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

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- Loser66

@Empty

- alekos

well, we could integrate both sides of the DE
\[\int\limits x'' = \int\limits ax' + \int\limits bx + \int\limits c \]

- Empty

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?

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## More answers

- Loser66

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

- Loser66

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

- Empty

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.

- Loser66

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

- Empty

Haha yeah no problem, I need to practice for sure

- Empty

Ok I see now.

- Empty

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. :)

- Loser66

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?

- Empty

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. :)

- Empty

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

- Loser66

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.

- Loser66

ok, thanks a ton. lalalala...

- Empty

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

- Loser66

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

- Loser66

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

- Empty

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.

- Loser66

wait, new or original one?

- alekos

i think they're asking for the original equation

- alekos

which is why you would need the initial conditions

- Empty

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

- Loser66

Thanks a lot. I am reading.

- IrishBoy123

.

- anonymous

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 $$\boxed{x(t)=-c/b+e^{at/2}\left((u+c/b)\cosh\left(\frac12 t\sqrt{4b+a^2}\right)\\\qquad\qquad\qquad+(2vb-au-ac/b)\sinh\left(\frac12 t\sqrt{4b+a^2}\right)\right)}$$

- Loser66

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?

- anonymous

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

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