To solve homogeneous differential equation 1st order, use substitution U \u003d y / x, that is, U is a new unknown function depending on the IKSA. Hence y \u003d UX. The derivative y 'find using the product differentiation rule: y' \u003d (ux) '\u003d u'x + x'u \u003d u'x + u (since x' \u003d 1). For another form of recording: DY \u003d UDX + XDU. After a substitution, the equation is simplifying and come to the equation with separating variables.

Examples of solving homogeneous differential equations of the 1st order.

1) solve equation

We check that this equation is homogeneous (see how to determine uniform equation). Making the replacement u \u003d y / x, from where y \u003d ux, y '\u003d (ux)' \u003d u'x + x'u \u003d u'x + u. We substitute: u'x + u \u003d u (1 + ln (UX) -LNX). Since the logarithm of the work is equal to the sum of logarithms, Ln (UX) \u003d LNU + LNX. From here

u'X + U \u003d U (1 + LNU + LNX-LNX). After bringing similar terms: U'X + U \u003d U (1 + LNU). Now reveal brackets

u'X + U \u003d U + U · LNU. In both parts, it is U, from here U'X \u003d U · LNU. Since U is a function from IKSA, U '\u003d DU / DX. We substitute

Received equation with separating variables. We divide the variables for which both parts multiply on DX and divide on x · u · lnu, provided that the product x · u · lnu ≠ 0

We integrate:

In the left side is a tabular integral. In the right - we make a replacement t \u003d lnu, from where dt \u003d (lnu) 'du \u003d du / u

ln│t│ \u003d ln│x│ + c. But we have already discussed that in such equations instead, with more convenient to take Ln│c. Then

ln│t│ \u003d ln│x│ + ln│c│. By the property of logarithms: ln│t│ \u003d ln │cx. Hence T \u003d CX. (under the condition, x\u003e 0). It's time to make a replacement: lnu \u003d cx. And another reverse replacement:

By the property of logarithms:

This is a common integral of the equation.

Remember the condition of the product x · u · lnu ≠ 0 (and therefore x ≠ 0, u ≠ 0, lnu ≠ 0, from where u ≠ 1). But x ≠ 0 from the condition, U ≠ 1 remains where X ≠ Y is left. Obviously, y \u003d x (x\u003e 0) included in common decision.

2) Find a private integral of the Y '\u003d X / Y + Y / X equation that satisfies the initial conditions y (1) \u003d 2.

First, we check that this equation is homogeneous (although the presence of the 3-X and X / Y's terms already indirectly indicates this). Then we make the replacement u \u003d y / x, from where y \u003d ux, y '\u003d (ux)' \u003d u'x + x'u \u003d u'x + u. We substitute the obtained expressions to the equation:

u'X + U \u003d 1 / U + U. Simplify:

u'X \u003d 1 / U. Since U is a function from IKS, U '\u003d DU / DX:

Received equation with separating variables. To divide the variables, we multiply both parts on DX and U and divide on x (x ≠ 0 by condition, from here U ≠ 0 also, it means that the solutions do not occur).

We integrate:

and since there are table integrals in both parts, we immediately get

Perform a reverse replacement:

This is a common integral of the equation. We use the initial condition Y (1) \u003d 2, that is, we substitute to the resulting solution y \u003d 2, x \u003d 1:

3) Find a general integral of a homogeneous equation:

(x²-Y²) DY-2XYDX \u003d 0.

