Equations of the first order unsolved by derivatives.
We consider the equation
F (x, y, y’) = 0 (1)
and suppose that it allows the parametric representation
х=ϕ(u, v)
у=ψ(u, v) (2)
у'=α(u, v) so that F(ϕ(u, v) ,ψ(u, v) ,α(u, v) ) ≡0, for all values of the parameters u and v . We assume, that the functions ϕ(u, v) , ψ(u, v) , α(u, v) are differentiable. Using the basic relation between the differentials and the derivative along the integral curves of the 1-st order dy=у’dx we find the connection between the parameters u and v . In fact, we have
dx
=
du+
dv
dy=
du+
dv
у’=α(u,
v)
Thus, we
obtain
du+
dv
= α(u,
v)[
du+
dv
]
(3)
Equation (3) is the equation solved by derivative.
In the equation (3) variables u and v are equal. Taking, for example, u as independent variable, and integrating equation (3), we obtain v = ω(u, c) which is general solution of (3).
6) Method of introducing a parameter. Lagrange and Clairaut equations.
Lets consider the equation
F (x, y, y’) = 0 (1)
and suppose that it allows the parametric representation
х=ϕ(u, v)
у=ψ(u, v) (2)
у'=α(u, v)
so that F(ϕ(u, v) ,ψ(u, v) ,α(u, v) ) ≡0, for all values of the parameters u and v . We assume, that the functions ϕ(u, v) , ψ(u, v) , α(u, v) are differentiable. Using the basic relation between the differentials and the derivative along the integral curves of the 1-st order dy=у’dx we find the connection between the parameters u and v .
In fact, we have
dx
=
du+
dv,
dy =
du+
dv
у’=α(u,
v)
.
Thus, we obtain
du+ dv = α(u, v) *( du+ dv ) (3)
Equation (3) is the equation solved by derivative.
In the equation (3) variables u and v are equal. Taking, for example, u as independent variable, and integrating equation (3), we obtain v = ω(u, c) which is general solution of (3). Then x = ϕ(u,ω(u, c)), y = ψ(u,ω(u, c)) is general solution of (1) in parametric form.
The practical application of this method involves overcoming two difficulties:
1) find a parametric representation of the equation (1);
2) integration of the equation (3).
The first difficulty is easily overcome, if (1) is solved for x or y, i.e. has the form
y = ξ(x, y’ ) (4)
x = η( y, y ’ ) (5)
Equation (4) admits a parametric representation
x = x, y’ = p, y = ξ(x, p) (6)
and (5) the parametric representation as
y = y, y’ = p, x = η( y, p) (7)
Both difficulties can be overcome by integrating the equations of Lagrange and Clairaut. Equation of the form
y = ϕ( y’ )x + ψ( y’ ), (8)
where ϕ( y’ ) and ψ( y’ ) are continuous functions is called the equation of Lagrange.
If ϕ( y’ ) = y’ , then (8) is called the equation of Clairaut.