31.Differential equations in full differentials .Integral multiplier.

Differential equation M(x,y)dx+N(x,y)dy=0 (1)

is called a differential equation in full differential, if the left part of a complete differential some smooth functions , i.e. if

M(x,y)= (x,y) N(x,y)= (x,y)

 A necessary and sufficient condition for the existence of such a function is of the form:

= (2)

To solve a differential equation in diferentials need to find a functionF(x,y)

Then the General solution of a differential equation can be written in the formF(x,y)=C

for an arbitrary constantC.

The integrating multiplier for differential equationsM(x,y)dx+N(x,y)dy=0

the name of the function g(x,y), after multiplication by which the differentialthe equation is transformed into full diferential. If functions M and N in the equation have continuous partial derivatives and do not vanish simultaneously, the integrating factor exists. However, a General method for finding does not exist.

32.Integration methods. Homogeneous and linear differential equations of the first order.

F (x, y, y’) = 0 (1)

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 .

Infact, we have

dx =(∂ϕ/du) du+(∂ϕ/dv) dv,dy =(∂ψ/du)du + (∂ψ/dv) dv,

Thus, weobtain

(∂ψ/ du) du + (∂ψ/dv) dv=α(u, v)[(∂ϕ/ du) du+(∂ϕ/dv) 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) integrationoftheequation (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.

33.Incomplete equations . Equations assuming reduction of order.

A second order differential equation is written in general form as

F(x,y,y’,y”)=0

where F is a function of the given arguments.  If the differential equation can be resolved for the second derivative y'', it can be represented in the following explicit form:

y”=f(x,y,y’)

In special cases the function f in the right side may contain only one or two variables. Such incomplete equations  include 5 differenttypes:

y’’=f(x), y’’=f(y), y’’=f(y’), y’’=f(x,y’) , y’’=f(y,y’)

With the help of certain substitutions, these equations can be transformed into first order equations.  In the general case of a second order differential equation, its order can be reduced if this equation has a certain symmetry. We consider some cases of equations such as:

Case 1. Equation of type y''= f (x)

For an equation of type y'' = f(x), its order can be reduced by introducing a new function p(x), such that y' = p(x).As a result, we obtain the first order differential equation

P’=f(x)

Solving it, we find the function p(x). Then we solve the second equation

Y’=p(x)

and obtain the general solution of the original equation.

Case 2. Equation of type y''= f (y)

The right-hand side of the equation depends only on the variable y. We introduce a new function p(y), setting y' = p(y). Then we can write:

Y’’= = = * = p

So the equation becomes:

p=f(y)

Solving it, we find the function p(y). Then we find the solution of the equation y' = p(y), that is, the function y(x).