### Solving pure-time differential equations through integration

\begin{align*} \diff{F}{t} = f(t) \end{align*}

Goal: determine the solution $F(t)$.

$F(t)$: the antiderivative of $f(t)$.

A function has one derivative.

A function has many antiderivatives:
$F(t) + C$ for any constant number $C$

Write antiderivative in terms of an integral: $$\int f(t)dt = F(t) + C$$

Example: \begin{align*} \diff{F}{t} = 4t \end{align*}

Antiderivative: {}

$$\int 4t\,dt =2 t^2 + C$$

### Integration: derivative rules in reverse

Power rule for derivatives: $\displaystyle \diff{t^{n+1}}{t} = (n+1)t^n$   $\Rightarrow$   $\displaystyle \diff{}{t}\left(\frac{t^{n+1}}{n+1}\right)=t^n$

Power rule for integrals: $\displaystyle \int t^n dt = \frac{t^{n+1}}{n+1} + C$       (need $n \ne -1$)

Example:

$$\int t^2 dt = \frac{t^{2+1}}{2+1}+C = \frac{t^3}{3} + C$$

Back to first example: $\displaystyle \diff{F}{t} = 4t$

$\displaystyle F(t) = \int 4t \, dt$ $\displaystyle = 4 \int t \, dt$ $\displaystyle = 4 \int t^1 \, dt$

$\displaystyle = 4 \frac{t^2}{2} +C = 2t^2 +C$

Example: $\displaystyle \int (2x^3-4x+1)dx$

$\displaystyle = 2 \int x^3dx - 4 \int x\,dx + \int 1 \, dx$ $\displaystyle = 2 \int x^3dx - 4 \int x\,dx + \int x^0 \, dx$

$\displaystyle = 2 \frac{x^4}{4} - 4 \frac{x^2}{2} + \frac{x^1}{1} + C$

$\displaystyle = x^4/2 - 2x^2 + x + C$

### Exponential rule for integration

Recall derivative:

$\displaystyle \diff{}{t} e^t = e^t$

Reverse for integration:

$\displaystyle \int e^t dt = e^t +C$

$\displaystyle \diff{}{t} e^{3t} = 3 e^{3t}$

$\displaystyle \int 3 e^{3t} dt = e^{3t} +C$

$\displaystyle \diff{}{t} e^{kt} = k e^{kt}$

$\displaystyle \int k e^{kt} = e^{kt} +C$

More examples:

$\displaystyle \int (3+ 10e^{5t}) dt$ $\displaystyle = \int 3 dt + 2\int 5e^{5t} dt$ $\displaystyle =3t + 2e^{5t} +C$

$\displaystyle\int e^{7x} dx$ $\displaystyle = \frac{1}{7}\int 7e^{7x} dx$ $\displaystyle = \frac{1}{7} e^{7x} +C$

### Integration rule for 1/t

Power rule for $n \ne -1$:

$\displaystyle \int t^n dt = \frac{t^{n+1}}{n+1} + C$.

What about $n=-1$? $\displaystyle \int \frac{1}{t} dt = ?$

Recall derivative of logarithm:

$\displaystyle \diff{}{t} \ln(t) = \frac{1}{t}$

Reverse: $\displaystyle \int \frac{1}{t} dt = \ln(t) + C$

Logarithm: defined only for $t > 0$.
Integral: symmetric for $t<0$.

Add absolute value to fix rule:

$$\int \frac{1}{t} dt = \ln(|t|) + C$$