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Loser66
 one year ago
1)Reduce to a linear system of first order
\(\dfrac{d^2x}{dt^2}=a\dfrac{dx}{dt}+bx+c\); \(x(0) =u\); \(\dfrac{dx}{dt}(0) =v\)
2) Solve, using Laplace transform
Please, help
Loser66
 one year ago
1)Reduce to a linear system of first order \(\dfrac{d^2x}{dt^2}=a\dfrac{dx}{dt}+bx+c\); \(x(0) =u\); \(\dfrac{dx}{dt}(0) =v\) 2) Solve, using Laplace transform Please, help

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alekos
 one year ago
Best ResponseYou've already chosen the best response.0well, we could integrate both sides of the DE \[\int\limits x'' = \int\limits ax' + \int\limits bx + \int\limits c \]

Empty
 one year ago
Best ResponseYou've already chosen the best response.4I don't think that's the way to go here, can you use the rules for laplace transforms or what level of sophistication are you allowed to use here?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1It becomes \(x' = ax + bx^2/2 + cx\) which is not linear!!

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1\(x" ax'bx =c\) . Can we solve the homogeneous part to get one of the solution, then use reduction of order to get the required form?

Empty
 one year ago
Best ResponseYou've already chosen the best response.4I don't know, I am not too familiar with reduction of order I'm afraid but what you described sounds good. Actually you could potentially turn this into a system of equations and solve it that way, it's been a while so I'm a little rusty.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Let me think more. Thanks for being here.:)

Empty
 one year ago
Best ResponseYou've already chosen the best response.4Haha yeah no problem, I need to practice for sure

Empty
 one year ago
Best ResponseYou've already chosen the best response.4We need to turn it into a system of 1st order equations. We do this by doing a standard technique of doing this substitution: \[u=x\]\[v=x'\] Take the derivative of both of these to get: \[u'=x'\]\[v'=x''\] Now we combine them together for the first equation, and then we take our differential equation \(x''=ax'+bx+c\) and substitute these in to get the second equation: \[u'=v\]\[v'=av+bu+c\] So that completes part a. :)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1You assume u and v are not constant, right? but we have x(0) = u, that is a constant, not a function to take derivative. Am I right?

Empty
 one year ago
Best ResponseYou've already chosen the best response.4They're not constants, specifically they depend on t. \[u(t)=x(t)\]\[v(t)=x'(t)\] If they were constants, taking their derivatives would also have given us zero above when I differentiated both of them. :)

Empty
 one year ago
Best ResponseYou've already chosen the best response.4The main point is to get a system of equations where you have a derivative equal to things that aren't derivatives, like I've shown above. For example I could have written: \[v'=au'+bu+c\] but then that would mess up the system of equations and make it not so pretty, we want that \(u'=v\) so we can write just \[v'=av+bu+c\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Let me try other way, if we meet at the result, we are done Let \(y_1(t) = x(t)\\y_2(t) = x'(t)\) then \( y_1'(t) =y_2(t)\) and \(y_2' = ay_2+by_1+c\) Ah!! the same with you.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1ok, thanks a ton. lalalala...

Empty
 one year ago
Best ResponseYou've already chosen the best response.4Awesome good job! :) Now how I would solve this is to plug this in a matrix but I don't think that's what they want you to do

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Now adding the condition, \(y_1(0) = x(0) = u\\y_2(0) = x'(0) =v\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1for part 2) they ask me to solve by Laplace transform, not matrix.

Empty
 one year ago
Best ResponseYou've already chosen the best response.4Well should be possible now, go ahead and try to solve the new system and I can try to help you if you need it.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1wait, new or original one?

alekos
 one year ago
Best ResponseYou've already chosen the best response.0i think they're asking for the original equation

alekos
 one year ago
Best ResponseYou've already chosen the best response.0which is why you would need the initial conditions

Empty
 one year ago
Best ResponseYou've already chosen the best response.4Here, this is exactly the method I would use to solve this, go ahead and see if you can figure it out from here and I can help you fill in the details if you're still struggling. http://math.stackexchange.com/questions/1068473/howtosolvealinearsysteminmatrixformusinglaplacetransform

