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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 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 $z_0=(z_2+z_1)/2$, $R- |z_2-z_1|/2$ ...

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We have initial condition at point $xt_0=0$, and solution blows up as $t=-\l(2)$. While we can consider solution, given by this formula as $t<-ln(2)$ it is disconnected from intial condition. Our solution blew up as $t=-\ln (2)$.