# Question #f913b

Oct 7, 2017

A simple first ordinary differential equation: $y \left(t\right) ' + y \left(t\right) = 3$, the analytical solution reads: $y \left(t\right) = \left({y}_{0} - 1\right) {e}^{- t} + 1$

#### Explanation:

Introduction
Differential equations can be classified essentially on: ordinary and partial. Furthermore, they can be classified into: linear and nonlinear.

Fundamentaly. linear differential equations do not have variables in their coefficients. The simplest example are ordinary differential equations (ODEs).

The general form for linear first order ODEs:

$a \left(t\right) \cdot y \left(t\right) ' + b \left(t\right) \cdot y \left(t\right) = g \left(t\right)$

They all have analytical solutions.

A simple Linear ordinary differential equations

$y \left(t\right) ' + y \left(t\right) = 3$

Applying the following strategy, we can obtain the solution for any equation of this shape.

Multiply both side by $\mu \left(t\right)$, where $\mu \left(t\right)$ is a generic function:

$\mu \left(t\right) y \left(t\right) ' + \mu \left(t\right) y \left(t\right) = 3 \mu \left(t\right)$

See that it takes us to conclude:

$\mu \left(t\right) ' = \mu \left(t\right)$

By basic calculus, we can find:

$\mu \left(t\right) = C {e}^{t}$, take $C = 1$

Moreover, we can conclude that:

$\left({e}^{t} \cdot y \left(t\right)\right) ' = 3 {e}^{t}$, take $C = 1$

By calculus, we can find the solution:

$y \left(t\right) = \left({y}_{0} - 1\right) {e}^{- t} + 1$

${\lim}_{t \to \infty} y \left(t\right) = {y}_{0}$