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Thanks a lot. I am reading.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to reduce to a linear system of first order, you just need to decouple it into two linear (NOT affine) equations, one in \(x'=x+c/b\) and one in \(y=dx/dt\):$$\dfrac{d^2x}{dt^2}=a\dfrac{dx}{dt}+bx+c\implies \left\{\begin{array}{ll}\frac{dx}{dt}=y\\\frac{dy}{dt}=bx'+ay\end{array}\right.$$ which gives:$$\begin{bmatrix}0&1\\b&a\end{bmatrix}\begin{bmatrix}x'\\y\end{bmatrix}=\frac{d}{dt}\begin{bmatrix}x'\\y\end{bmatrix}$$in matrix form; we can take the Laplace transform rowbyrow, with \(X(s)=\mathcal{L}\{x\},Y(s)=\mathcal{L}\{y\}\)$$\begin{bmatrix}0&1\\b&a\end{bmatrix}\begin{bmatrix}X\\Y\end{bmatrix}=s\begin{bmatrix}X\\Y\end{bmatrix}\begin{bmatrix}x'(0)\\y(0)\end{bmatrix}\\s\begin{bmatrix}X\\Y\end{bmatrix}\begin{bmatrix}0&1\\b&a\end{bmatrix}\begin{bmatrix}X\\Y\end{bmatrix}=\begin{bmatrix}x'(0)\\y(0)\end{bmatrix}\\\begin{bmatrix}s&0\\0&s\end{bmatrix}\begin{bmatrix}X\\Y\end{bmatrix}\begin{bmatrix}0&1\\b&a\end{bmatrix}\begin{bmatrix}X\\Y\end{bmatrix}=\begin{bmatrix}x'(0)\\y(0)\end{bmatrix}\\\begin{bmatrix}s&1\\b&sa\end{bmatrix}\begin{bmatrix}X\\Y\end{bmatrix}=\begin{bmatrix}x'(0)\\y(0)\end{bmatrix}\\\begin{bmatrix}X\\Y\end{bmatrix}=\begin{bmatrix}s&1\\b&sa\end{bmatrix}^{1}\begin{bmatrix}x'(0)\\y(0)\end{bmatrix}\\\begin{bmatrix}X\\Y\end{bmatrix}=\frac1{s(sa)b}\begin{bmatrix}sa&b\\1&s\end{bmatrix}\begin{bmatrix}x'(0)\\y(0)\end{bmatrix}\\\begin{bmatrix}X\\Y\end{bmatrix}=\begin{bmatrix}\frac{sa}{s(sa)b}&\frac{b}{s(sa)b}\\\frac1{s(sa)b}&\frac{s}{s(sa)b}\end{bmatrix}\begin{bmatrix}u+c/b\\v\end{bmatrix}=\begin{bmatrix}(u+c/b)\frac{sa}{s(sa)b}+v\frac{b}{s(sa)b}\\(u+c/b)\frac1{s(sa)b}v\frac{s}{s(sa)b}\end{bmatrix}$$so let's look at these expressions in more detail: $$\frac1{s(sa)b}=\frac1{s^2asb}\\\quad=\frac1{(sa/2)^2(b+a^2/4)}$$so now recall $$\mathcal{L}^{1}\left\{\frac{\sqrt{b+a^2/4}}{(sa/2)^2(b+a^2/4)}\right\}=e^{at/2}\sinh\left(t\sqrt{b+a^2/4}\right)=S(t)\\\mathcal{L}^{1}\left\{\frac{sa/2}{(sa/2)^2(b+a^2/4)}\right\}=e^{at/2}\cosh\left(t\sqrt{b+a^2/4}\right)=C(t)$$so it follows that $$\frac1{(sa/2)^2(b+a^2/4)}\to \frac2{\sqrt{4b+a^2}}S(t)\\\frac{s}{(sa/2)^2(b+a^2/4)}\to C(t)+\frac{a}{\sqrt{4b+a^2}}S(t)\\\frac{sa}{(sa/2)^2(b+a^2/4)}\to C(t)\frac{a}{\sqrt{4b+a^2}}S(t)$$ so we find that the inverse transform gives us: $$x'=\left(u+c/b\right)C(t)+\frac{2vba(u+c/b)}{\sqrt{4b+a^2}}S(t)$$and then $$\boxed{x(t)=c/b+e^{at/2}\left((u+c/b)\cosh\left(\frac12 t\sqrt{4b+a^2}\right)\\\qquad\qquad\qquad+(2vbauac/b)\sinh\left(\frac12 t\sqrt{4b+a^2}\right)\right)}$$

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Thanks for the solution. I have question on it. 1)...which gives .. \[\begin{bmatrix}0&1\\b&a\end{bmatrix}\begin{bmatrix}x'\\y\end{bmatrix}=\frac{d}{dt}\begin{bmatrix}x'\\y\end{bmatrix}\] I think there is mistake there at the right hand side. It should be \(\dfrac{d}{dt}\left[\begin{matrix}x\\y\end{matrix}\right]\), not x', right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is \(x'\) because \(x'=x+c/b\) to convert the system into a linear one
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