Kathleen Rausch

2022-01-18

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

${P}_{0}^{{}^{\prime}}\left(t\right)=\alpha {P}_{1}\left(t\right)-\beta {P}_{0}\left(t\right)$

${P}_{1}^{{}^{\prime}}=\beta {P}_{0}\left(t\right)-\alpha {P}_{1}\left(t\right)$

We also know${P}_{0}\left(t\right)+{P}_{1}\left(t\right)=1$

We also know

Jimmy Macias

Beginner2022-01-19Added 30 answers

Note that from the equation you have

${P}_{0}^{{}^{\prime}}\left(t\right)=\alpha {P}_{1}\left(t\right)-\beta {P}_{0}\left(t\right)=-{P}_{1}^{{}^{\prime}}\left(t\right)$

which gives us${P}_{0}^{{}^{\prime}}\left(t\right)+{P}_{1}^{{}^{\prime}}\left(t\right)=0$ which gives us ${P}_{0}\left(t\right)+{P}_{1}\left(t\right)=c$ . We are given that $c=1$ . Use this now to eliminate one in terms of the other.

For instance,${P}_{1}\left(t\right)=1-{P}_{0}\left(t\right)$ and hence we get,

${P}_{0}^{{}^{\prime}}\left(t\right)=\alpha (1-{P}_{0}\left(t\right))-\beta {P}_{0}\left(t\right)\Rightarrow {P}_{0}^{{}^{\prime}}\left(t\right)=\alpha -(\alpha +\beta ){P}_{0}\left(t\right)$

Let${Y}_{0}\left(t\right)={e}^{(\alpha +\beta )t}{P}_{0}\left(t\right)\Rightarrow {Y}_{0}^{{}^{\prime}}\left(t\right)$

$={e}^{(\alpha +\beta )t}[{P}_{0}^{{}^{\prime}}\left(t\right)+(\alpha +\beta ){P}_{0}\left(t\right)]=\alpha {e}^{(\alpha +\beta )t}$

Hence,${Y}_{0}\left(t\right)=\frac{\alpha}{\alpha +\beta}{e}^{(\alpha +\beta )t}+k$ i.e.

$P}_{0}\left(t\right)=\frac{\alpha}{\alpha +\beta}+k{e}^{-(\alpha +\beta )t$

$P}_{1}\left(t\right)=1-{P}_{0}\left(t\right)=\frac{\beta}{\alpha +\beta}-k{e}^{-(\alpha +\beta )t$

which gives us

For instance,

Let

Hence,

Stuart Rountree

Beginner2022-01-20Added 29 answers

Use ${P}_{0}\left(t\right)+{P}_{1}\left(t\right)=1$ to turn it into

${P}_{0}^{{}^{\prime}}\left(t\right)=\alpha -(\alpha +\beta ){P}_{0}\left(t\right)$ , 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${y}^{\prime}(x)=Ay(x)$ When A is a matrix with constants, the solution can be written in terms of the exponent matrix ${e}^{Ax}$ .

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My steps:

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