How Do I Solve This Initial Value Problem Using a Given Solution Function?

AI Thread Summary
To solve the initial value problem for the linear system dY/dt = [1, -1; 1, 3] Y with the given solution Y(t) = [te^(2t), -(t+1)e^(2t)], the initial condition Y(0) = [0, 2] can be directly applied. By substituting t=0 into the solution function, the specific values at that time are obtained, resulting in Y(0) = [0, 2e^(2*0)] = [0, 2]. There is no need for additional matrix multiplication or substitutions, as the solution function inherently accounts for the time variable. This straightforward approach confirms that the initial condition is satisfied. The process is simplified by directly evaluating the solution at the specified initial time.
BlkDaemon
Messages
15
Reaction score
0
I'm hacking at this particular linear system:

dY/dt = [1, -1; 1, 3] Y

I've already found myself a solution using the following function:

Y(t) = [ te^(2t), -(t+1)e^(2t) ]

That was fun, actually, once I figured out what the hell I was doing.

Here's my question: the next part of the problem asks for a solution to the intial value problem of the above differential equation, where

Y(0) = [0, 2]

I'm not sure how to attack the problem. Do I sub that matrix in for both values of Y, on both sides? If I do, then where do I put Y(t) so that this works out correctly? Or is there some other matrix multiplication step that I'm missing?

Any ideas?

Thanks in advance.

BlackDaemon
 
Physics news on Phys.org
You have a solution for the system. Generally, you will have a superpostion of two linearly independent solutions and you will adjust their coefficients to match the initial values.
 


First of all, great job on finding a solution to the given linear system! Now, to solve the initial value problem, you can simply plug in the values of Y(0) = [0, 2] into your solution function Y(t) = [te^(2t), -(t+1)e^(2t)]. This will give you the specific values of Y at time t=0. So, your solution to the initial value problem would be Y(t) = [0, 2e^(2t)]. This means that at time t=0, the values of Y are Y(0) = [0, 2], just as given in the problem.

You do not need to do any matrix multiplication or substitution, as your solution function already takes into account the values of Y at different times. Just plug in the values and you will have your solution to the initial value problem.

I hope this helps clarify the process. Keep up the good work!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top