Proof of first order ODE nonhomogenuous

Question

Find the general solution of:

$$x'(t)=f(t)x(t)+g(t)$$

How do I prove the general solution for this? What are the intuitive steps to solve something like this? What do I mean by that is that for:

$$x'(t)=x(t)f(t)$$

you just divide by x(t) on both sides and you get:

$$\frac{x'(t)}{x(t)}=f(t)$$

and then you integrate both sides and you get the general solution of:

$$x(t)=ke^{\int f(t)dt}$$

Which is enough for me to understand this proof, how about for the nonhomogenous? How about for the second or higher order differential equations?

Answer 1

Write the equation as $x'(t)-f(t)x(t)=g(t)$. Multiply by (the so-called integrating factor) $e^{-h(t)}$ where $h(t)$ is an anti-derivative of $f(t)$. We get $\frac d {dt} (e^{-h(t) x(t)})= e^{-h(t)} (x'(t)-f(t)x(t))=e^{-h(t)}g(t)$. Integrating this you can write down the general solution of the DE as $x(t)=e^{h(t)}(c+\int_0^{t} e^{-h(s)}g(s)\, ds)$ where $c$ is a constant.

Answer 2

Dividing by $x(t)$ is not such a great idea, as you do not know whether it vanishes. Instead:

Step 1. Multiply the whole equation by the integrating factor $$ m(t)=\exp\left(-\int_{t_0}^t f(s)\,ds\right), $$ which is positive, and hence no loss of information. Pick first a point $t_0$ in the interval where $f,g$ are continuous. Then $m'=-fm$ and hence $$ m(x'-fx)=mg\quad\Longrightarrow\quad mx'+m'x=m(x'-fx)=mg \quad\Longrightarrow\quad (mx)'=mg $$ Then integrate, over $[t_0,t]$ and obtain $$ m(t)x(t)-m(t_0)x(t_0)=\int_{t_0}^t m(s)g(s)\,ds $$ or $$ x(t)=\exp\left(\int_{t_0}^t f(s)\,ds\right)x(t_0)+\int_{t_0}^t \exp\left(\int_s^t f(\sigma)\,d\sigma\right)g(s)\,ds $$