Question

Question #f913b

Answer 1

A simple first ordinary differential equation: `y(t)' + y(t)=3`, the analytical solution reads: `y(t)=(y_0 - 1)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(t)*y(t)' + b(t)*y(t)=g(t)`

They all have analytical solutions.

A simple Linear ordinary differential equations

`y(t)' + y(t)= 3`

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

Multiply both side by `mu(t)`, where `mu(t)` is a generic function:

`mu(t)y(t)' + mu(t)y(t)= 3mu(t)`

See that it takes us to conclude:

`mu(t)'= mu(t)`

By basic calculus, we can find:

`mu(t)= Ce^(t)`, take `C=1`

Moreover, we can conclude that:

`(e^(t)*y(t))'= 3e^(t)`, take `C=1`

By calculus, we can find the solution:

`y(t)= (y_0 - 1)e^(-t) + 1`

Some words about the solution

It does not matter the initial solution, it will always converge to it; it is a nice property since as long as you give enough time, the system will always come back to the initial state.

`lim_(t->oo)y(t)=y_0`

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