The method of variation of post-linear heterogeneous equalities. Method of variation of prevіlnyh fast

The method of variation is fairly constant, and the Lagrange method is another way of developing linear differential equations of the first order and the Bernoulli equation.

Linear differential alignment of the first order - equal to the form y + p (x) y = q (x). How the right side should be zero: y'+p(x)y=0, ce - linear uniformly equal to the 1st order. Obviously, equal to the non-zero right part, y'+p(x)y=q(x), heterogeneous linear alignment of the 1st order.

Method of variation of fairly constant (Lagrange method) polygaє in the offensive:

1) It’s possible to solve the homogeneous equalization y+p(x)y=0: y=y*.

2) In the final solution, Z is important not as a constant, but as a function in xu: C \u003d C (x). It is known that the verbal solution (y *) 'and in the mind of the mind is represented by the subtraction of viraz for y* and (y*)'. From the taken equality, we know the function (x).

3) In a global solution of a homogeneous equality, knowledge of C(x) virases is presented.

Let's take a closer look at the method of variation of fairly constant. Vіzmemo y yourself zavdannya, scho y u povnyaєmo hіd dіshennya і perekonaєmosya, scho otmanі vіdpovіdі zbіgayutsya.

1) y'=3x-y/x

Let's rewrite the level of the standard look (on the basis of the Bernoulli method, we need to write down the form of the bula only for that, schobachit, that the level is linear).

y'+y/x=3x (I). Now diemo behind the plan.

1) It is more uniformly equal y+y/x=0. The price is equal to the changes that are divided. We represent y'=dy/dx, we represent: dy/dx+y/x=0, dy/dx=-y/x. Offenses of parity are multiplied by dx and divisible by xy≠0: dy/y=-dx/x. Integrable:

2) In the omnipotent solution of homogeneous equality, C is taken into account not by a constant, but by a function like x: C=C(x). Zvіdsi

Otrimanі virazi podstavlyaєmo for mind (I):

Integrating insults to parts of equality:

here C is already a new constant.

3) In the final solution of the homogeneous alignment y=C/x, demi took into account C=C(x), so y=C(x)/x, instead of C(x), we represent the knowledge of viraz x³+C: y=(x³ + C)/x or y=x²+C/x. They took the same vіdpovіd, like and pіd hоvіshennya by the Bernoulli method.

Suggestion: y=x²+C/x.

2) y'+y=cosx.

Here the level is already recorded in the standard look, it is not required to change it.

1) Varying uniformly linear alignment y'+y=0: dy/dx=-y; dy/y=-dx. Integrable:

To take a better form of entry, the exhibitor in the world C is accepted as new:

Tse pereprechennya vykonali, schob better know pokhіdnu.

2) In the omnipotent solution of the linear homogeneous alignment, C is important not as a constant, but as a function of x: C = C(x). For qєї wash

Otrimanі virazi y and y' is represented as mind:

Let's multiply the hurt parts of jealousy by

Integrating the insulting parts equal to the formula for integrating the parts, we take:

Here it is not a function, but a constant constant.

3) At the far end of the same line

Substituting the found function С(x):

They took the same vіdpovіd, like and pіd hоvіshennya by the Bernoulli method.

The method of variation is fairly constant and stagnant for cherry.

y'x+y=-xy².

Aligned to the standard look: y+i/x=-y² (II).

1) It is more uniformly equal y+y/x=0. dy/dx=-y/x. Multiply the offending parts of the equal by dx and divide by y: dy/y=-dx/x. Now integrable:

Submitting to take away the virazi for the mind (II):

Let's just say:

We took away the equalization of the change of money C і x:

Here C is already a constant. At the process of integration, they wrote zam_st (x) simply Z, so as not to change the record. And for example, they turned to C (x), so as not to stray C (x) from the new C.

3) For the final solution of the uniform alignment y=C(x)/x, the function С(x) can be found:

They took away the same conclusion as in case of execution by the Bernoulli method.