Replacing u \u003d y / x, from where y \u003d ux, dy \u003d xdu + udx. We substitute:

(X²- (UX) ²) (XDU + UDX) -2Ux²dx \u003d 0. We endure x² for brackets and divide both parts on it (under the condition X ≠ 0):

x² (1-UDX) (XDU + UDX) -2UX²DX \u003d 0

(1-Uq) (XDU + UDX) -2UDX \u003d 0. Reveal brackets and simplify:

xDU-U²DU + UDX-U³DX-2UDX \u003d 0,

xDU-U²DU-U³DX-UDX \u003d 0. Grouping terms with DU and DX:

(X - U²X) DU- (U³ + U) DX \u003d 0. We carry out general multipliers for brackets:

x (1-U²) DU-U (U² + 1) DX \u003d 0. We share variables:

x (1-U²) DU \u003d U (U² + 1) DX. To do this, both part of the equation divide on Xu (U² + 1) ≠ 0 (respectively, add the requirements X ≠ 0 (already noted), U ≠ 0):

We integrate:

In the right part of the equation - a tabular integral, a rational fraction on the left side, lay out on simple multipliers:

(Or in the second integral, it was possible to make a replacement T \u003d 1 + U², DT \u003d 2UDU under the sign of the differential, DT \u003d 2UDU - who likes what kind of way). We get:

According to the properties of logarithms:

Reverse replacement

Remember the condition U ≠ 0. Hence y ≠ 0. When C \u003d 0 y \u003d 0, it means that the loss of solutions does not occur, and y \u003d 0 enters the common integral.

Comment

You can get a record of the solution in another form if left to leave the term with X:

The geometrical meaning of the integral curve in this case is a family of circles with centers on the Oy axis and passing through the origin of the coordinates.

Tasks for self-test:

1) (x² + y²) dx-xydy \u003d 0

1) We check that the equation is homogeneous, after which we make the replacement u \u003d y / x, from where y \u003d ux, dy \u003d xdu + udx. We substitute in the condition: (x² + x²u²) dx-x²u (xdu + udx) \u003d 0. Dividing both parts of the equation on x² ≠ 0, we obtain: (1 + U²) DX-U (XDU + UDX) \u003d 0. Hence DX + U²DX-Xudu-U²DX \u003d 0. Similar, we have: dx-xudu \u003d 0. Hence the XUDU \u003d DX, UDU \u003d DX / X. We integrate both parts:

Stop! Come on, let's try to figure out this cumbersome formula.

In the first place should go the first variable to the degree with some coefficient. In our case, it

In our case, it is. As we found out, it means here the degree at the first variable - converges. And the second variable in the first degree - on the spot. Coefficient.

We have it.

The first variable to the degree, and the second variable in the square, with the coefficient. This is the last member of the equation.

As you can see, our equation is suitable for determination in the formula.

Let's look at the second (verbal) part of the definition.

We have two unknown and. It converges here.

Consider all the terms. In them, the sum of the degrees of unknowns should be the same.

The amount of degrees is equal.

The amount of degrees is equal to (when and when).

The amount of degrees is equal.

As you can see, everything converges !!!

Now let's practice in the definition of homogeneous equations.

Determine which equations are homogeneous:

Uniform equations - equations under the numbers:

Consider the equation separately.

If we divide each alkalis on decomposition every term, we get

And this equation completely falls under the definition of homogeneous equations.

How to solve homogeneous equations?

Example 2.

We split the equation on.

In our condition, Y cannot be equal. So we can safely divide on

By replacing, we get a simple quadratic equation:

Since this is a given square equation, we use the Vieta Theorem:

By returning replacement, get the answer

Answer:

Example 3.

We divide the equation on (by condition).

Answer:

Example 4.

Find if.

Here you need not to divide, but multiply. Multiply all the equation for:

We will replace and solve a square equation:

By raising replacement, we get the answer:

Answer:

Solution of homogeneous trigonometric equations.

The solution of homogeneous trigonometric equations is no different from the methods of solution described above. Only here, among other things, you need to know a little trigonometry. And be able to decide trigonometric equations (You can read the section).

Consider such equations on the examples.

Example 5.

Decide equation.

We see a typical homogeneous equation: and are unknown, and the sum of their degrees in each slighter is equal.

Such homogeneous equations are not solved, but before splitting the equations on, consider the case when

In this case, the equation will take the form: it means. But the sinus and cosine cannot simultaneously be equal, because by the main trigonometric identity. Therefore, you can safely divide:

Since the equation is given, then on the Vieta Theorem:

Answer:

Example 6.

