First Order, Linear

A differential equation of the form

$$\frac{d p(T)}{dT} + g(T) p(T) = q(T)$$

where g(T) and q(T) are known, i.e., given functions, and p(T) is the desired solution, a function of T, can be solved using integration factors. We note that multiplying by a factor (so far unknown) h(T), one would have

$$h(T)\frac{d p(T)}{dT} + h(T)g(T) p(T) = h(T)q(T)$$

which could be rewritten as

$$\frac{d h(T)p(T)}{dT}- p(T)\frac{d h(T)}{dT} + h(T)g(T) p(T) = h(T)q(T)$$ (5)

If the second and third terms could be made to cancel, i.e., if

$$p(T)\frac{d h(T)}{dT} = h(T)g(T) p(T)$$ (6)

the resulting differential equation (5) would read

$$\frac{d [h(T)p(T)]}{dT}= h(T)q(T)$$

which would be integrable. But p(T) cancels from both sides of the equation (6)

$$p(T)\frac{d h(T)}{dT} = h(T)g(T) p(T)$$

$$\frac{d h(T)}{dT} = h(T)g(T)$$

so, dividing both sides by h(T) and multiplying both sides by dT and integrating one has

$$\int \frac{1}{h(T)}\frac{d h(T)}{ dT} dT= \int g(T) dT$$

Integration of the left hand side gives

$$\ln h(T) = \int g(T) dT$$

that is

$$h(T) = e^{\int g(T) dT}$$

which is, acutally, a formula for h(T).

As an example consider the first-order equation

$$\frac{dp(T)}{dT} + 2 T p(T) = T \,\,;\,\, p(1)=3$$

where g(T) = 2T and q(T) = T. One has

$$\int 2T dT = T^2$$

so

$$h(T) = e^{\int 2T dT} = e^{T^2}$$

so

$$\frac{d[p(T)e^{T^2}]}{dT} -2T e^{T^2} p(T) +2T e^{T^2} p(T) = e^{T^2} T$$

Cancelling, one has

$$p(T) e^{T^2} = \int Te^{T^2} dT = \frac{e^{T^2}}{2} + C$$

Finally, solving for p(T) produces

$$p(T) = \frac{1}{2} + Ce^{-T^2}$$

One now uses the initial condition to finish. Thus p(1) = 3

$$3 = \frac{1}{2} + C e^{-1^2}$$

giving an equation for C.