interval of the function is labelled. I want to know if there is better or more accurate language I could use.

Monotonicity of vector fields
If I have a vector field (x,y,z) on a one dimensional line (x-axis) and if I have to check its monotonicity between two intervals. Will it be monotonic if:
(1) only one of the components of vector field is monotonic in that interval. OR (2) two of the components of vector field are monotonicaly increasing and the other is monotonicaly decreasing in that interval. OR (3) all the components of vector field are either monotonicaly increasing or monotonicaly decreasing in that interval. .

Asymptotes for a function
$f(x)=\frac{x^2-<!--\; -\; -->2ax}{x-<!--\; -\; -->a}$
where a is positive.
The question asks to give equations for the 2 asymptotes. The first is obvious: x=a. The other asymptote is y=x−a but I am unsure how to derive this.

If h has positive derivative and $\mathrm{\phi <!--\; \phi \; -->}$ is continuous and positive. Where is increasing and decreasing f
The problem goes specifically like this:
If h is differentiable and has positive derivative that pass through (0,0), and $\mathrm{\phi <!--\; \phi \; -->}$ is continuous and positive. If:
$f(x)=h({\int <!--\; \int \; -->}_0^{\frac{x^4}{4}-<!--\; -\; -->\frac{x^2}{2}}\mathrm{\phi <!--\; \phi \; -->}(t)dt).$
Find the intervals where f is decreasing and increasing, maxima and minima.
My try was this:
The derivative of f is given by the chain rule:
$f^{\prime }(x)=h^{\prime }({\int <!--\; \int \; -->}_0^{\frac{x^4}{4}-<!--\; -\; -->\frac{x^2}{2}}\mathrm{\phi <!--\; \phi \; -->}(t)dt)\mathrm{\phi <!--\; \phi \; -->}(\frac{x^4}{4}-<!--\; -\; -->\frac{x^2}{2})$
We need to analyze where is positive and negative. So I solved the inequalities:
$\frac{x^4}{4}-<!--\; -\; -->\frac{x^2}{2}>0\wedge <!--\; \wedge \; -->\frac{x^4}{4}-<!--\; -\; -->\frac{x^2}{2}<0$
That gives: $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->\surd 2)\cup <!--\; \cup \; -->(\surd 2,\mathrm{\infty <!--\; \infty \; -->})$ for the first case and $(-<!--\; -\; -->\surd 2,\surd 2)$ for the second one. Then (not sure of this part) $h^{\prime }({\int <!--\; \int \; -->}_0^{\frac{x^4}{4}-<!--\; -\; -->\frac{x^2}{2}}\mathrm{\phi <!--\; \phi \; -->}(t)dt)>0$ and $\mathrm{\phi <!--\; \phi \; -->}(\frac{x^4}{4}-<!--\; -\; -->\frac{x^2}{2})>0$ if $x\in <!--\; \in \; -->(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->\surd 2)\cup <!--\; \cup \; -->(\surd 2,\mathrm{\infty <!--\; \infty \; -->}).$ Also if both h′ and $\mathrm{\phi <!--\; \phi \; -->}$ are negative the product is positive, that's for $x\in <!--\; \in \; -->(-<!--\; -\; -->\surd 2,\surd 2)$.
The case of the product being negative implies:
$x\in <!--\; \in \; -->[(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->\surd 2)\cup <!--\; \cup \; -->(\surd 2,\mathrm{\infty <!--\; \infty \; -->})]\cap <!--\; \cap \; -->(-<!--\; -\; -->\surd 2,\surd 2)=[(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->\surd 2)\cap <!--\; \cap \; -->(-<!--\; -\; -->\surd 2,\surd 2)]\cup <!--\; \cup \; -->[(\surd 2,\mathrm{\infty <!--\; \infty \; -->})\cap <!--\; \cap \; -->(-<!--\; -\; -->\surd 2,\surd 2)]=\mathrm{\varnothing <!--\; \varnothing \; -->}.$
So the function is increasing in $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->\surd 2),(-<!--\; -\; -->\surd 2,\surd 2),(\surd 2,\mathrm{\infty <!--\; \infty \; -->})$. So the function does not have maximum or minimum. Not sure of this but what do you think?

