The method of dovіlnih fast. Variation Method Social transformation. State that church
The method of variation of fairly recent zastosovuetsya on the variance of heterogeneous differential equalities. Tsey lesson of appointments for students, even though more or less are well oriented in topics. For some reason, you start to know the DC, tobto. є teapot, I recommend almost the first lesson: Differential alignment of the first fret. Apply solution. And as soon as you finish it, be kind, let me know that the folding method is possible. Because wine is simple.
Do some types of habits have the method of variation of fairly fast ones?
1) The method of variation of fairly constant can be beaten with linear inhomogeneous DC of the 1st order. Yakshcho equal to the first order, those became (constant) tezh one.
2) The method of variation of fairly late vicory for cherry blossoms linear heterogeneous equalities of a different order. Here two posts (constants) are varied.
It is logical to admit that the lesson is composed of two paragraphs. Axis writing tsyu propozitsiyu, і hvilin ten bolіsno thinking, how could I add more reasonable crap for a smooth transition to practical applications. Ale, I think of thoughts, since there are no holy ones, wanting nothing and not evil of anything. That's why I take it for the first paragraph.
Method of variation of fairly constant
for linear non-homogeneous alignment of the first order
Before looking at the method of variation of a fairly constant bazhano, we should know from the article Linear differential alignment of the first order. At that lesson we were trained first way heterogeneous DC of the 1st order. Whose first way of accomplishment, I guess, is called replacement method or Bernoulli method(don't mess with peers of Bernoulli!!!)
At once we can see another way of seeing- the method of variation of a fairly long time. I will bring only three butts, moreover, I will take them from the wisest lesson. Why so few? That is why the solution in another way will be more similar to the solution in the first way. In addition, for my warnings, the method of variation of fairly recent ones is more likely to be replaced by the replacement method.
butt 1
(Diffur from Example No. 2 lesson Linear heterogeneous DC of the 1st order)
Solution: Dane is equal to linear heterogeneous and may know looking: 
At the first stage, it is necessary to repeat the simple alignment: 
Tobto is stupidly nullified to the right part - write zero instead.
Rivnyannia I will name additional relatives.
For this butt, it is necessary to virishite the foot additionally equal:
Before us equal to the changes that are divided, I can’t imagine any solution for you: 
In this manner: 
- Zagalne solution of additional equivalence.
On the other side instead of deacoyu constant Poki sho unknown function, how to deposit in the form of "iks":
Zvіdsi i naming the method - variable constant. As an option, the constant can be a function, as we can know at once.
At weekend heterogeneous equal let's replace:
Let's imagine 
at the river :
control moment - two dodanki in the left part of the rush. I don’t know what else, next shukati pardon more.
As a result, the replacements were eliminated equal to the changes that are divided. We share changes and integrate.
Like grace, exponentials are also fast:
Before the found function, we add a “normal” constant:
At the final stage, we will decide our replacement:
The function is well known!
In such a rank, a bold decision: 
Suggestion: more solution: 
If you see two methods of solution, then you can easily remember that in both cases we knew your own integrations. Vіdminnіst is more than an algorithm solution.
Now it’s more foldable, I’ll comment on another butt as well:
butt 2
Know the global solution of differential alignment 
(Diffur from Example No. 8 lesson Linear heterogeneous DC of the 1st order)
Solution: Let's get straight to the point :
Let's nullify the right part and virishimo dopomіzhne equal:
The main solution of the additional level:
We will replace the heterogeneous equal:
Behind the rule of differentiation is creation: 
Let's imagine the output is not uniform:
Two warehouses in the left part are rushing, meaning we are on the right way: 
Integrable by parts. The letter z of the formula for integrating parts is tasty for us, but it’s already a problem for the decision, for example, the letters “a” and “be” are victorious:
Now the replacement has been carried out:
Suggestion: more solution:
І one butt for independent vision:
butt 3
To know more about the solution of the differential equation, which supports the tasks of the cob mind.
