# Solve the following intial value problem

## 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?

## The Attempt at a Solution

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SammyS
Staff Emeritus
Homework Helper
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y1' = y2 → y1'' = y2'

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

D2y1 + 4Dy1 + 5y1 = 0

Can you take it from there?

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

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

SammyS
Staff Emeritus
Homework Helper
Gold Member
It's an 'operator' which just represents d/dx .

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

$$y_1^{\prime} = y_2$$ implies that $$y_1^{\prime \prime} = y_2^{\prime}$$
The second equation you gave can thus be rewritten in terms of just y_1 and its derivatives,
$$y_2^{\prime} = y_1^{\prime \prime}= -5y_1 - 4y_1^{\prime} \rightarrow y_1^{\prime \prime} + 4y_1^{\prime}+5y_1 = 0$$
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

y_1(t) = e^{-2t}(c_1 \sin(t) + c_2 \cos(t))

Use your initial conditions to determine the constants.