How would I solve these differential equations? Thanks so much for the help!P0′(t)=αP1(t)−βP0(t)P1′=βP0(t)−αP1(t)We also know...

Kathleen Rausch

Answered question

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

Answer & Explanation

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

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

Skilled2022-01-24Added 366 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 ${e}^{Ax}$.