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**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 $y=0$

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Because in addition to solutipons you found, there is another one, namelu $y=0$

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You should have it in Calculus II

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

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Simple observation is enough

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You ca refer to the fact that straight lines in $\mathbb{C}$ are also straight lines in $\mathbb{R}^2$ and conversely

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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

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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

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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.

$\cos (4\pi/9)+i\sin (4\pi/9)$ belongs to the first quadrant and pretty close to $i$. Just draw a little picture.

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Read W2L1 handout. THere is an explanation

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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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I believe question was not about Linear Equations but more general ones

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You need to wait lectures ; there will be shown some recepies to find integrating factors, but there is no general algorithm

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You should read an announcement: you should not do anything, or rather as much as you need to ensure that in the real Quiz 1 (not in the Mock Quiz) you do not do anything stupid.

Also you should read description of this board and not to post here unrelated questions.

Also you should read description of this board and not to post here unrelated questions.

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You need to derive it. Even the following argument would be insufficient for a full mark:

We know that it will be a circle, we know that its center will be on the real axis (left from $z=0$), so find two points of intersection of this circle with $x$-axis:

$(z_1-4)= -4z_1$ and $(z_2-4)= 4z_1$. Finding $z_{1,2}$ (even if you do it) and setting $z_0=(z_2+z_1)/2$, $R- |z_2-z_1|/2$ ...

We know that it will be a circle, we know that its center will be on the real axis (left from $z=0$), so find two points of intersection of this circle with $x$-axis:

$(z_1-4)= -4z_1$ and $(z_2-4)= 4z_1$. Finding $z_{1,2}$ (even if you do it) and setting $z_0=(z_2+z_1)/2$, $R- |z_2-z_1|/2$ ...