Determine the monotonic intervals of the functions
i need to determine the monotonic intervals of this function $y=2x^3-<!--\; -\; -->6x^2-<!--\; -\; -->18x-<!--\; -\; -->7$. I tried the below but i am not sure if i am doing it right.
My work: $\begin{array}{rl}y=2x^3-<!--\; -\; -->6x^2-<!--\; -\; -->18x-<!--\; -\; -->7& ⟺<!--\; ⟺\; -->6x^2-<!--\; -\; -->12x-<!--\; -\; -->18=0\\ & ⟺<!--\; ⟺\; -->6(x^2-<!--\; -\; -->2x-<!--\; -\; -->3)=0\\ & ⟺<!--\; ⟺\; -->(x-<!--\; -\; -->3)(x+1)\\ & ⟺<!--\; ⟺\; -->x-<!--\; -\; -->3=0x+1=0\\ & ⟺<!--\; ⟺\; -->x=3,x=-<!--\; -\; -->1\end{array}$
so my function increases when $x\in <!--\; \in \; -->[3,+\mathrm{\infty <!--\; \infty \; -->}[$ and decreases when $x\in <!--\; \in \; -->[-<!--\; -\; -->1,3]\cup <!--\; \cup \; -->]-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->1]$.
Please i want to know how to solve this problem any help with explanation will be appreciated. thanks in advanced

The effect of changing sample sizes on outliers
I know that the size of a sample is inversely proportional to the width of a confidence interval, and that outliers tend to increase the width of the interval as well. So that must mean that increasing the sample size reduces the effect of outliers on a confidence interval, and decreasing the sample size amplifies the effect, correct?
How can I show this using formulae instead of words for, say a confidence interval for a one-sample z-test?
Also as a side note, does changing the sample size change how outliers affect the p-value of a hypothesis test? I'm inclined to say yes, but I'm not sure how to justify that conclusion.

Confusion between the definition of increasing function and a theorem regarding it.
The definition of increasing function given in my school maths text book is
Let I be an open interval contained in the domain of a real valued function f. Then f is said to be increasing on I if $a<<!--\; <\; -->b⟹<!--\; ⟹\; -->f(a)\le <!--\; \le \; -->f(b)$ for all $a,b\in <!--\; \in \; -->I$.
And a theorem given after it is
f is increasing in I if $f^{\prime }(x)><!--\; >\; -->0\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->I$.
Shouldn't it be $f^{\prime }(x)\ge <!--\; \ge \; -->0$.
Is constant function an increasing function? Or a function like $f(x)=x^3$ where $f^{\prime }(x)=0$ at some or all points which are increasing according to definition.
Edit: The definition of decreasing function given is
f is decreasing on I if $a<<!--\; <\; -->b⟹<!--\; ⟹\; -->f(a)\ge <!--\; \ge \; -->f(b)$.
A constant function follow this definition too.

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

Sketch $f(x)=\sin <!--\; \; -->x+\frac{1}{x}$ finding local maxima and minima, intervals of increase and decrease. I'm trying to use differentiation to draw this picture and find critical points.
So, I get $f^{\prime }(x)=\cos <!--\; \; -->x-<!--\; -\; -->\frac{1}{x^2}$. However, I'll have to deal with inequality $f^{\prime }(x)>0$, and $f^{\prime }(x)=0$, I feel I lack an ability to solve an equation like this. So is there any better way to find local maximun and minumum for this function?

What are the problems that may arise by the application of Euler method?
2 problems are described. One of them:
1.It is written that from from the convergence order it can be concluded that halving the error means halving the step-size and thus doubling the number of grid points. This leads to the doubling of computational costs which might cause an enormous effort.
Do we halve the step-size in Euler method? For my imagination for Euler method we just define a step size as we wish. As smaller we take as more precise will be our approximation. But where does "halving" come from? Where is my confusion?

I am given the function $x(t)=4\arctan <!--\; \; -->t$ and told that routine computations will show that $x(0)=0$ and $x(1)=\mathrm{\pi <!--\; \pi \; -->}$.
I must determine a differential equation for $x(t)$ of the form $x^{\prime }(t)=f(t,x)$.
I must then use $f$ to approximate $\mathrm{\pi <!--\; \pi \; -->}$.
I will use Euler's Method to find minimum number of iterations ($n$) needed to yield the approximation $\mathrm{\pi <!--\; \pi \; -->}\approx 3.1400$, such that $n-1$ iterations is strictly less than 3.1400.
How do I find $n$? It seems to me that it can't be found given the instruction set that I am provided.

