Initial value problems with matrices

[2RIP]

Given y'=[matrix]y + [matrix of constants]. y(0)=[initial conditions] Would I treat the matrix with variation of parameters? Or should I ignore the constants on right hand side and find the general solution. Then when it comes to solving the initial conditions, then do the calculations with it?

Thanks in advance.

 

[gabbagabbahey]

Well, if you had y' -Ay=B, and y(0)=C ,where A,B,C are known constants, what would you do?...Why not do the same thing with matrices?

 

[2RIP]

Yeah, I supposed I would've just done a separable equation with +B. Then initial condition will also be involved with B. So a constant wouldn't require me to use other methods of solving.

Thanks for explaining in an apprehensive way.

 

[2RIP]

Wait sorry, after trying the question out. How should I treat the matrix of constants from the beginning? I just ignored it and continued calculating for a general solution, then when it was time to calculate for initial conditions, I couldn't reason to where those constants would've come into play.

 

[gabbagabbahey]

Perhaps you should show me what your general solution for the question I posted earlier (the one without matrices) is.

 

[2RIP]

I ended up y= (e^At+CA+B)/A where C is the constant to be solved by initial condition.

 

[gabbagabbahey]

Hmm... that doesn't look quite right: using that solution, y(0)=(e^AC+B)/A...not C...I think you'd better show me your steps.

 

[2RIP]

y'-Ay=B
dy/dt=B+Ay
$$\int dy/(B+Ay)$$=$$\int dt$$
1/A(ln|B+Ay|)=t+C
B+Ay=e_A(t+C)
y=(e^A(t+C)-B)/A

Oops, it was -B sorry. I didn't solve for C because you said to solve for general solution? So that would mean only leave it with unidentified constants?

um.. but if I solved for C, then the solution is
y=(e^{A(t+(ln(B+AC)/A))}-B)/A

 

[HallsofIvy]

For any equation of the form Y'= Ay+ C, you can first solve Y'= AY to get Y(t)= De^At, with D an unknown constant matrix, and then look for a "specific solution" which will be a constant: if Y is a constant then Y'= 0 so AY+ C= 0 and Y= A^-1C (assuming A has an inverse).
The general solution to the entire equation is Y= De^At+ A^-1C and you can find D by using the initial condition.