### Author Topic: Euler equations  (Read 2316 times)

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##### Euler equations
« on: October 14, 2019, 10:59:12 AM »
Hi, I have two questions respectively about 3.3 problem 34 and 35.
1) In problem 34, we are asked to make a change of variables from $t$ to $x=\ln{t}$. This is fine - I substituted in my equations for $\frac{d^2y}{dt^2}$ and $\frac{dy}{dx}$ and got the desired result. However, I am not exactly clear on what happens to the term $by$, which doesn't change even though the variable $y$ is represented in technically does. Is it okay to leave it like that since we essentially end up back substituting when writing the roots of the characteristic equation?
2) In problem 35, finding the roots of the characteristic equation yields $r_1 = i$ and $r_2 = -i$. Is it sufficient to write the equation $y(t) = C_1t^i + C_2t^{-i}$, or should I use the same justification as is given in section 3.3 to write it as $y(t) = C_1\cos{ln(t)} + C_2\sin{ln(t)}$ so that we have real solutions?
« Last Edit: October 15, 2019, 04:13:14 PM by nadia.chigmaroff »

#### Victor Ivrii

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##### Re: Euler equations
« Reply #1 on: October 15, 2019, 11:02:08 AM »
1) Since $y$ is the same thing it should not change

2) Better to write as $y(t)=C_1\cos(\ln(t))+C_2\sin(\ln(t))$

You need to escape all names of functions : \cos, \ln to produce them upright and provide a correct spacing

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##### Re: Euler equations
« Reply #2 on: October 15, 2019, 04:14:56 PM »
What do you mean, $y$ is the same thing? I feel like it becomes a sort of composition of functions, which is why I'm confused.

#### Victor Ivrii

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##### Re: Euler equations
« Reply #3 on: October 17, 2019, 12:37:51 AM »
Indeed, $y$ is the same same in the sense that $y[t]=y(x(t))$. However $\dfrac{dy}{dt}\ne \dfrac{dy}{dx} (x(t))$ but contains an extra factor.