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

Jimmy Macias

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,
${P}_{0}^{{}^{\prime }}\left(t\right)=\alpha \left(1-{P}_{0}\left(t\right)\right)-\beta {P}_{0}\left(t\right)⇒{P}_{0}^{{}^{\prime }}\left(t\right)=\alpha -\left(\alpha +\beta \right){P}_{0}\left(t\right)$
Let ${Y}_{0}\left(t\right)={e}^{\left(\alpha +\beta \right)t}{P}_{0}\left(t\right)⇒{Y}_{0}^{{}^{\prime }}\left(t\right)$
$={e}^{\left(\alpha +\beta \right)t}\left[{P}_{0}^{{}^{\prime }}\left(t\right)+\left(\alpha +\beta \right){P}_{0}\left(t\right)\right]=\alpha {e}^{\left(\alpha +\beta \right)t}$
Hence, ${Y}_{0}\left(t\right)=\frac{\alpha }{\alpha +\beta }{e}^{\left(\alpha +\beta \right)t}+k$ i.e.
${P}_{0}\left(t\right)=\frac{\alpha }{\alpha +\beta }+k{e}^{-\left(\alpha +\beta \right)t}$
${P}_{1}\left(t\right)=1-{P}_{0}\left(t\right)=\frac{\beta }{\alpha +\beta }-k{e}^{-\left(\alpha +\beta \right)t}$

Stuart Rountree

Use ${P}_{0}\left(t\right)+{P}_{1}\left(t\right)=1$ to turn it into
${P}_{0}^{{}^{\prime }}\left(t\right)=\alpha -\left(\alpha +\beta \right){P}_{0}\left(t\right)$, which you should be able to solve.

alenahelenash

There is a general method to solve such equations, if we view them as a linear system of equation${y}^{\prime }\left(x\right)=Ay\left(x\right)$When A is a matrix with constants, the solution can be written in terms of the exponent matrix ${e}^{Ax}$.