Apply for self-verification:

1. Let's rewrite the equals of the standard look: y'-2y = x.

1) We diverge uniformly y'-2y = 0. y'=dy/dx, stars dy/dx=2y, we multiply the offset of the equal parts by dx, divisible by y and integrable:

Zvіdsi known y:

Virazi for y and y' is represented in the mind (for the style of life, C replace C (x) and C' replace C "(x)):

For the value of the integral at the right part, we use the formula for integration by parts:

Now we substitute u, du and v y with the formula:

Here Z = const.

3) Now it is presented at the top of the same

Let's take a look at the linear non-homogeneous differential equalization of the first order:
(1) .
There are three ways to untie this equal:

method of constant variation (Lagrange).

Let's look at the solution of the linear differential alignment of the first fret by the Lagrange method.

Variation method postiynoy (Lagrange)

The method of post-study variation is evidently equal in two stages. At the first stage, we can easily see if it’s equal and it’s more or less equal. From the other side of the stage, we will replace the post-integration, the elimination of the first stage of the solution, into the function. After all, it’s a shameful decision of the weekend.

Let's look at the alignment:
(1)

Krok 1

Shukaєmo solution of uniform alignment:

The price is equal to the changes that are divided

Divide change - multiply by dx, divide by y:

Integrable:

Integral over y-tabular:

Todi

Potentially:

Replace constant e C with C and take away the sign of the modulus, which can be multiplied by constant ±1, yaku is inclusive in C:

Krok 2 Replace post C with function

Now let's replace the constant C with the function x:
c → u (x)
Tobto, shukatimemo decision of the weekend (1) at the sight:
(2)
We know I'll go.

According to the rule of differentiation of folding functions:
.
Behind the rule of differentiation is creation:

.
Presented at the weekend (1) :
(1) ;

.
Two dicks rush:
;
.
Integrable:
.
Presented in (2) :
.
As a result, we obsessively solve the linear differential equation of the first order:
.

The butt of the development of a linear differential alignment of the first order by the Lagrange method

Virishiti Rivnyannia

Solution

Virishuemo uniformly equal:

We share changes:

Let's multiply by:

Integrable:

Table integrals:

Potentially:

Replace e C with C and remove the signs of the module:

Zvіdsi:

Let's replace the constant C with the function x:
c → u (x)

We know I’ll go:
.
Presented at the exit is equal:
;
;
Abo:
;
.
Integrable:
;
Virishennya rivnyannia:
.

Method of variation of prevіlnyh fast

The method of variation of sufficient time for inducing the development of a linear non-uniform differential alignment

a n (t)z (n) (t) + a n − 1 (t)z (n − 1) (t) + ... + a 1 (t)z"(t) + a 0 (t)z(t) = f(t)

polagaє at the replacement of prevіlnyh fasting c k for a deep solution

z(t) = c 1 z 1 (t) + c 2 z 2 (t) + ... + c n z n (t)

uniform uniform alignment

a n (t)z (n) (t) + a n − 1 (t)z (n − 1) (t) + ... + a 1 (t)z"(t) + a 0 (t)z(t) = 0

for additional functions c k (t) , similar to those that satisfy the linear systems of algebra

The designator of the system (1) is the Wronskian functions z 1 ,z 2 ,...,z n .

As is the first for , taken when fixing constant values of integration, then the function

є solutions of outward non-homogeneous linear differential alignment. The integration of a heterogeneous equivalence for the manifestation of a wild rozvyazannya of a similar uniform equivalence is made, in such a rank, to quadratures.

The method of variation of sufficient constants for stimulating the solution of the system of linear differential equations in the normal vector form

swear by a private decision (1) by looking

de Z(t) - the basis of rozv'azkіv vіdpovіdnogo odnorodnogo іvnyannja, records y vyglyadі matrixes, and the vector function, which replaced the vector of prevіlnyh postіynyh, is assigned to spіvvіdnennyam. Shukane private solution (with zero cob values at t = t 0 may look