- Separable equations
- Unique solutions
- Enveloping curve of a beam of curves
- Homogeneous functions. Application to differential equations
- Exact 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
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.
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