Hey friend, I want to share my experience, which is to be tutor to my friend. She is Anna Susanti (07305141027), my partner in this moment who learning my explanation. On Friday, 9th January 2009 at 2 pm, in Anna’s boarding home, I try to explain the mathematics topic about Homogeneous Equations. As reference, I use the book from Shepley L. Ross with title “Differential Equations”.
First, I explain about the definition of Homogeneous Equations. Then, I explain an example, theorem and another example to find the solution of Homogeneous Equations.
The first-order differential equation M (x, y) dx + N (x, y) dx = 0 is said to be homogeneous if, when written in the derivative form (dy / dx) = f (x, y), there exists a function g such that f (x, y) can be expressed in the form g (y / x).
The differential equation (x^2 – 3y^2) dx + 2xy dy = 0 is homogeneous. To see this, we first write this equation in the derivative form
dy / dx = (3y^2 – x^2) / 2xy
now observing that
(3y^2 – x^2) / 2xy = (3y / 2x) – (x / 2y) = (3/2) (y / x) – (1/2) (1 / (y / x) )
we see that the differential equation under consideration may be written as
dy / dx = (3/2) (y / x) – (1/2) (1/ (y / x) )
in which the right member is of the form g (y / x) for a certain function g.
M (x, y) dx + N (x, y) dy = 0 …………….(2.24)
Is a homogeneous equation, then the change of variables y = vx transforms (2.24) into a separable equatio0n in the variables v and x.
Proof. Since M(x, y) dx + N (x, y) dy = 0 is homogeneous, it may be written in the form
dy / dx = g (y / x)
Let y = vx. Then
dy / dx = v + x dv / dx
and (2.24) becomes
v + x dv / dx = g (v)
[v – g (v)] dx + x dv = 0
This equation is separable. Separating the variables we obtain
dv / (v – g (v)) + dx /x = 0 ………………(2.25)
Thus to solve a homogeneous differential equation of the form (2.24), we let y = vx and transform the homogeneous equation into a separable equation of the form (2.25).
From this, we have
int dv / (v – g (v)) + int dx / x = c,
where c is an arbitrary constant. Letting F(v) denote
int dv / (v – g (v))
and returning to the original dependent variable y, the solution takes the form
F (y / x) + ln (x) = c
Solve the equation
(x^2 – 3y^2) dx + 2xy dy = 0
We have already observed that this equation is homogeneous. Writing it in the form
dy / dx = - (x / 2y) + (3y / 2x)
and letting y = vx, we obtain
v + x dv / dx = - (1/ 2v) + (3v / 2),
x dv / dx = - (1/ 2v) + (v / 2)
x dv / dx = (v^2 - 1) / 2v
This equation is separable. Separating the variable, we obtain
2v dv / (v^2 - 1) = dx / x
Integrating, we find
ln (v^2 - 1) = ln x + ln c
(v^2 - 1) = cx
where c is an arbitrary constant. The reader should observe that no solutions were lost in the separation process. Now, replacing v by y / x we obtain the solutions in the form
(y^2 / x^2) – 1 = cx
(y^2 – x^2) = (cx) x^2
If y > x > 0, then this may be expressed somewhat more simply as
y^2 – x^2 = cx^3
And the last, I give Anna an exercise. She can solves that exercise successfully. But when I explained to Anna, I find some difficult, it’s because I explained it in English, so she had a few difficult to understand my explanation. But because this activity, I think now I and Anna become more understands about Homogeneous Equations.
For Anna, Thank you for your attention . . .