Question

Solve these linear equations in the form y=yn+yp with yn=y(0)e^at.

a. y'-4y=-8

b. y'+4y=8

which one has a steady state?

Answer 1

a)
`y(t) = y_(0)e^(4t) + 2`. It does not have a steady state

b)
`y(t) = y_(0)e^(-4t) + 2`. It has a steady state.

Explanation:

a)
`y' -4y = -8`

The first step is finding
`y_(n)(t)`. So:

`y' - 4y = 0`

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

`r - 4 = 0`

`r = 4`

So:

`y_(n)(t) = y_(0)e^(4t)`

Since this differential equation has a positive eigenvalue, it does not have a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

`y_(p)(t) = C`

So

`(y_(p))' -4(y_(p)) = -8`

`(C)' - 4C = -8`

C is a constant, so (C)' = 0.

`-4C = -8`

`4C = 8`

`C = 2`

The solution in the form is

`y(t) = y_(n)(t) + y_(p)(t)`

`y(t) = y_(0)e^(4t) + 2`

b)
`y' +4y = 8`

The first step is finding
`y_(n)(t)`. So:

`y' + 4y = 0`

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

`r + 4 =`

`r = -4`

So:

`y_(n)(t) = y_(0)e^(-4t)`

Since this differential equation does not have a positive eigenvalue, it has a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

`y_(p)(t) = C`

So

`(y_(p))' +4(y_(p)) = 8`

`(C)' + 4C = 8`

C is a constant, so (C)' = 0.

`4C = 8`

`C = 2`

The solution in the form is

`y(t) = y_(n)(t) + y_(p)(t)`

`y(t) = y_(0)e^(-4t) + 2`