• Support PF! Buy your school textbooks, materials and every day products Here!

Solve the following intial value problem

  • Thread starter Rubik
  • Start date
  • #1
97
0

Homework Statement


Solve the following inital value problem

y1' = y2
y2' = -5y1-4y2

with inital conditions y1(0) = 1, y2(0) = 0

Homework Equations



(A-[itex]\lambda[/itex]I)x = 0

The Attempt at a Solution


So I first started by setting out like this

y2' = -5y1 - 4y2
which then suggest 0 = -[itex]\lambda[/itex](-4-[itex]\lambda[/itex]) + 5
0 = [itex]\lambda[/itex]2+4[itex]\lambda[/itex] + 5
But this is going to give me complex roots.. so I was just wondering have I made a mistake in my characteristic equation?

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,237
962
y1' = y2 → y1'' = y2'

So, the second equation becomes: y1'' = -5y1 - 4y1' .

D2y1 + 4Dy1 + 5y1 = 0

Can you take it from there?
 
  • #3
97
0
No sorry I don't understand why you have a D in there..
 
  • #4
867
0
D is the differential operator, basically another way of writing d/dx. D2y = d2y/dx2 = y''
 
  • #5
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,237
962
It's an 'operator' which just represents d/dx .

I get the same characteristic equation as you did, so not much help.
 
  • #6
11
0
\begin{equation}y_1^{\prime} = y_2 \end{equation} implies that \begin{equation}y_1^{\prime \prime} = y_2^{\prime} \end{equation}
The second equation you gave can thus be rewritten in terms of just y_1 and its derivatives,
\begin{equation}y_2^{\prime} = y_1^{\prime \prime}= -5y_1 - 4y_1^{\prime} \rightarrow y_1^{\prime \prime} + 4y_1^{\prime}+5y_1 = 0\end{equation}
You just have to solve the equation. Think of the physical parts of the equation. This should be decaying exponentially, but oscillating as well. Try guessing something like
\begin{equation}
y_1(t) = e^{-2t}(c_1 \sin(t) + c_2 \cos(t))
\end{equation}
Use your initial conditions to determine the constants.
 

Related Threads for: Solve the following intial value problem

Replies
6
Views
2K
Replies
6
Views
1K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
2
Views
2K
Replies
1
Views
1K
Replies
1
Views
751
Replies
1
Views
2K
Top