Solving 1st order lin. diff. equations with continuous coefficient

Let be a function of class on an interval .

As well, let be a continuous function and any function.

Let be a linear differential equation or order with continuous coefficient, and its associated homogeneous equation:

Solving the homogeneous equation

is solution, but considering as a non-zero function, the following function will be an homogeneous solution of :

Solving the general equation

Function will be specific solution of :

We will have as a total solution of , the addition of these two solutions:

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Demonstration

Let be a function of class on an interval .

As well, let be a continuous function and any function.

Solving a first order linear differential equation with continuous coefficient consists of finding functions , which are solutions of an equation of type:

We can first find solutions to the homogeneous equation :

If is solution of the homogeneous equation , and is specific solution to the equation , then:

By performing the operation :

Thanks to the linearity of the derivative, we notice that remains solution of , because:

As a result, will be a total solution of :

Solving the homogeneous equation

We call homogeneous equation , the equivalent of the equation but without any right side member.

Thus, est solution évidente, but let us consider as a non-zero function.

From , let us try to find a function which corresponds:

So, calculating its antiderivative,

We then have as an homogeneous solution for :

There are indeed an infinite number of solutions for , corresponding to ignorance of the value of the constant . he latter can be set subsequently based on the existence of initial conditions.

1. Verification of the homogeneous solution

If is solution of , then:

Now, we apply the product of two functions:

is definitely a homogeneous solution of .

Solving the general equation

Knowing a solution to the homogeneous equation , we can start from this to find specific solutions of .

1. The method of the variations of parameters

Indeed, when looking for a specific solution de type :

By applying the method of the variations of parameters (or Lagrange's method), we do have:

Given the following system:

Thanks to both equations and , we deduce an equation to solve, to determine :

Let us antiderivate each member:

We inject into :

We then have as a specific solution for :




2. Verification of the specific solution

if is solution of , then:

We again apply the product of two functions:

So,

is definitely a specific solution of .




3. Obvious solutions

In simple cases, we can directly try to find solutions, in specific by identifying the coefficients.

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Examples

Here are several examples of how to resolve a having the following general form:

1. With a constant coefficient

Let us take a constant coefficient , and a function , then the equation now becomes:




1. Solving the homogeneous equation

We then have a homogeneous solution :

2. Solving the general equation

We are looking for a specific solution of the type:

To do this, we can use the method of the variations of parameters, or use the coefficient identification method.




1. By the method of the variations of parameters

Using this method, we do have:

Given the following system:

Thanks to both equations and , we deduce an equation to solve, to determine :

We antiderivate each side to determine :

We perform an integration by parts with the following choice of and :

We perform another an integration by parts:

Now, we inject into :




2. By the coefficient identification method

Given the relative simplicity of the equation :

We can look for a specific solution of type second degree polynomial:

By injecting into , we do have:

Which leads us to the following system:

So, we directly obtain solutions for :

And therefore a solution for :

3. Addition of solutions

The total solution of is the addition of both solutions and , so:

4. Verification of the total solution

If is a solution of , then:

Let us check it.

is definitely a total solution of .

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2. With a continuous coefficient

Let us take a continuous coefficient , and a function , then the equation becomes:




1. Solving the homogeneous equation

We then have a homogeneous solution :

2. Solving the general equation

We are looking for a specific solution of the type:




1. By the method of the variations of parameters

Using this method, we do have:

Given the following system:

Thanks to both equations and , we deduce an equation to solve, to determine :

We antiderivate each side to determine :

Now, injecting into :

3. Addition of solutions

The total solution of s the addition of the two solutions and , so:

4. Verification of the total solution

If is solution of , then:

Let us check it.

is definitely a total solution of .