Expert Answer to "How would I solve these differential equations? Thanks..."

Kathleen Rausch

Answered question

2022-01-18

How would I solve these differential equations? Thanks so much for the help!

P_{0}^{^{'}} (t) = α P_{1} (t) - β P_{0} (t)

P_{1}^{^{'}} = β P_{0} (t) - α P_{1} (t)

We also know

P_{0} (t) + P_{1} (t) = 1

Jimmy Macias

Beginner2022-01-19Added 30 answers

Note that from the equation you have

P_{0}^{^{'}} (t) = α P_{1} (t) - β P_{0} (t) = - P_{1}^{^{'}} (t)

which gives us

P_{0}^{^{'}} (t) + P_{1}^{^{'}} (t) = 0

which gives us

P_{0} (t) + P_{1} (t) = c

. We are given that

c = 1

. Use this now to eliminate one in terms of the other.
For instance,

P_{1} (t) = 1 - P_{0} (t)

and hence we get,

P_{0}^{^{'}} (t) = α (1 - P_{0} (t)) - β P_{0} (t) \Rightarrow P_{0}^{^{'}} (t) = α - (α + β) P_{0} (t)

Let

Y_{0} (t) = e^{(α + β) t} P_{0} (t) \Rightarrow Y_{0}^{^{'}} (t)

= e^{(α + β) t} [P_{0}^{^{'}} (t) + (α + β) P_{0} (t)] = α e^{(α + β) t}

Hence,

Y_{0} (t) = \frac{α}{α + β} e^{(α + β) t} + k

i.e.

P_{0} (t) = \frac{α}{α + β} + k e^{- (α + β) t}

P_{1} (t) = 1 - P_{0} (t) = \frac{β}{α + β} - k e^{- (α + β) t}

Stuart Rountree

Beginner2022-01-20Added 29 answers

Use

P_{0} (t) + P_{1} (t) = 1

to turn it into

P_{0}^{^{'}} (t) = α - (α + β) P_{0} (t)

, which you should be able to solve.

alenahelenash

Expert2022-01-24Added 556 answers

There is a general method to solve such equations, if we view them as a linear system of equation

y^{'} (x) = A y (x)

When A is a matrix with constants, the solution can be written in terms of the exponent matrix

e^{A x}

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