Decide equation.

As in the example, you need to divide the equation on. Consider the case when:

But the sinus and cosine cannot simultaneously be equal, because by the main trigonometric identity. Therefore.

We will replace and solve a square equation:

We will make a replacement and find and find:

Answer:

Solution of homogeneous indicative equations.

Uniform equations are solved in the same way as discussed above. If you forgot how to solve indicative equations - See the appropriate section ()!

Consider several examples.

Example 7.

Decide equation

Imagine as:

We see a typical homogeneous equation, with two variables and the amount of degrees. We divide the equation to:

As you can see, making a replacement, we obtain a given square equation (this does not need to be fear of fission on zero - always strictly more zero):

On the Vieta Theorem:

Answer: .

Example 8.

Decide equation

Imagine as:

We divide the equation on:

We will replace and solve a square equation:

The root does not satisfy the condition. Producing reverse replacement and find:

Answer:

Uniform equations. AVERAGE LEVEL

First, on the example of one task I remind what is homogeneous equations and that the solution of homogeneous equations is.

Solve the task:

Find if.

Here you can see a curious thing: if you divide each person on, we get:

That is, now there are no individual and, now the variable in the equation is the desired value. And this is a conventional square equation that is easy to solve using the Vieta's theorem: the product of the roots is equal, and the amount is numbers and.

Answer:

View equations

called homogeneous. That is, this is an equation with two unknown, in each consideration of which the same amount of the degrees of these unknowns. For example, in the example above, this amount is equal. The solution of homogeneous equations is carried out by dividing on one of the unknown to this extent:

And subsequent replacement of variables :. Thus, we obtain equation of degree with one unknown:

Most often, we will meet the equations of the second degree (that is, square), and we can decide:

Note that it is possible to divide (and multiply) all the equation to the variable only if we are convinced that this variable cannot be zero! For example, if we are asked to find, we immediately understand that, since it is impossible to divide. In cases where this is not so obvious, it is necessary to separately check the case when this variable is zero. For example:

Decide equation.

Decision:

We see here a typical homogeneous equation: and are unknown, and the sum of their degrees in each alkali is equal.

But, before splitting on and get a square equation regarding, we must consider the case when. In this case, the equation will take the form: it means that. But the sinus and cosine can not be simultaneously equal to zero, because by the main trigonometric identity :. Therefore, you can safely divide:

I hope this solution is completely clear? If not, read the section. If it is not clear where it came from, you need to return even earlier - to the section.

Share myself:

1. Find if.
2. Find if.
3. Decide equation.

Here I will briefly write directly the solution of homogeneous equations:

Solutions:

Answer:.

And here you have to do not divide, but multiply:

Answer:

If the trigonometric equations you have not passed yet, this example can be skipped.

Because here we need to share on, make sure of the first, one hundred he equal to zero.:

And it is impossible.

Answer:.

Uniform equations. Briefly about the main thing

The solution of all homogeneous equations is reduced to one of the unknown to the degree and further substitute for variables.

Algorithm:

Well, the topic is finished. If you read these lines, then you are very cool.

Because only 5% of people are able to master something on their own. And if you read to the end, then you got into these 5%!

Now the most important thing.

You figured out the theory on this topic. And, I repeat, it ... it's just super! You're better than the absolute majority of your peers.

The problem is that this may not be enough ...

For what?

For the successful passing of the USE, for admission to the Institute on the budget and, most importantly, for life.

I will not convince you anything, I'll just say one thing ...

People who received a good education earn much more than those who did not receive it. These are statistics.

But it is not the main thing.

The main thing is that they are happier (there are such research). Perhaps because there are much more opportunities in favor of them and life becomes brighter? I do not know...

But, think myself ...

What you need to be sure to be better than others on the exam and be ultimately ... happier?

Fill a hand by solving tasks on this topic.

You will not ask the theory on the exam.

You will need solve tasks for a while.

And if you did not solve them (a lot!), You definitely be a foolishly mistaken or just do not have time.