,
(Diffur from Example No. 4 lesson Linear heterogeneous DC of the 1st order)
Solution:
Dane DK is linearly heterogeneous. Vikoristovuєmo method of variation of rather fast. Virishimo additionally equal:
Let's change and integrate: 
Solution: 
We will replace the heterogeneous equal: 
We accept substitution: 
In such a rank, a bold decision: 
We know a private solution that supports the tasks of the mind of the mind: 
Suggestion: private solution:
Solutions for a lesson can be clear for a final design of a task.
Method of variation of prevіlnyh fast
for linear inhomogeneous alignment of a different order
with constant coefficients
It often occurred to me a little thought that the method of variation of fairly recent equals of a different order is not an easy thing. Ale, I let my feet go: it’s better for everything, the method of riches is important, the shards are not sharpened so often. But really, there are no special foldings - the overflow of the decision is clear, insightful, insightful. I handsome.
In order to master the method, it is necessary to understand the difference between different orders in a different order by choosing a private solution to look at the right part. This method is reportedly reviewed in the article Heterogeneous DC of the 2nd order. Guessing that linearly heterogeneous equal to another order with constant coefficients can be seen:
The method of picking, which, looking at the figure of the uroci, is less likely to pass in the surroundings of the lower slopes, if there are rich segments, exponenti, sine, cosine in the right part. Ale scho robiti, if on the right, for example, drіb, logarithm, tangent? In such a situation, the method of post-variation still comes to the rescue.
butt 4
Know the global solution of the differential equation of a different order 
Solution: At the right side this equal If you know more, then you can say that the method of choosing a private solution does not work. Vikoristovuєmo method of variation of rather fast.
There is no threat, the cob of solution is quite significant:
We know better solution special homogeneous equal: 
It is folded and virіshimo characteristically equal: 
- otrimano pov'yazane complex root, to that zagalne solution:
To give respect to the recording of a sparkling solution - as if it were arches, they were opened.
Now we can practically use the same trick as for first-order equality: varying constants, replacing them with unknown functions. Tobto, the solution of the heterogeneous ryvnyannya will be shukati at the sight:
De - Poki sho unknown functions.
It looks like a haunt of butovyh vіdhodіv, but at the same time, everything is sorted out.
Like unknown functions appear. Our meta is to know better, moreover, known worse things are to be satisfied with the first and the other equal system.
Are the sounds taken by "igreeks"? Їх bring lelek. We marveled at the otriman earlier than the final solution, and we write it down:
Let's know the fun:
Іz the left parts were taken apart. What is right-handed?
- all the rights of a part of the vihіdny equal, to this particular type:
Coefficient - the same coefficient for other expenses:
Really may be sure, and our butt is not a blame.
Everything cleared up, now you can fold the system: 
Ring out the system by Cramer's formulas, vikoristovuyuchi standard algorithm. The only difference lies in the fact that the replacement of numbers can be a function.
We know the head of the system: 
Like a bully, like a “two-by-two” vyznachnik opens up, go back to the lesson How to count? Posilannya vede on the doshka ganbi =)
Otzhe, otzhe, the system can be one solution.
We know I’ll go: 
And yet, not all, until we know I’ll lose more.
The function itself follows the integration:
Chosen with another function: 
Here we add a "normal" constant
At the final stage, the solution is guessing, how did we joke about the solution of a heterogeneous equalization? This one has:
Needed functions are well known! 
The vikonati substitution was lost and it was written down:
Suggestion: more solution:
For principle, for vidpovidi, it is possible to open the arches.
The new re-verification is based on the standard scheme, which was seen at the lesson Heterogeneous DC of the 2nd order. But the re-verification will not be easy, the little ones may know how to do important things and carry out a cumbersome substitution. It’s unacceptable singularity, if you sing such difuri.
butt 5
Razv'yazati differential equalization by the method of variation of sufficient postal
This is an example of an independent solution. In fact, at the right side of the tezh drib. Let's guess the trigonometric formula, її, before speech, it will be necessary to stop in the solution process.
