Boundary and Initial Conditions

The general solution to a 1st Order DE always contains one undefined constant of integration, like $c$. A boundary (for x) or initial (for t) condition is given, typcially y(x) at a given x value.

Example: T goes from 90$^\circ$c to 70$^\circ$c in 10 minutes(t), the room temperature is 20$^\circ$c(T$_0$), find T after 20 minutes.

$$\frac{dT}{dt} = -\alpha(T-T_0)$$

$$\int \frac{dT}{T-T_0} = -\alpha \int dt$$

$$\ln(T-T_0) = -\alpha t + c$$

$$T-T_0 = e^{-\alpha t + c} = Ae^{-\alpha t}$$

$$A = e^c$$

We can now use our initial conditions to solve for T(20)

$$\textrm{When } t = 0$$

$$T-T_0 = 90 - 20 = 70$$

$$A = 70$$

$$\textrm{When } t = 10$$

$$T = T_0 + Ae^{-\alpha t}$$

$$70 = 20 + 70e^{-10\alpha }$$

$$e^{-10\alpha} = \frac{70-20}{70} = \frac{5}{7}$$

$$\textrm{When } t = 20$$

$$T = 20 + 70e^{-20\alpha}$$

$$(e^{-10\alpha})^2 = e^{-20\alpha}$$

$$T = 20 + 70\left(\frac{5}{7}\right)^2 = 55.7^\circ c$$