It's like in sport - you need to repeat many times to win for sure.

Find where you want a collection, necessarily with solutions, detailed analysis And decide, decide, decide!

You can use our tasks (not necessarily) and we, of course, we recommend them.

In order to fill the hand with the help of our tasks, you need to help extend the life to the textbook youcever, which you are reading now.

How? There are two options:

1. Open access to all hidden tasks in this article - 299 rub.
2. Open access to all hidden tasks in all 99 articles of the textbook - 499 rub.

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Find the task and decide!

Homogeneous

In this lesson, we will look at the so-called uniform first-order differential equations. As well as equations with separating variables and linear inhomogeneous equations This type of Du is found in almost any control work on the topic of diffuses. If you went to a page from the search engine or are not very confidently oriented in differential equations, I first strongly recommend to work out the introductory lesson on the topic - Differential equations of first order. The fact is that many principles for solving homogeneous equations and the technical techniques used will be exactly the same as for the simplest equations with separating variables.

What is the difference between homogeneous differential equations from other types of DU? This is the easiest way to immediately explain on a specific example.

Example 1.

Decision:
what first of all should be analyzed when solving anyone Differential equation first order? First of all, it is necessary to check if it is impossible to immediately divide the variables using "school" actions? Usually, such an analysis is made mentally or trying to divide variables on the draft.

IN this example variables can not be divided (You can try to across the components from the part into part, to increase the multipliers for brackets, etc.). By the way, in this example, the fact that variables cannot be divided is quite obvious due to the presence of a multiplier.

The question arises - how to solve this diffur?

Need to check, and is this uniform equation? The check is simple, and the verification algorithm itself can be formulated as follows:

In the original equation:

instead We substitute instead We substitute derivative do not touch:

Lambda letter is conditional parameterAnd here he plays the following role: if the transformation will be able to "destroy" all lambdes and obtain the initial equation, then this differential equation is homogeneous.

Obviously, lambda is immediately reduced in an indicator:

Now in the right part we endure lambda for brackets:

and both parts divide on this very lambda:

As a result everything Lambda disappeared as a dream as morning fog, and we obtained the original equation.

Output: This equation is homogeneous

How to solve a homogeneous differential equation?

I'm very good news. Absolutely all homogeneous equations can be solved with the same single (!) Standard replacement.

The function "Igarek" follows replace work Some function (also depending on "X") and "IKSA":

Almost always written in short:

We find out what the derivative turns into such a replacement, we use the derivation of the product. If, then:

We substitute in the original equation:

What will this replacement give? After this replacement and simplified, we guaranteed We obtain an equation with separating variables. Remember Like first love :) and, accordingly,.

After the substitution, we carry out maximum simplifications:

Since this is a function depending on "X", its derivative can be recorded by a standard fraction :.
In this way:

We share variables, while in the left part you need to collect only "TE", and in the right part - only "Xers":

Variables are separated, integrate:


According to my first technical Council From the article Differential equations of first order Constant In many cases, it is advisable to "arrange" in the form of logarithm.

After the equation is integrated, you need to spend replacement, it is also standard and the only one:
If, then
In this case:

In 18-19 cases of 20, the solution of a homogeneous equation is recorded as a common integral.

Answer: General integral:

Why is almost always the answer of a homogeneous equation given in the form of a common integral?
In most cases, it is impossible to express "igrek" explicitly (obtain a general solution), and if it is possible, the general solution is most often obtained by bulky and cored.

For example, in the considered example, the general decision can be obtained, inspired logarithms on both parts of the total integral:

- Well, wherever nothing went. Although, you agree, it's still curved.

By the way, in this example I did not quite "decently" recorded a general integral. It's not a mistakeBut in the "good" style, I remind you, the common integral is accepted in the form. To do this, immediately after the integration of the equation, the constant should be recorded without any logarithm (Here is the exception to the rule!):

And after the reverse replacement, get a common integral in the "classic" form:

The received answer can be checked. To do this, you need to train a common integral, that is, to find derived from the function specified implicitly:

Get rid of fractions, multiplying each part of the equation on:

The initial differential equation was obtained, it means that the solution was found correctly.

