Calculus 2 Equations: Expert Solutions and More

Transition in solving differential equations using Method of Undetermined Coefficient
${\displaystyle x^{2}y{}^{″}-3x{y}^{\prime }+3y=\ln x}$

To find the solution to a differential equation of the form
${\displaystyle x^{\prime }=Ax}$
where A is a matrix.

There exists a bijective ${\displaystyle f:\mathbb{N}\to \mathbb{N}}$ such that:
${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{(-1)}^{f\left(n\right)}\ln \frac{f\left(n\right)+1}{f\left(n\right)}=\ln 2010}$

First order diferential equation with two integrating factors can be solved by separation of variables
${\displaystyle M(x,y)dx+N(x,y)dy=0}$
if f(x) and g(y) are integrating factors then the diferential equation can be solved by separation of variables.

Finding Hamilton function which equations do I have to solve to find ${\displaystyle v(\theta ,\varphi ,\dot{\theta },\varphi )}$?
${\displaystyle L(\theta ,\varphi ,\dot{\theta },\dot{\varphi })=\frac{m{b}^2}{2}({\dot{\theta }}^2+{\sin }^2\left(\theta \right){\dot{\varphi }}^2)+mbg\cos \left(\theta \right)}$

Finding non-zero function such that ${\displaystyle f\left(2x\right)=f^{\prime }\left(x\right)\cdot f{}^{″}\left(x\right)}$.
I am thankful if someone can help me or show me the clue. As honestly as possible, I got stuck on this problem.
I need some help in finding a ${\displaystyle f\left(x\right)\ne 0}$ such that
${\displaystyle f\left(2x\right)=f^{\prime }\left(x\right)\cdot f{}^{″}\left(x\right),}$
where f', f" are the first and second derivatives, respectively. It is not an ordinary differential equation.
My last try was to put ${\displaystyle f\left(x\right)=k{e}^{ax}}$, and then
${\displaystyle k{e}^{2ax}=ka{e}^{ax}\cdot k{a}^2{e}^{ax}.}$
Now by canceling ${\displaystyle e^{ax}}$, we have
${\displaystyle k=k^2{a}^3.}$
This shows ${\displaystyle k=1,a=1}$ and finally ${\displaystyle f\left(x\right)=e^x}$. But I want to solve the problem analytically.

Find the solution of the system ${\displaystyle x^{\prime }=x-2y\text{ and }{y}^{\prime }=4x+5y}$.

How to prove
${\displaystyle \sum _{n=p+1}^{\mathrm{\infty }}\frac{1}{n^2-p^2}=\frac{1}{2p}(1+\frac{12}{+}\cdots +\frac{1}{2p})}$

Use the Intermediate Value Theorem to show that there is a root of equation
$x^3+2x-<!--\; -\; -->\frac{1}{x}=0$
in $(\frac{1}{4},1)$. Find this root to two decimal places.

Use the Intermediate Value Theorem to prove that the equation
$-<!--\; -\; -->x^3+4\sin <!--\; \; -->(x)+4{\cos }^2<!--\; \; -->(x)=0$
has at least two solutions. You should carefully justify each of the hypothesis of the theorem.

How do I know that at least two exist because I know you're meant to look for a change in sign.

On a recent problem, I received the following scenario: An object moves back and forth on a straight track. During the time interval $0\le <!--\; \le \; -->t\le <!--\; \le \; -->30$ minutes, the object's position, $x$, and velocity, $v$, are continuous functions; some of their values are shown in the table (which I have reproduced below).

$\begin{array}{|ccc|}\hline t\mathbf{\text{ (min)}}& x(t)\mathbf{\text{ (feet)}}& v(t)\mathbf{\text{ (feet/min)}}\\ 0& \text{12}& -<!--\; -\; -->20\\ 10& \text{50}& \text{20}\\ 15& \text{18}& \text{3}\\ 25& \text{60}& -<!--\; -\; -->2\\ 30& \text{60}& \text{10}\\ \hline\end{array}$

The question was, for $0<t<30$, does there exist a time t when $v(t)=-<!--\; -\; -->22$?

I tried to apply the intermediate value theorem, and concluded that the answer was "not necessarily." I reasoned that because the values of $v(t)$ in the table ranged between −20 (at the beginning) and 10 (at the end), and −22 wasn't between those two values, we couldn't be sure that such a t exists.

The teacher gave me 3/4 points on the question. Their comment was that I should have considered the mean value theorem, too. They did not write anything about my intermediate value theorem analysis, but I still have grave doubts about my application of the intermediate value theorem ... and, as for the mean value theorem, I have no idea how to proceed.

Would anyone here be able to shed some light on how the intermediate value and mean value theorems could be used to determine whether there exists a t where $v(t)=-<!--\; -\; -->22$? Thanks so much.

Suppose that $f:[a,b]\to <!--\; \to \; -->\mathbb{R}$ is continuous and that $f(a)<0$ and $f(b)>0$. By the intermediate-value theorem, the set $S=\{x\in <!--\; \in \; -->[a,b]:f(x)=0\}$ is nonempty. If $c=supS$, prove that $c\in <!--\; \in \; -->S$.

My first thought was to show that $S$ is finite therefore $c\in <!--\; \in \; -->S$, but there is no guarantee that $f$ doesn't have infinitely many zeros.

A thought I have now is to show that $c>max(S)$ can not be true, but I do not know how to show this, is this even the correct thing to show? Thank you in advance for any input.

For each natural number n and each number ${\displaystyle x\in [0,1]}$ let
${\displaystyle f_n\left(x\right)=\frac{x}{nx+1}}$

For a differential equation say ${\displaystyle y^{\prime }=t{y}^3}$ with ${\displaystyle y\left(0\right)=1}$ how would we know if the sequence from Euler's method exists or not?

Fix a positive number ${\displaystyle \alpha }$ and consider the series
${\displaystyle \sum _{k=1}^{\mathrm{\infty }}\frac{1}{(k+1){\left[\ln (k+1)\right]}^{\alpha }}}$

Find the value of 'x' if
${\displaystyle \left(\frac{1}{2^{{\log }_{x}4}}\right)\cdot \left(\frac{1}{2^{{\log }_{x}16}}\right)\cdot \left(\frac{1}{2^{{\log }_{x}256}}\right)\dots =2}$

Find the solution of this differential equation whose graph it is through the point (1, 3e).
Let ${\displaystyle y_1\left(x\right)=e^x}$ and ${\displaystyle y_2\left(x\right)=x^2+1+e^x}$ be solutions of the differential equation
${\displaystyle a\left(x\right)y^{\prime }+b\left(x\right)y=c\left(x\right)}$
where a, b, c are continuous functions on ${\displaystyle \mathbb{R}}$ and the function a has no zeros. Find the solution of this differential equation whose graph it is through the point (1,3e).
I know that difference between ${\displaystyle y_1\left(x\right)\text{and}{y}_2\left(x\right)}$ is a solution of homogeneous differential equation. So solution of a homogeneous differential equation is
${\displaystyle y_h=A(x^2+1)}$
I do not know how to get the particular solution. If I use variation constant for ${\displaystyle y_p=z\left(x\right)(x^2+1)}$, where is z(x) a function, then I get
${\displaystyle z=\int \frac{c\left(x\right)}{a\left(x\right)}{e}^{\int \frac{b\left(x\right)}{a\left(x\right)}dx}dx}$