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What the Hell. Differential Eq

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[SOLVED] What the Hell. Differential Eq

Okay, so I used Method of undetermined coefficients for this one. I got a solution that when differentiated and plugged back in almost works, bit is off by a little.


Picture1-8.png



When I plug back into y'=4x+3y+(-5t+6)
I get y'=........+9 not 6. So I have a feeling it has to do with my D=11/7 term..... but I have no idea what.

Anyone see what I am doing wrong? Is my matrix incorrect? Or my assumption of Xp? Or neither?

Thanks
 

Answers and Replies

2,981
2
Sorry, I forgot to change the matrix....this is what I used to get A,B,C,D

[tex]\left[\begin{array}{cccc}6 & 1& 0 & 0\\4 & 3 & 0 & 0\\ 1 & 0 & 6 & 1\\
0 & -1 & 4 & 3\end{array}\right]*\left[\begin{array}{c}A\\B\\C\\D\end{array}\right]=\left[\begin{array}{c}-3\\5\\0\\0\\
\end{array}\right][/tex]
 
MathematicalPhysicist
Gold Member
4,139
149
well after solving:
t: 6A+B+3=0
4A+3B-5=0
you get 14A+14=0
B=3 after solving for the next equation of t^0 you get that:
C=0 and D=-1.
I think in this simple case it's better to just solve it without matrix.
 
HallsofIvy
Science Advisor
Homework Helper
41,738
898
Your matrix is not correct. For example, your third row says "A+ 6C+ D= 0" but the corresponding equation (A= 6At+ 6C+ Bt+ D+ 3t with t= 000) is "A= 6B+ D" which is "-A+ 6C+ D= 0".
 
exk
119
0
have you tried taking the laplace x-form of your system? (sometimes that simplifies the problem)
 
2,981
2
well after solving:
t: 6A+B+3=0
4A+3B-5=0
you get 14A+14=0
B=3 after solving for the next equation of t^0 you get that:
C=0 and D=-1.
I think in this simple case it's better to just solve it without matrix.
Right. I actually just use a CAS to solve the matrix for me...so that is easier. I just could not figure out where my matrix was wrong, but I have got the same numbers as you now.

Now I am having trouble getting his to plug back in again!
 
2,981
2
GOT IT! Thanks!
 

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