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 P0(t)+P1(t)=1

Name: 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 P0(t)+P1(t)=1
Uploaded: 2022-02-23T17:22:57+00:00
Description: 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 P0(t)+P1(t)=1

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Jimmy Macias · Accepted Answer

Note that from the equation you haveP0′(t)=αP1(t)−βP0(t)=−P1′(t)which gives us P0′(t)+P1′(t)=0 which gives us P0(t)+P1(t)=c. We are given that c=1. Use this now to eliminate one in terms of the other.For instance, P1(t)=1−P0(t) and hence we get,P0′(t)=α(1−P0(t))−βP0(t)⇒P0′(t)=α−(α+β)P0(t)Let Y0(t)=e(α+β)tP0(t)⇒Y0′(t)=e(α+β)t[P0′(t)+(α+β)P0(t)]=αe(α+β)tHence, Y0(t)=αα+βe(α+β)t+k i.e.P0(t)=αα+β+ke−(α+β)tP1(t)=1−P0(t)=βα+β−ke−(α+β)t

Stuart Rountree · Accepted Answer

Use P0(t)+P1(t)=1 to turn it intoP0′(t)=α−(α+β)P0(t), which you should be able to solve.

alenahelenash · Accepted Answer

There is a general method to solve such equations, if we view them as a linear system of equationy′(x)=Ay(x)When A is a matrix with constants, the solution can be written in terms of the exponent matrix eAx.

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...

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