The method of variation of more recent ones is the most universal method. You can virishiti be-like equal, like virishuetsya by the method of selecting a private solution for the appearance of the right part. Wait for food, but why not beat the method of variation of pretty fasts? The conclusion is obvious: the choice of a private decision, which, looking at the lesson Heterogeneous equal to another order, signifi- cantly speed up the decision and short-term record - no fucking with the variables and integrals.
Let's look at two butts to the managers of Kosh.
butt 6
Find out more about the solution of the differential equation
, 
Solution: I'm calling that exhibitor at the cirque city.
Vikoristovuєmo method of variation of rather fast.
We know better solution special homogeneous equal: 
- otrimano razne dіysne korіnnya, zagalne sluchennya:
A radical solution to the heterogeneous equal joking at the sight:, de - Poki sho unknown functions.
Let's build the system:
In this view:
,
We know the following: 
, 
In this manner: 
The system can be seen behind Cramer's formulas:
Again, the system is only one solution.
Introducing the integration function:
There are vikoristany method.
Inform a friend of the integration function: 
Such an integral fails replacement method:
From the very replacement it can be seen:
In this manner: 
The goal integral can be known by the method of seeing a perfect square, but in butts with diffusers, I will lay out the drib by the method of unimportant coefficients:
Offending functions found:
As a result, a radical solution of a heterogeneous alignment:
We know a private solution that satisfies the mind of the mind .
Technically, the search for a solution is developed in a standard way, as seen from the article Heterogeneous differential equalities of a different order.
Tremite, now we know I’ll go away in the face of the found infamous solution:
The axis is such a dissimilarity. Asking yoga is not obov'yazkovo, it's easier to put together a system of equals. Vіdpovіdno to cob minds :
Imagine knowing the values of the constants 
at the top solution:
You can pack the logarithms for different types of logarithms.
Suggestion: private solution: 
Like a bachite, difficulties can be blamed in integrals and similar ones, but not in the algorithm itself, the method of variation of rather recent ones. Why didn’t I zalyakav you, why everything is Kuznetsov’s collection!
For relaxation, the rest, more simple butt for self-reliance:
butt 7
Virishiti zavdannya Koshi
, 
The butt is clumsy, but creative, if you fold the system, it’s respectful to look at it, first of all virishuvati ;-),
As a result, a radical solution:
We know a private solution that helps the minds of the cob .
Imagine knowing the values of the constants in the global solution:
Suggestion: private solution:
The method of derivation of linear non-homogeneous differential equations of higher orders from constant coefficients by the method of variation of constant Lagrange is considered. Lagrange's method is also used for the solution of any linear non-homogeneous alignments, as a fundamental system of homogeneous alignment solutions.
ZmistDiv. also:
Lagrange's method (variation of post)
Let's look at the linear non-homogeneous differential alignment with the constant coefficients of the sufficient n-th order:
(1)
.
The method of variation of the post, considered by us as equal to the first order, is also similar to the equal to the higher orders.
The decision is made in two stages. At the first stage, we see the right part of it and we see it equally. As a result, we will take a decision to take revenge on the pretty n who are late. At another stage, we change the post. Therefore, it is important to note that these functions are independent of the changeable x and it is important to look at these functions.
I want to see here the equivalence of the constant coefficients, but Lagrange's method is also stagnant and for solving whether there are linear heterogeneous equalities. For whom, however, there can be a fundamental system of rozv'yazkіv uniform equalization.
Krok 1
Like and in times equal to the first order, we joke about a more blatant solution of a homogeneous equal, equating the right to a heterogeneous part to zero:
(2)
.
A more serious solution to such a jealousy can be seen:
(3)
.
Here - pretty fast; - n linearly independent distributions of a homogeneous alignment (2), as they establish the fundamental system of distributions of the same alignment.
Krok 2. Variation of post functions - replacement of post functions
At another stage, we will deal with the variation of the former. In other words, we will replace the constant on the function in the form of an independent change x:
.
That's why we're looking for a solution to the exit equation (1) for such a look:
(4)
.
As far as we can represent (4) (1), then we take one differential equation for n functions. We can connect these functions with additional equals. If you see n equals, you can assign n functions to that number. Dodatkovі rivnyannya can be folded in a different way. Ale mi tse robimo so that the solution is a little simplest looking. For this, when differentiating, it is necessary to equate the members to zero in order to avenge the worse functions. Let's demonstrate.
