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

$$P'_0(t) = \alpha P_1(t) - \beta P_0(t)$$ $$P'_1(t) = \beta P_0(t) - \alpha P_1(t)$$

We also know $P_0(t)+P_1(t)=1$

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3 Answers 3

up vote 4 down vote accepted

Note that from the equation you have $$P'_0(t) = \alpha P_1(t) - \beta P_0(t) = -P'_1(t)$$ 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, $$P'_0(t) = \alpha (1-P_0(t)) - \beta P_0(t) \Rightarrow P'_0(t) = \alpha - (\alpha + \beta)P_0(t)$$

Let $Y_0(t) = e^{(\alpha + \beta)t}P_0(t) \Rightarrow Y'_0(t) = e^{(\alpha + \beta)t} \left[P'_0(t) + (\alpha + \beta) P_0(t) \right] = \alpha e^{(\alpha + \beta)t}$

Hence, $Y_0(t) = \frac{\alpha}{\alpha + \beta}e^{(\alpha + \beta)t} + k$ i.e. $$P_0(t) = \frac{\alpha}{\alpha + \beta} + k e^{-(\alpha+\beta)t}$$ $$P_1(t) = 1 - P_0(t) = \frac{\beta}{\alpha + \beta} - k e^{-(\alpha+\beta)t}$$

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Thank you so much! Much appreciated. –  icobes Mar 19 '11 at 2:09
    
One question: I don't follow how you get a + ke^−(α+β)t term for P_0(t). Wasn't Y_0(t) a product? Thanks. –  icobes Mar 19 '11 at 2:15
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Use $P_0(t)+P_1(t)=1$ to turn it into $P_0'(t)=\alpha-(\alpha+\beta)P_0(t)$, which you should be able to solve.

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Could you give me a hint on how to solve P_0′(t)=(1−α−β)P_0(t)? I've never learned how to solve differential equations before and this was a question for a probability class so that is why I am completely lost. Thanks so much! –  icobes Mar 19 '11 at 2:05
    
Note I corrected the algebra. You might look at en.wikipedia.org/wiki/Linear_differential_equation for a start. –  Ross Millikan Mar 19 '11 at 2:09
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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 $e^{Ax}$.

More details can be found here.

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