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**Chapter 4 / Re: 4.2 question 28**

« **on:**October 25, 2020, 01:13:50 PM »

you need to write it, if you hope for any answer

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you need to write it, if you hope for any answer

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Indeed, instead of $\log(\pm w)$ with $+2\pi mi$ we write $\log(w)$ with $+\pi mi$ since $\log (-w)=\log(w)+\i i$

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Of **some** fundamental set (remember a constant factor!)

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Yes, some of them are geometric series, and some of $e^{z}$, $\sin(z)$, $\sinh(z)$ and so on. However some can be derived from those, ether by substitution (f.e. $z^2$ instead of $z$), some by integration, differentiation, multiplication by $z^m$ or combination of both. F.e. consider geometric $\dfrac{1}{1-z}$. Integratinfg we can get power series for $-\Log (1-z)$, diffeerentiating for $\frac{1}{(1-z)^m}$ ,...

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Hi there, I think there might be some problem with Q28. After I found the integrating factor and then solve for the solution, it's impossible to find the solution of my h(y) in this case (see the last line of my writing).If you found an integrating factor correctly there should be no problem to $h(y)$. There is a problem with

In the textbook, it is e^(-2y). Maybe this one is correct.Both are correct

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is du/dx+i dv/dx? Why can't we take the derivative with respect to y?You can take also derivative by $y$ but you need to multiply it by $i$ (think why)

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Try to solve it by yourself (it follows right from the definition)

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why the complex plane is both open and closed? why did it close? Because by definition, a set is called closed if it contains its boundary. And I don't see the complex plane contains its boundary.And what is the boundary of $\mathbb{C}$?

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You should remember that each segment of the polygonal curve may be served not by two discs, but several discs (think why)

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This "meaning" has no sense. You should worry about math, not about this crasp

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Almost done, but you need also define interval, on which solution exists; it must include $0$

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That's right but it is a custom to denote by $\varepsilon,\delta$ something small. In this case capital letters, say $R,M$ would be better, for arbitrarily large

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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.

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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 isWe 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 perfectnotexact, 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 iswhy 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?