While studying Fourier analysis last semester, I saw an interesting identity:
$\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{1}{n^2-<!--\; -\; -->{\mathrm{\alpha <!--\; \alpha \; -->}}^2}=\frac{1}{2{\mathrm{\alpha <!--\; \alpha \; -->}}^2}-<!--\; -\; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2\mathrm{\alpha <!--\; \alpha \; -->}\tan <!--\; \; -->\mathrm{\pi <!--\; \pi \; -->}\mathrm{\alpha <!--\; \alpha \; -->}}$
whenever $\mathrm{\alpha <!--\; \alpha \; -->}\in <!--\; \in \; -->\mathbb{C}\setminus <!--\; \setminus \; -->\mathbb{Z}$, which I learned two proofs using Fourier series and residue calculus.
More explicitly, we can deduce the theorem using Fourier series of $f(\mathrm{\theta <!--\; \theta \; -->})=e^{i(\mathrm{\pi <!--\; \pi \; -->}-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->})\mathrm{\alpha <!--\; \alpha \; -->}}$ on $[0,2\mathrm{\pi <!--\; \pi \; -->}]$ or contour integral of the function $g(z)=\frac{\mathrm{\pi <!--\; \pi \; -->}}{(z^2-<!--\; -\; -->{\mathrm{\alpha <!--\; \alpha \; -->}}^2)\tan <!--\; \; -->\mathrm{\pi <!--\; \pi \; -->}z}$ along large circles.
But these techniques, as long as I know, wasn't fully developed at Euler's time.
So what was Euler's method to prove this identity? Is there any proof at elementary level?

The problem I have is the initial value problem
$y^{‴}=x+y$
with
$y(1)=3,y^{\prime }(1)=2,y^{″}(1)=1$
that should be solved with Eulers method using the step length, $h=\frac{1}{2}$.

The iteration step for Eulers method is $y_{n+1}=y_n+hf(x_n,y_n)$. So I should need a system of equations of my initial values I have. I started with substituting:
$U_1=y,U_2=y^{\prime },U_3=y^{″}$
and then
$U_1^{\prime }=U_2,U_2^{\prime }=U_3,U_3^{\prime }=x+U_1$
And it's here I'm stuck, I don't understand how I should start iterate with the step-length from here, I have only encounteblack first-order problem with Eulers method so I would love if someone could point me in the right direction.

Increasing and decreasing intervals
Find a polynomial f(x) of degree 4 which increases in the intervals $(-<!--\; -\; -->\infty ,1)$ and (2,3) and decreases in the intervals (1,2) and $(3,\infty )$ and satisfies the condition $f(0)=1$.
It is evident that the function should be $f(x)=a{x}^4+b{x}^3+c{x}^2+dx+1$. I differentiated it. Now, I'm lost. I tried putting $f^{\prime }(1)>0$, $f^{\prime }(3)-<!--\; -\; -->f^{\prime }(2)\ge <!--\; \ge \; -->0$, and $f^{\prime }(2)-<!--\; -\; -->f^{\prime }(1)\le <!--\; \le \; -->0$. Am I doing correct? Or is there another method?

Find interval of increase and decrease
So im supposed to find the interval of decrease and increase here. Ive gotten up to taking the derivative which is $-<!--\; -\; -->4x(x^2-<!--\; -\; -->1)$ and then setting it to 0 i got (-1,0,1) Im lost at what to do now?
Im supposed to take it for this below:
$f(x)=7+2x^2-x^4$

Question about strictly increasing and continuous functions on an interval
If $F,\mathrm{\phi <!--\; \phi \; -->}:[a,b]\to <!--\; \to \; -->\mathbb{R}$ with F continuous, $\mathrm{\phi <!--\; \phi \; -->}$ strictly increasing and $(F+\mathrm{\phi <!--\; \phi \; -->})(a)>(F+\mathrm{\phi <!--\; \phi \; -->})(b)$ is $F+\mathrm{\phi <!--\; \phi \; -->}$ decreasing on some interval in (a,b)?
This is clearly true if $\mathrm{\phi <!--\; \phi \; -->}$ has finitely many points of discontinuity in the interval but I am unsure if this statement is true if there are infinitely many points of discontinuity.