Differential equations are useful for modelling problems.
A differential equation $dy/ty=f(t,y)$ which can cleanly be written as the product of a function $p(t)$ of only the dependent variable with a function $q(y)$ of only the independent variable, $$ \frac{dy}{dt} = f(t,y) = p(t)q(y), $$ is called separable. Suppose that $q(y)$ is non-zero for $y$ value(s) of interest. Then the solution of this differential equation satisfies $$ \int q^{-1}(y)dy = \int p(t)dt. $$
Suppose $dy/dt=5t^4 y$ and $y(0)=18$. Letting $p(t)=5t^4$ and $q(y)=y$, $$ \int y^{-1}dy = \int 5t^4 dt \Rightarrow \ln y = t^5 + C \Rightarrow e^{\ln y} = y = e^{t^5 + C} = Ce^{t^5}. $$ Using the initial conditions, $$ y(0) = 18 = Ce^{0^5} = Ce^0 = C \Rightarrow C = 18, $$ and so the solution is $y(t) = 18e^{t^5}$.
| $f(t)=\mathcal{L}^{-1}[F(s)]$ | $F(s)=\mathcal{F}[f(t)]$ |
|---|---|
| 1 | $1/s$ (for $s > 0$) |
| $e^{at}$ | $1/(s-a)$ (for $s > a$) |
| $t^n$ (for integer $n > 0$) | $n!/s^{n+1}$ (for $s > 0$) |
| $t^p$ (for $p > -1$) | $\Gamma(p+1)/s^{p+1}$ (for $s > 0$) |
| $\sin(at)$ | $a/(s^2+a^2)$ (for $s > 0$) |
| $\cos(at)$ | $s/(s^2+a^2)$ (for $s > 0$) |
| $\sinh(at)$ | $a/(s^2-a^2)$ (for $s > \lvert a\rvert$) |
| $\cosh(at)$ | $s/(s^2-a^2)$ (for $s > \lvert a\rvert$) |
$$ e^{ix} = \cos x+i\sin x $$