Question

Turn y'' + y = t^(3/2), y(0) = 1, y'(0) = 0 into a first order system

Answer 1

Let \(\large u(t) = y^{\prime}(t)\). Then \(\large u^{\prime}(t) = y^{\prime\prime}(t)\). Thus, \(\large y^{\prime\prime} + y = t^{3/2} \xrightarrow{u=y^{\prime}}{} u^{\prime}=t^{3/2}-y\). From here, it shouldn't be too difficult to see what the resulting first order ODE system is. I hope this helps! :-)

Answer 2

I see that the bottom portion of the system would be -1 and t^(3/2), but I don't understand where to top comes from

Answer 3

I'm sorry, the bottom portion would be -1 and 0 correct?

Answer 4

I'm sorry, but I'm not exactly sure what you mean by the bottom portion. Are you possibly referring to the initial conditions?

Answer 5

for u' = P(t)y + g(t), the bottom portion of P

Answer 6

[ I'm not sure what goes here]
[ -1, 0]

Answer 7

Ah, so P is supposed to be a matrix?

Answer 8

yes

Answer 9

I'm guessing we could say that\[y_2'= -y_1 + t^(3/2), y_2 = y_1'\]

Answer 10

Ok, So, if you make the substitution that I mentioned previously, the second order ODE transforms into the following system (as equations, not a matrix):
\[\large\left\{\begin{aligned} y^{\prime} &= u\\ u^{\prime} &= t^{3/2} - y\end{aligned} \right.\]
It turns out that this is actually a non-homogeneous system due to the \(t^{3/2}\) term. Now, the objective is to write the equation in the form \(\mathbf{x}^{\prime}(t) = A\mathbf{x}(t) + \mathbf{f}(t)\) where \(A\) is the coefficient matrix, \(\mathbf{x}(t)\) is the solution vector you'd hope to find and \(\large\mathbf{f}(t)\) is the vector that contains the non-homogeneous terms of the each equation in the system.
We now see that as a matrix equation,
\[\large\left\{\begin{aligned} y^{\prime} &= u\\ u^{\prime} &= t^{3/2} - y\end{aligned} \right.\implies \begin{bmatrix}y^{\prime}\\ u^{\prime}\end{bmatrix} = \begin{bmatrix}0 & 1\\ -1 & 0\end{bmatrix}\begin{bmatrix}y \\ u\end{bmatrix} + \begin{bmatrix}0 \\t^{3/2}\end{bmatrix}\]
where the initial condition would be \(\large\mathbf{x}(0) = \begin{bmatrix}y(0) \\ u(0)\end{bmatrix} = \begin{bmatrix}1 \\ 0\end{bmatrix}\)
Does this clarify things? :-)

Answer 11

Ok that makes sense, thanks a lot

Answer 12

wow!! amazing, be your fan now @ChristopherToni XD