1st-order ordinary differential equations

Separable equations

The separable differential equations are those of the form where can be expressed either as the quotient or as the product .

In the first case, , while in the second case

Example 1: Solve the differential equation

Let's take so that and therefore the solution will be given implicitly by

Example 2: Find all the solutions of the equation

Let's and let's integrate each function: Solutions are Note that the function is also a solution of the equation, but it is not included in the general solution that is to say that for all C we will have . We call solutions not included in the general solution singular solutions of the differential equation.

Unique solutions

In example 2 we have seen that we can have solutions that are not included in the general solution, called singular solutions. When will this happen and how can we find those other solutions? Let's remember what we have seen in the post Ordinary differential equations: introduction, section "Bundle of curves and differential equations of the first order": the general solution of the ordinary differential equation represents a bundle of curves such that each curve of that beam satisfies the equation, that is, for each point the slope of any of the curves at that point fulfills .

Enveloping curve of a bundle of curves and singular solution

Two one-beam lines (blue and red) and beam envelope curve (green)

Given a bundle of curves the curve is called the beam envelope curve such that it is tangent to all the curves of the beam.

Example 3: The image shows two lines of the line bundle , specifically the two lines corresponding to the values in red and in blue, and the enveloping curve of the beam, which is tangent to the first line at a point close to and the second line about .

Remember that we can associate to any bundle of curves a differential equation such that the beam is the general solution of the equation: for each value of C, we have a curve of the beam such that its derivative verifies. But we have seen that the enveloping curve is tangent to all the curves of the beam, that is, its derivative at each point coincides with the derivative of one of the curves of the beam; this is why the enveloping curve is also a solution of the differential equation of the beam.

Example 4.Consider the parabolal bundle , all curves are tangent to the X axis at some point; the image depicts three of the beam curves, tangent to 0 at points 1.

Parabola beam 4y=(x + C)²

What is the differential equation of the curve beam? We derive with respect to x the equation of the beam: , and eliminate the constant C using this equality and the equation of the beam:

Note that the line , being tangent to all the curves of the beam, has the same derivative as the curves of the beam at each point on the X axis; therefore, it also verifies the differential equation of the beam. In general: if there exists an enveloping curve that is tangent to all the curves of a beam, then that curve will be a singular solution of the differential equation of the beam.

Calculating the envelope of a beam of curves

Given any value , the envelope will have the same derivative at that point as some curve of the beam, that is, there will be a constant C, dependent on , such that the derivative of the beam curve will have the same value at that point. We dispense with : in general for each x we will have a C(x). Let us derive with respect to x the expression of the beam, taking into account that y also depends on x, using the rule of the chain and the implicit derivation:

We clear y' to obtain the derivative at any point of any curve of the beam:

We now turn to the enveloping curve; at the contact points between each curve and the envelope the derivatives match the curves of the beam; in each given value x we will have determined a C(x), as the value y(x) also coincides with that of the beam, we will have that at the point of contact is fulfilled. We derive again with respect to x:

We clear and':

The two expressions for the derivative y' must be equal:

We see that it must be: that has two possible solutions, the trivial and the condition The first is an unimportant solution because it tells us that the constant C will not depend
of x (degenerate curve beam: all curves are equal), while the second is the general condition to be met by the envelope.

Example 5: Find the expression of the enveloping curve of the line bundle of example 3,

We express it in the form to derive with respect to C and equal to zero:

We substitute this value in the expression of the beam to remove C and get the equation of the envelope:

The envelope is

Example 6: Calculate the envelope of the curve beam of example 4.

Homogeneous functions. Application to differential equations.

A function is homogeneous of degree n if for every parameter Per example, the function is homogeneous of degree 2, since

In the special case that is homogeneous of degree 0 we have , and gives us a method of solving differential equations in which the F is homogeneous of degree 0: with the change of variable they become equations of separate variables. Indeed:

where we have used the x instead of the as a parameter of the homogeneous. Then

Example 7: Solve

then it is homogeneous of degree 0. We make the change to get it to be of separate variables:

Then In the first integral we make the variable change that makes it:

that it is of a rational type; we decompose into simple fractions:

We equate the two integrals, and undo the changes to obtain the general solution implicitly:

Reduction to homogeneous

The rational functions of form

they are only homogeneous of degree 0 when , however if we think of the numerator and denominator of the function as if they were two lines , we can try to find its intersection point , if it exists, then the variable change (which is a translation of the xy axes to the XY axes, taking as the coordinate origin the point ), converts the function into homogeneous of degree 0.

Example 8: Solve

We find the intersection of the lines:

we make the change which transforms the equation into:

containing a homogeneous function of degree 0; we make the change to separate the variables:

We decompose the first integral:

We equalize integrals and undo the change:

Special case: the lines are parallel

In the special case that the two lines have no intersection, then they are parallel, with proportional director vectors: ; but then we can do

with variable change becomes

which is of separate variables.

Example 9:

The lines and have no intersection, they are parallel; we make the change to obtain:

Exact equations

The equations of the form or equivalently, , we will say that they are exact as long as M and N are continuous functions and
check the condition

The solution will be given implicitly by where

Example 10: The equation is exact, because As will be and also equalizing the two expressions we have so the general solution is

Example 11: Solve the differential equation

As the equation is exact. Then

and when we equalize the two expressions we get

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