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Messages - Victor Ivrii

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In two examples were made errors.

A. Consider equation
y'= -\frac{y}{x}+y^2.
It is Bernoulli equation. We solve it by the method of variation of the constant. Consider first corresponding linear homogeneous equation
y'= -\frac{y}{x}.
It has a solution
y= Cx^{-1}
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}).
(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
where $c=\mathsf{const}$. Then  $C=-\dfrac{1}{\ln (x)+c}$ and plugging into (\ref{3}) we get
\boxed{ y=-\frac{1}{x(\ln (x)+c)}.}

B. Consider equation
y'- \tan(x) y =\cos(x).
We solve it by the method of integrating factor. Multiplying (\ref{4}) by unknown yet factor $\mu=\mu(x)$ we get
\mu y' - \mu \tan (x) y = \mu \cos(x).
We want the left hand expression to be $\mu y'+\mu 'y$, which means
\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)).
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
\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.
You must know simple trigonometric formulae. Then
\boxed{y= \frac{1}{2}(x+2C)\sec(x)+ \frac{1}{2}\sin (x).}

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 $j$?

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

Home Assignment 6 / Re: Laplace Fourier Transform S5.3.P Q1
« on: March 19, 2019, 02:14:40 PM »
I think we need to make assumption that the Fourier transformation u will be 0 as y goes to infinity,that’s what my TA did in tutorial
Or, at least, does not grow exponentially as $|k|\to \infty$

Home Assignment 6 / Re: Laplace Fourier Transform S5.3.P Q1
« on: March 12, 2019, 06:25:23 PM »
There was an explanation why one of the solutions in half-plane should be rejected (does not satisfy condition at infinity)

APM346--Misc / Re: Spring 2018 Midterm Problem 3 (Both Main and Late)
« on: February 26, 2019, 04:15:46 PM »
Misprints. Corrected

APM346--Misc / Re: Scope of Term Test 1
« on: February 15, 2019, 02:00:37 AM »
Ch 1--3

Home Assignment 3 / Re: S2.3 Problem 8
« on: February 05, 2019, 09:14:53 PM »
[quoute]Hello professor, could u give me a hint as to why we have to impose the restriction for absolute value of x to be smaller or equal to 1?[/quote]
Draw characteristic lines and plot $t=x^2/2$

Home Assignment 3 / Re: HW2.3 Problem 6
« on: February 02, 2019, 08:17:10 PM »
In 6(4) How to ensure the general solution is continuous as r = 0 since the denominator includes r and in the numerator there are 2 arbitrary fuctions? Thanks!
Find conditions to these functions which are necessary and sufficient for the continuity

Home Assignment 3 / Re: Problem2(17) even or odd?
« on: February 01, 2019, 04:56:18 PM »
If you change $t\mapsto -t$, equation does not change, and $u|_{t=0}$ does not change, but $u_t|_{t=0}$ acquires sign "$-$". However, since $u|_{t=0}=0$, if you replace in addition $u\mapsto -u$, then nothing changes. Thus $u(x,t)=-u(x,-t)$.

If on the other hand, $h(x)=0$ then $u$ would be even with respect to $t$

Home Assignment 3 / Re: S2.3 problem2(17)(18)
« on: January 31, 2019, 12:16:47 PM »
No, you draw lines according to conditions: $x=1+ct$, $x=-1+ct$, $x=1-ct$ and $x=-1-ct$.

And normally one need to consider all 9 domains. However, our problems have two symmetries and it is sufficient to consider only 4 domains intersecting with the 1st quadrant and extend solution to the remaining 5, using the fact that solution is either odd or even with respect to $t$ (you need to understand, what is the case), and also solution is either odd or even with respect to $x$ (you need to understand, what is the case).

Home Assignment 3 / Re: S2.3 problem2(17)(18)
« on: January 31, 2019, 07:19:41 AM »
Since in this problem nothing is said about $t>0$ the complete solution should cover all cases. However, since in the problems either $g(x)=0$, or $h(x)=0$ then $u(x,t)$ is odd or even with respect to $t$, respectively. At least in some problems you can observe that solution must be even or odd with respect to $x$ as well.

In such problem, as (17), we have several regions. But we need to work out only some of them and extend to the rest by above arguments.

Home Assignment 3 / Re: S2.3 problem2(17)(18)
« on: January 31, 2019, 03:47:08 AM »
duplicate removed.

Yes, you need to consider all cases and it is recommended to draw a plane and different regions there

Home Assignment 3 / Re: HW2.3 Problem 6
« on: January 31, 2019, 03:41:57 AM »
"and thereby solve the spherical wave   equation." So, find $u$ from expression for $v$

I have difficulties obtaining the general solution of the equation $u_{tt} - c^2u_{xx} = 0$. From the online textbook Section2.3, it mentions $v = u_t +cu_x$, and it gets the result $v_t - cv_x = 0$ by chain rule.
Not by a chain rule. Just from equation
But when I expand $v_t - cv_x = 0$, I get an extra term $x'(t)$ when applies chain rule to $v_t$ because I think $u_{t}$ in $v$ can be two separated into to parts which are $u_t(t)$ and $u_t(x(t))$. Then applying chain rule, it becomes $v_t = u_{tt} + x'(t) +c(u_{xt} + x'(t))$ where the term $x'(t)$ can not be cancelled. I wonder where is the problem of my thought.
Several days ago it was rewritten to make it more clear. Read online version

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