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Messages - Victor Ivrii

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1
Term Test 1 / Re: reduction of order
« on: October 20, 2019, 07:41:48 PM »
Shouldn't we multiplying the integral by $y_1$?
Indeed

2
Only solutions to Tet 1 should be posted here--and only after I say so (sometimeTuesday )

Everything else will be removed

3
Term Test 1 / Re: Cheat Sheet for Term tests
« on: October 20, 2019, 06:24:01 PM »
Are we allowed to have a handwritten cheat sheet for the term tests or is the provided  table of antiderivatives the only allowed "Cheat sheet" on the test ?
Using handwritten will lead you to a chat in the Office of Academic Integrity. The answer was given in Quercus discussions

4
Term Test 1 / Re: reduction of order
« on: October 20, 2019, 06:21:20 PM »
Use may use this formula
$$
y_2 = y_1\int \frac{W}{y_1^2}\,dx
$$

5
Term Test 1 / Re: Past Mid Terms
« on: October 20, 2019, 02:26:03 PM »
Go to MAT244--2018F (f.e. or 2018S and so on) >> MAT244--Tests >> Term Test 1.

Surprise, surprise!

6
Chapter 3 / Re: Euler equations
« on: October 17, 2019, 12:37:51 AM »
Indeed, $y$ is the same same in the sense that $y[t]=y(x(t))$. However $\dfrac{dy}{dt}\ne \dfrac{dy}{dx} (x(t))$ but contains an extra factor.

7
Chapter 3 / Re: Euler equations
« on: October 15, 2019, 11:02:08 AM »
1) Since $y$ is the same thing it should not change

2) Better to write as $y(t)=C_1\cos(\ln(t))+C_2\sin(\ln(t))$

You need to escape all names of functions : \cos, \ln to produce them upright and provide a correct spacing

8
Quiz-1 / MOVED: MAT24f4 TUT5103 Quiz2
« on: October 07, 2019, 05:32:24 AM »

9
Quiz-1 / MOVED: MAT24f4 TUT5103 Quiz2
« on: October 07, 2019, 05:32:12 AM »

10
Chapter 1 / Re: TUT0402 question
« on: October 07, 2019, 05:30:33 AM »
1) This integral is calculated by a trivial substitution $u=x^2+y^2$ (remember, $y$ is constant here)

2) \[ \] and \( \) are not math delimiters on this forum (use single or double dollars) since they interfere with the forum's markdown language

11
Chapter 3 / Re: Initial conditions evaluated at different $t_0$'s?
« on: October 07, 2019, 05:27:07 AM »
As david pointed out, an initial value problem is 1-point problem. Otherwise it is not an initial value problem  and usually, even for higher order equations, there are only two points, which are the ends of the interval; then it is called a boundary value problem which is covered in the end of the textbook (if its title is ... with boundary value problems) but not in our class. On the other hand, BVPs for ODE are important in PDE class (like APM 346) and covered there.

Important is that BVP are not always solvable uniquely.

12
MAT244--Lectures & Home Assignments / Existence and Uniqueness Theorem
« on: September 24, 2019, 10:56:17 AM »
Answering question on discussions

Definitely the role of the Existence and Uniqueness Theorems are much more important than Limits in Manual Computations (and honestly, I have no idea what "Manual Computation" means). However this role is more theoretical both in ODEs and PDEs. Still, if we are talking about numerical solutions (taught in different classes, we skip Chapter 8, and briefly look at section 2.7) we need to be pretty sure that the object we are trying to find exist and we find all of them.

For centuries from I. Newton (who introduced ODEs) mathematicians did not care much about existence, because they were looking for solutions of real life problems and believed in existence and also because the rigorous apparatus of Real Analysis which allows to prove such theorems came into existence only in 19th century. You may want to look at very sketchy Lecture_Note_to_Section_2.8_Existence-Uniqueness_Theorem.

Uniqueness is a different matter: mathematicians observed that the solution to the Cauchy problem is not necessarily unique (remember, that the general solution to the 1st order ODE is $x=\varphi(t;C)$ or $\Phi(x, t; C)=0$ in the explicit and implicit form correspondingly and we need to specify one solution one needs to impose an extra condition; f.e. $x(t_0)=x_0$. They discovered that there could be a singular solution which is not a regular solution which means that it cannot be obtained from the general solution by freezing $C$ but which in each point coincides with some (depending on the point) regular solution. In more details see Lecture_Note_to_Chapter_2_Singular_Solutions.

Both of these lecture notes are optional

13
In two examples were made errors.

A. Consider equation
\begin{equation}
y'= -\frac{y}{x}+y^2.
\label{1}
\end{equation}
It is Bernoulli equation. We solve it by the method of variation of the constant. Consider first corresponding linear homogeneous equation
\begin{equation}
y'= -\frac{y}{x}.
\label{2}
\end{equation}
It has a solution
\begin{equation}
y= Cx^{-1}
\label{3}
\end{equation}
with constant $C$ (do it by yourself!). Now consider (\ref{3}) with $C$ which is not a constant (variation!) and plug it into (\ref{1}).
\begin{gather*}
(Cx^{-1})' =-x^{-1}(Cx^{-1})+(Cx^{-1})^2\implies C'x^{-1}-Cx^{-2}=-Cx^{-2} + C^2x^{-2}\implies C' =C^2x^{-1}\\
\implies  \frac{dC}{C^2}=\frac{dx}{x}\implies -\frac{1}{C}=\ln(x)+c
\end{gather*}
where $c=\mathsf{const}$. Then  $C=-\dfrac{1}{\ln (x)+c}$ and plugging into (\ref{3}) we get
\begin{equation*}
\boxed{ y=-\frac{1}{x(\ln (x)+c)}.}
\end{equation*}

B. Consider equation
\begin{equation}
y'- \tan(x) y =\cos(x).
\label{4}
\end{equation}
We solve it by the method of integrating factor. Multiplying (\ref{4}) by unknown yet factor $\mu=\mu(x)$ we get
\begin{equation}
\mu y' - \mu \tan (x) y = \mu \cos(x).
\label{5}
\end{equation}
We want the left hand expression to be $\mu y'+\mu 'y$, which means
\begin{equation*}
\mu'=-\mu \tan(x) \implies \frac{d\mu}{\mu}= \tan(x)\,dx \implies \ln (\mu) =-\int \tan(x)\,dx = -\int\frac{\sin(x)}{\cos(x)}\,dx = \ln (\cos(x)).
\end{equation*}
You must know this integral. We do not need any constant since we need just one integrating factor.

So $\mu =\cos(x)$. (\ref{5}) is now
\begin{gather*}
\cos(x)y'-\sin (x)y =\cos^2(x)\implies \bigl(\cos(x)y\bigr)'= \cos^2(x)\implies
\cos(x)y = \int \cos^2(x)\,dx = \int \frac{1+\cos(2x)}{2}\,dx =\\
 \frac{x}{2}+\frac{\sin (2x)}{4}+C=
\frac{x}{2}+\frac{\sin (x)\cos(x)}{2}+C.
\end{gather*}
You must know simple trigonometric formulae. Then
\begin{equation*}
\boxed{y= \frac{1}{2}(x+2C)\sec(x)+ \frac{1}{2}\sin (x).}
\end{equation*}

14
APM346--Lectures & Home Assignments / Re: Exam Practice: Prob 69-72
« on: April 09, 2019, 08:08:35 PM »
You understand, that in the left there is summation with respect to $j$?

15
APM346--Misc / Re: Scope of FE
« on: April 02, 2019, 05:44:40 AM »
See practice problems and announcements

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