Solve the following intial value problem

[Rubik]

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-$\lambda$I)x = 0

The Attempt at a Solution

So I first started by setting out like this

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

 

[SammyS]

y₁' = y₂ → y₁'' = y₂'

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

D²y₁ + 4Dy₁ + 5y₁ = 0

Can you take it from there?

 

[Rubik]

No sorry I don't understand why you have a D in there..

 

[Bohrok]

D is the differential operator, basically another way of writing d/dx. D²y = d²y/dx² = y''

 

[SammyS]

It's an 'operator' which just represents d/dx .

I get the same characteristic equation as you did, so not much help.

 

[Octonion]

\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.