Question

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

Answer 1

@Empty

Answer 2

well, we could integrate both sides of the DE

\[\int\limits x'' = \int\limits ax' + \int\limits bx + \int\limits c \]

Answer 3

I 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?

Answer 4

It becomes \(x' = ax + bx^2/2 + cx\) which is not linear!!

Answer 5

\(x" -ax'-bx =c\) . Can we solve the homogeneous part to get one of the solution, then use reduction of order to get the required form?

Answer 6

I 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.

Answer 7

Let me think more. Thanks for being here.:)

Answer 8

Haha yeah no problem, I need to practice for sure

Answer 9

Ok I see now.

Answer 10

We 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. :)

Answer 11

You 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?

Answer 12

They'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. :)

Answer 13

The 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
\[v'=av+bu+c\]

Answer 14

Let 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.

Answer 15

ok, thanks a ton. lalalala...

Answer 16

Awesome 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

Answer 17

Now adding the condition, \(y_1(0) = x(0) = u\\y_2(0) = x'(0) =v\)

Answer 18

for part 2) they ask me to solve by Laplace transform, not matrix.

Answer 19

Well should be possible now, go ahead and try to solve the new system and I can try to help you if you need it.

Answer 20

wait, new or original one?

Answer 21

i think they're asking for the original equation

Answer 22

which is why you would need the initial conditions

Answer 23

Here, 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/how-to-solve-a-linear-system-in-matrix-form-using-laplace-transform

Answer 24

Thanks a lot. I am reading.

Answer 25

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Answer 26

to 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 row-by-row, with \(X(s)=\mathcal{L}\{x\},Y(s)=\mathcal{L}\{y\}\)$$\begin{bmatrix}0&1\\b&a\end{bmatrix}\begin{bmatrix}X\\Y\end{bmatrix}=s\begin{bmatrix}X\\Y\end{bmatrix}-\begin{bmatrix}x'(0)\\y(0)\end{bmatrix}\\s\begin{bmatrix}X\\Y\end{bmatrix}-\begin{bmatrix}0&1\\b&a\end{bmatrix}\begin{bmatrix}X\\Y\end{bmatrix}=\begin{bmatrix}x'(0)\\y(0)\end{bmatrix}\\\begin{bmatrix}s&0\\0&s\end{bmatrix}\begin{bmatrix}X\\Y\end{bmatrix}-\begin{bmatrix}0&1\\b&a\end{bmatrix}\begin{bmatrix}X\\Y\end{bmatrix}=\begin{bmatrix}x'(0)\\y(0)\end{bmatrix}\\\begin{bmatrix}s&-1\\-b&s-a\end{bmatrix}\begin{bmatrix}X\\Y\end{bmatrix}=\begin{bmatrix}x'(0)\\y(0)\end{bmatrix}\\\begin{bmatrix}X\\Y\end{bmatrix}=\begin{bmatrix}s&-1\\-b&s-a\end{bmatrix}^{-1}\begin{bmatrix}x'(0)\\y(0)\end{bmatrix}\\\begin{bmatrix}X\\Y\end{bmatrix}=\frac1{s(s-a)-b}\begin{bmatrix}s-a&b\\1&s\end{bmatrix}\begin{bmatrix}x'(0)\\y(0)\end{bmatrix}\\\begin{bmatrix}X\\Y\end{bmatrix}=\begin{bmatrix}\frac{s-a}{s(s-a)-b}&\frac{b}{s(s-a)-b}\\\frac1{s(s-a)-b}&-\frac{s}{s(s-a)-b}\end{bmatrix}\begin{bmatrix}u+c/b\\v\end{bmatrix}=\begin{bmatrix}(u+c/b)\frac{s-a}{s(s-a)-b}+v\frac{b}{s(s-a)-b}\\(u+c/b)\frac1{s(s-a)-b}-v\frac{s}{s(s-a)-b}\end{bmatrix}$$so let's look at these expressions in more detail: $$\frac1{s(s-a)-b}=\frac1{s^2-as-b}\\\quad=\frac1{(s-a/2)^2-(b+a^2/4)}$$so now recall $$\mathcal{L}^{-1}\left\{\frac{\sqrt{b+a^2/4}}{(s-a/2)^2-(b+a^2/4)}\right\}=e^{at/2}\sinh\left(t\sqrt{b+a^2/4}\right)=S(t)\\\mathcal{L}^{-1}\left\{\frac{s-a/2}{(s-a/2)^2-(b+a^2/4)}\right\}=e^{at/2}\cosh\left(t\sqrt{b+a^2/4}\right)=C(t)$$so it follows that $$\frac1{(s-a/2)^2-(b+a^2/4)}\to \frac2{\sqrt{4b+a^2}}S(t)\\\frac{s}{(s-a/2)^2-(b+a^2/4)}\to C(t)+\frac{a}{\sqrt{4b+a^2}}S(t)\\\frac{s-a}{(s-a/2)^2-(b+a^2/4)}\to C(t)-\frac{a}{\sqrt{4b+a^2}}S(t)$$ so we find that the inverse transform gives us: $$x'=\left(u+c/b\right)C(t)+\frac{2vb-a(u+c/b)}{\sqrt{4b+a^2}}S(t)$$and then $$\boxed{x(t)=-c/b+e^{at/2}\left((u+c/b)\cosh\left(\frac12 t\sqrt{4b+a^2}\right)\\\qquad\qquad\qquad+(2vb-au-ac/b)\sinh\left(\frac12 t\sqrt{4b+a^2}\right)\right)}$$

Answer 27

Thanks 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?

Answer 28

it is \(x'\) because \(x'=x+c/b\) to convert the system into a linear one