In order to present the transfer of the solution (4) to the output of the equation (1), it is necessary to know the similarities of the first n orders of the function, written as (4). Differentiation (4), zastosovuyuchi rules of differentiation sum and dobutku:
.
Group members. On the back of the head, we write down the members with the last ones, and then the members with the better ones:
.
We impose on the functions first of all:
(5.1)
.
Todi viraz for the first time mother will be more simple looking:
(6.1)
.
Tim in the very way I know a friend will die:
.
We impose on the functions of a friend of mine:
(5.2)
.
Todi
(6.2)
.
And so far. In advanced minds, we equate limbs that sweep away similar functions to zero.
In this order, to choose the next steps for the functions:
(5.k) ,
then the first similar in terms of matimut is the simplest form:
(6.k) .
Here.
We know the n-th trip:
(6.n)
.
Presented at the exit is equal (1):
(1)
;
.
It's a lie that all functions seem equal (2):
.
To the sum of members, to revenge, to give zero. At the result we take:
(7)
.
As a result, we took the system linear rivers for the hikers:
(5.1)
;
(5.2)
;
(5.3)
;
. . . . . . .
(5.n-1) ;
(7') .
Virishyuchi tsyu system, it is known virazi for similar functions x . Integrating, taking into account:
.
Here - already do not lie in the vіd x postіynі. Submitting (4), we will take a more serious decision of the weekend.
Respectfully, the designations of the values of the similar ones did not win over the fact that the coefficients a i є are constant. Tom Lagrange's method zastosovuetsya for vyshennya be-any linear inhomogeneous equalities, as seen in the fundamental system of rozvyazkіv uniform alignment (2).
Apply
Razvyazati rivnyannya by the method of variations of the post (Lagrange).
Application solution > > >
Verification of the higher orders by the Bernoulli method
Verification of linear inhomogeneous differential equations of higher orders with constant coefficients of linear substitution
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 decision
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
For a system with constant coefficients, the remaining viraz will be asked:
matrix Z(t)Z− 1 (τ) called Cauchy matrix operator L = A(t) .
Let's look now linearly heterogeneous alignment
. (2)
Let y 1 ,y 2 ,.., y n be a fundamental system of solutions, and - a fundamental solution of a uniform uniform equation L(y)=0. In a similar way, you are equal in the first order, shukatimemo is the solution of equal (2) at the sight
. (3)
Let's go over to the one who has a solution for such a look. For which we represent the equal function. For the installation of this function in the river, we know її pokhіdni. Persha is good 
. (4)
When the other part is calculated, the right part (4) will have some extra money, when the third part is calculated, the big extra money will appear, and so on. Therefore, for the clarity of the distant rahunka, the first dodanok (4) is equal to zero. From urakhuvannyam tsgogo, friend, good luck 
. (5)
Behind them, earlier, mirkuvan, in (5) we also take into account the first addition equal to zero. Nareshti, n-th good old day 
. (6)
Substituting the value of the similar ones at the end of the day, maybe 
. (7)
Another addendum (7) is equal to zero, since the functions y j , j=1,2,..,n, є are solutions of a similar uniform alignment L(y)=0. One by one, we take the system of equalizations of algebra for the value of functions C "j (x) 
(8)
The sign of the system є the sign of the Wronsk fundamental system is the solution y 1 ,y 2 ,..,y n of a similar uniform alignment L(y)=0 and that is not equal to zero. Also, the main solution of the system (8). Knowing this, we take away the functions C "j (x), j=1,2,…,n, and, also, i C j (x), j=1,2,…,n Substituting the values (3), we take solution of linear heterogeneous alignment.
The addition method is called the method of variation of a fairly constant chi by the Lagrange method.
