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Messages - Victor Ivrii

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It was explained in the lectures that solutions in the parametric form are also admissible. You could ask it on the tutorial.

Chapter 2 / Re: W3L3 Exact solutions to inexact equations
« on: September 29, 2020, 04:31:19 AM »
In week 3 lecture 3, we get the example $(-y\sin(x)+y^3\cos(x))dx+(3\cos(x)+5y^2\sin(x))dy=0$. We determine that this equation is not exact, but that we can make it exact by multiplying the equation by $y^2$. We then find the general solution to the new equation is $y^3\cos(x)+y^5\sin(x)=C$.

My question is why is this good enough? We didn't answer the original question. We answered a modified version which we chose specifically because it seems easier to us. Shouldn't we still find a solution to the original equation $(-y\sin(x)+y^3\cos(x))dx+(3\cos(x)+5y^2\sin(x))dy=0$? Is there some way we can "divide out" $y^2$ from $y^3\cos(x)+y^5\sin(x)=C$ to get it?
We have not modified equation, but simply multiplied it by an integrating factor. These two equations are equivalent, except as $y=0$, that means $C=0$. But $y=0$ os also a solution to the original equation. Checking this would give you a 100% correct solution, otherwise it is almost perfect

Chapter 1 / Re: how to solve question 19 in section 1.2 in the textbook
« on: September 27, 2020, 11:38:44 AM »

Darren, you may want to fix: replace $c_2$ by $c^2$.

Runbo, what  about division?

Chapter 1 / Re: how to solve question 19 in section 1.2 in the textbook
« on: September 27, 2020, 11:37:07 AM »
Darren, you may want to fix: replace $c_2$ by $c^2$.

Chapter 2 / Re: Question on Week3 Lec3 Integrating Factor
« on: September 26, 2020, 09:32:40 AM »
I could be wrong, but I think we just try each one until we get one that works :)

Chapter 2 / Re: Section 2.2 Question 3
« on: September 24, 2020, 01:37:33 AM »
Because in addition to solutipons you found, there is another one, namelu $y=0$

Chapter 2 / Re: Textbook 2.4, Example 2
« on: September 24, 2020, 01:36:34 AM »
You should have it in Calculus II

In this case usual cube of the sum would be the most efficient solution

Chapter 1 / Re: Section 1.3 Q15
« on: September 24, 2020, 01:33:32 AM »
Simple observation is enough

Chapter 1 / Re: Problems to 1.2 Q20
« on: September 23, 2020, 07:58:36 AM »
You ca refer to the fact that straight lines in $\mathbb{C}$ are also straight lines in $\mathbb{R}^2$ and conversely

Chapter 2 / Re: Mock Quiz Answer
« on: September 21, 2020, 12:28:01 PM »
It was not my intention to to check any math, so there could be that it is impossible recovering $y$ as a function of $x$. My purpose was to let you practice techical aspects

Chapter 2 / Re: Lec 0101 - 9/15 Question
« on: September 20, 2020, 07:49:47 PM »
One should remember that plugging $y=uy_1$ into inhimogeneous equation leaves $u'y_1$ in the left-hand expression. If you do not remember this, therefore you just do not understand the method of variations and you should reread previous slides

Chapter 1 / Re: Solving roots of complex numbers
« on: September 20, 2020, 07:46:18 PM »
The worst thing you can do is to use calculator to evaluate the value of, say, $\sin (4\pi/9)$ and $\cos (4\pi/9)$ numerically. But it may be useful to mention that
$\cos (4\pi/9)+i\sin (4\pi/9)$ belongs to the first quadrant and pretty close to $i$. Just draw a little picture.

Chapter 2 / Re: Lec5101 Question Mock Quiz 1
« on: September 20, 2020, 03:14:04 PM »
Read W2L1 handout. THere is an explanation

Chapter 2 / Re: section 2.1 practice problem 15
« on: September 20, 2020, 03:12:51 PM »
You need to solve equation and initial condition, you'll see that $(y+\frac{1}{2})^2 =x^2-\frac{1}{4}\implies $ solution exist only as $x\ge \frac{\sqrt{15}}{2}$. Point $x=\frac{\sqrt{15}}{2}$ is excluded as $y'=\infty$ there

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