Differential equations


Homogeneous first order differential equation

Homogeneous case sample

Inhomogeneous first order differential equation

Inhomogeneous case sample

Bernoulli differential equation

Bernoulli differential equation sample


In natural sciences and engineering sciences differential equations are very important as there are many natural laws expressed by them. A differential equation is the physical modelling of a natural phenomenon where the wanted function appears as function and its differentiation as well in one equation. The differentiation can appear in various orders (not only the first). The highest differentiation gives the order of the differential equation.

The simplest form of a differential equation is the homogeneous first order differential equation. So we start with that.


Chapters

Homogeneous first order differential equation

In the homogeneous first order differential equation there is just the wanted function and its differentiation and no additional term. Its formulation is:


DiffEqu

or


DiffEqu

To solve this equation for the wanted function y(x) a so called variable separation is carried out. That means the equation is multiplied by dx and divided by y to get x onto the right and y onto the left side:


DiffEqu

Now both sides must be integrated.


DiffEqu

The left side is integrated with y and the right side with x. Here the initial values x0 and y0 must be taken into account.

In a first step only the integration of left side is carried out:


DiffEqu

And so


DiffEqu

Now on both sides the exponential function is applied:


DiffEqu

and that’s the general solution for the homogeneous first order differential equation.




Homogeneous case sample

In the sample:


DiffEqu

There is


DiffEqu

and


DiffEqu

and with this


DiffEqu

and if x0 = 0


DiffEqu


Inhomogeneous first order differential equation

The inhomogeneous first order differential equation is more or less an extension of the homogeneous one. It has an additional term f(x) on the right side. Its formulation is:


DiffEqu


Where f(x) is called the disruptive term.

To solve this equation a so called variation of the constants is carried out.

Starting from the solution for the homogeneous case


DiffEqu


The constant y0 is replaced by a function c(x) and say


DiffEqu

and the differentiation of φ(x) is:


DiffEqu


and with this φ(x) the homogeneous case becomes


DiffEqu


and its differentiation


DiffEqu


Or with the formulation for φ’(x) from above


DiffEqu


Now. As


DiffEqu


we can replace c(x) * φ(x) by y and get


DiffEqu


If we compare this with the equation of the inhomogeneous case


DiffEqu


Its obvious that


DiffEqu


and that’s a homogeneous differential equation for c(x)


DiffEqu


and the solution for this is


DiffEqu


With this inserted in


DiffEqu


the solution for the inhomogeneous case becomes


DiffEqu


with φ(x) as we defined further above:


DiffEqu


Where c0 = y0


If this solution is written without brackets


DiffEqu


The part φ(x)* c(x) is the solution of the homogeneous case and the part on the very right is the extension to the inhomogeneous case.



Inhomogeneous case sample

In the sample of an inhomogeneous differential equation:


DiffEqu

There is


DiffEqu

and so


DiffEqu

and


DiffEqu

DiffEqu

And with this


DiffEqu

The solution :-)




Bernoulli differential equation

A differential equation like


DiffEqu


is called Bernoulli differential equation.

To solve such an equation the approach is the substitution


DiffEqu


Where z shall be a function.

For the substitution of y’(x) the chain rule for differentiations must be applied (see Differential calculus). If z is a function f(x), zk is a chain like


DiffEqu


Where


DiffEqu


and with the chain rule


DiffEqu


we get


DiffEqu


And therefore


DiffEqu


With this the equation becomes


DiffEqu


or divided by k * zk-1:


DiffEqu


Now, if k is defined as


DiffEqu


Then


DiffEqu


And we get a linear inhomogeneous differential equation


DiffEqu


That can be solved with the approach from above with


DiffEqu


or with the n of yn


DiffEqu


Bernoulli differential equation sample


The sample


DiffEqu


Is a Bernoulli differential equation.

To solve this, it must be brought into the form of the Bernoulli differential equation by dividing the whole equation by x3.


DiffEqu


Now we have


DiffEqu


And as n = 3


DiffEqu


With this


DiffEqu


With the solution for the homogeneous case


DiffEqu

DiffEqu


and the inhomogeneous part


DiffEqu


With these the solution for z becomes:


DiffEqu


And as we made the substitution y = zk, we have to undo that to get the solution for y and get:


DiffEqu


That’s it :-)