Example number 1. Let's look at the overall solution y" + 4y + 3y = 9e -3 x characteristic alignment r 2 + 4r + 3 = 0 check out -1 i -3. Therefore, the fundamental system of rozv'yazkіv uniform equalization is composed of functions y 1 = e - x and y 2 = e -3 x. Razvyazannya heterogeneous equalization can be seen at a glance y \u003d C 1 (x) e - x + C 2 (x) e -3 x. For the significance of the later C "1, C" 2, we fold the equalization system (8)
C′ 1 ·e -x +C′ 2 ·e -3x =0
-C′ 1 e -x -3C′ 2 e -3x =9e -3x
virishyuchi yaku, we know, integrating otrimani functions, maybe 
Residual
Example number 2. Verification of linear differential equalization of a different order from constant coefficients by the method of variation of sufficient constants: 
y(0) =1 + 3ln3
y'(0) = 10ln3
Solution:
This differential equalization is brought up to linear differential equalities with constant coefficients.
Razvyazannya ryvnyannya will be shukati at the sight of y = e rx. For which warehouse is the characteristic alignment of a linear homogeneous differential alignment with constant coefficients:
r 2 -6 r + 8 = 0
D = (-6) 2 - 4 1 8 = 4
The root of the characteristic alignment: r 1 = 4, r 2 = 2
Again, the fundamental solution system is composed by functions: y1 = e4x, y2 = e2x
The final solution of a homogeneous alignment can be seen: y \u003d C 1 e 4x + C 2 e 2x
Search for a private solution in a way of variation of a pretty fast.
For the recognition of the worst C "i, we fold the equalization system:
C′ 1 e 4x +C′ 2 e 2x =0
C′ 1 (4e 4x) + C′ 2 (2e 2x) = 4/(2+e -2x)
Virazimo C" 1 from the first line:
C" 1 \u003d -c 2 e -2x
and imaginable on a friend. At the result we take:
C" 1 \u003d 2 / (e 2x + 2e 4x)
C" 2 \u003d -2e 2x / (e 2x + 2e 4x)
Integrating functions C" i:
C 1 = 2ln(e -2x +2) - e -2x + C * 1
C 2 = ln(2e 2x +1) – 2x+ C * 2
Shards y \u003d C 1 e 4x + C 2 e 2x
C 1 = (2ln(e -2x +2) - e -2x + C * 1) e 4x = 2 e 4x ln(e -2x +2) - e 2x + C * 1 e 4x
C 2 = (ln(2e 2x +1) – 2x+ C * 2)e 2x = e 2x ln(2e 2x +1) – 2x e 2x + C * 2 e 2x
In such a rank, a radical solution of the differential equivalence may look:
y = 2 e 4x ln(e -2x +2) - e 2x + C * 1 e 4x + e 2x ln(2e 2x +1) – 2x e 2x + C * 2 e 2x
or
y = 2 e 4x ln(e -2x +2) - e 2x + e 2x ln(2e 2x +1) – 2x e 2x + C * 1 e 4x + C * 2 e 2x
We know a private decision for the mind:
y(0) =1 + 3ln3
y'(0) = 10ln3
Substituting x \u003d 0, we know the equality, we take:
y(0) = 2 ln(3) - 1 + ln(3) + C * 1 + C * 2 = 3 ln(3) - 1 + C * 1 + C * 2 = 1 + 3ln3
I know I’m going to die in the face of the taken away infamous solution:
y' = 2e 2x (2C 1 e 2x + C 2 -2x +4 e 2x ln(e -2x +2)+ ln(2e 2x +1)-2)
Substituting x = 0, we take:
y'(0) = 2(2C 1 + C 2 +4 ln(3)+ ln(3)-2) = 4C 1 + 2C 2 +10 ln(3) -4 = 10ln3
We take the system out of two equals:
3 ln(3) - 1 + C * 1 + C * 2 = 1 + 3ln3
4C 1 + 2C 2 +10ln(3) -4 = 10ln3
or
C * 1 + C * 2 = 2
4C1 + 2C2 = 4
or
C * 1 + C * 2 = 2
2C1 + C2 = 2
Stars: C 1 = 0, C * 2 = 2
Private decision to sign up as:
y = 2e 4x ln (e -2x +2) - e 2x + e 2x ln (2e 2x +1) - 2x e 2x + 2 e 2x
Lecture 44 Variation Method Linear heterogeneous alignment of a different order with constant coefficients. (special rights part).
Social transformation. The state is the church.
The social policy of the Bilshoviks was richly dictated by their class approach. Decree on 10 leaf fall 1917. I’m setting up the system, skasovanized pre-revolutionary ranks, titles and rewards. The choice of courts has been established; carried out the secularization of the civilian camps. A cost-free education and medical services were established (decree dated July 31, 1918). Women were given their rights as human beings (decrees dated 16 and 18 December 1917). Decree on the Shlyub zaprovadzhuvav іnstitute Hromadyansky Shlyub.
By the Decree of the RNC dated 20 September 1918, the Church of the Church was founded as a state power and as a system of illumination. Most of the church lane was confiscated. Patriarch of Moscow and All Russia Tikhin (announcement on the 5th leaf fall of 1917) on September 19, 1918 anathema to the fate Radianska Vlad and calling to fight against the bіshovikіv.
Look at the linear non-homogeneous alignment of a different order
The structure of the savage division of such a relationship is determined by the following theorem:
Theorem 1. Severe solution of heterogeneous equivalence (1) is served as the sum of any okremnogo solution of that equivalence and savage solution of a uniform uniform equivalence
Bringing. It is necessary to bring, scho sum
Іsnuє zagalne rіshennya rivnyannya (1). Let us know that the function (3) is equal to the solution (1).
Submitting the sum in rivnyannia (1) deputy at, let's mother
Shards are the solution equal (2), then the viraz, which stands at the first arches, is also equal to zero. Oskіlki є solution rivnyannya (1), then viraz, what is worth at the other arms, dorіvnyuє f(x). Otzhe, equivalence (4) is the sameness. In this rank, the first part of the theorem is finished.
We bring another firmness: viraz (3) є blatantly rozvyazannya rivnyannia (1). It is our duty to bring that it is enough to fast, that you can enter into this viraz, you can choose so that you are satisfied with the mind of the cob:
if there were no numbers x 0 , y 0 i (abi x 0 Bulo taken from tієї galuzі, de funktsії a 1 , a 2і f(x) uninterrupted).
Note what can be shown in the form. Todi on the basis of minds (5) matimemo
Virishimo tsyu system i significantly Z 1і Z 2. Let's rewrite the system in the view:
Respectfully, that the chief of the system is the chief of Vronsky for the functions 1і at 2 at the point x = x 0. Shards and functions behind the mind are linearly independent, then Vronsky’s signpost is not equal to zero; the same system (6) can be solved Z 1і Z 2, then. understand such meaning Z 1і Z 2, For which formula (3) stands for solution (1), which satisfies these cob minds. What did it take to bring.
Let's move on to the infamous method of finding private solutions of heterogeneous equalization.
Let's write a more serious solution of a homogeneous equalization (2)
Shukatimemo private solution of heterogeneous alignment (1) in form (7), looking at Z 1і Z 2 yak deyaki still unknown functions X.
Differentially equanimity (7):
Optimum required functions Z 1і Z 2 so that jealousy was victorious
Yakshto vrahuvati tsyu dodatkovu mind, then the first one is good for looking ahead
Differentiation now ce viraz, we know:
Substituting in equal (1), we take
Virazi, who stand at the first two arms, turn to zero, shards y 1і y2- Virishennya odnorodnogo rivnyannya. Otzhe, the rest of the jealousy looks like
In this way, function (7) will be the culmination of heterogeneous equality (1) in that case, as functions Z 1і Z 2 satisfy the equals (8) and (9). We add the system of equals from equals (8) and (9).
Oskіlki vyznachnik tsієї system є vronsky's vyznachnik for linearly independent solutions y 1і y2 equalization (2), vin is not equal to zero. Later, violating the system, we know how the functions of the X:
Virishing the system, we know, signs as a result of integration are taken. Let's imagine a well-known function formula, otrimuemo zagalne solution of heterogeneous equivalence, de-doubtfully fast.
Enthusiasm...