It is advisable to always check. But homogeneous equations are unpleasant to the fact that it is usually difficult to check their common integrals - this requires a very and very decent differentiation technique. In the examination examined, during the inspection, it was not necessary to find not the simplest derivatives (although the example itself is quite simple). If you can check - check!

Example 2.

Check the equation for homogeneity and find its common integral.

Answer in the form

This is an example for self-decide - So that you are mastered in the actions algorithm itself. Check on leisure, because Here it is quite complex, and I did not even bring it, otherwise you will no longer come to such a maniac :)

And now promised important momentmentioned at the very beginning of the topic
Select fatty black letters:

If during the transformations we "reset" the multiplier (not a constant) In the denominator, then risk of losing solutions!

And in fact, we encountered this in the first example. introductory lesson on differential equations. In the process of solving the igrek equation, it turned out to be in the denominator: but, obviously, the decision of the Du and as a result of a non-uniform transformation (division) is all the chances of losing it! Another thing is that it entered the overall solution at a zero value of the constant. Reset "IKSA" in the denominator can also be taken into account, because Does not satisfy the source diffuse.

A similar story with the third equation of the same lesson, during the decision of which we "dropped" to the denominator. Strictly speaking, here should be verified, and is not the solution to this Diffur? It is! But here "everything cost", because this function entered the general integral at.

And if such often with "separating" equations is often;) "rolls", then with homogeneous and some other diffusers can "not ride". With a high probability.

We analyze the problems that have been contemporated: in Example 1 There was a "reset" of the ICA, but cannot be a solution to the equation. But B. Example 2 We divided by But this also "left hands": because, they could not lose decisions, they simply do not have them here. But "happy cases", of course, made a specially, and it is not a fact that in practice they will be:

Example 3.

Solve differential equation

Is it really a simple example? ;-)

Decision: The homogeneity of this equation is obvious, but still - in the first step Be sure to check whether the variables cannot be divided. For the equation is also homogeneously, but the variables in it are quietly divided. Yes, there are such!

After checking on "separation", we replace and simplify the equation as much as possible:

We share variables, we collect "TE" on the left, right - "Xers":

And here is the stop. When dividing on we risks to lose two functions at once. Since, these are functions:

The first function is obviously the solution of the equation . We check the second - we substitute it and its derivative in our diffuse:

- The right equality is obtained, it means that the function is a solution.

AND these solutions we risk losing.

In addition, the "X" found in the denominator, however, the replacement implies that it is not equal to zero. Remember this fact. But! Be sure to checkIs the solution of the initial differential equation. No, it's not.

Take all this note and continue:

I must say, with the integral of the left part lucky, it happens much worse.

We collect on the right side a single logarithm, and discard the shackles:

And only now reverse replacement:

Multiply all the components on:

Now it should be checked - dangerous solutions entered the common integral . Yes, both solutions entered the overall integral at a zero value of the constant: therefore they do not need to additionally indicate in answer:

general integral:

Check. Not even check, but solid pleasure :)

The initial differential equation was obtained, it means that the solution is found true.

For self solutions:

Example 4.

Perform an inspection for homogeneity and solve the differential equation

General integral Check differentiation.

Complete solution And the answer at the end of the lesson.

Consider a pair of examples when a homogeneous equation is specified with ready-made differentials.

Example 5.

Solve differential equation

This is very interesting example, straight a whole thriller!

Decisionwe will get used to execute compact. First, mentally either on the draft make sure that the variables can not be divided here, after which we carry out an inspection for homogeneity - it is usually not carried out on finishing (if not specifically required). Thus, almost always the solution begins with the record: " This equation is homogeneous, we will replace: ...».

If a homogeneous equation contains finished differentials, it can be solved by a modified substitution:

But I do not advise you to use such a substitution, as it turns out the Great Wall of Differentials, where the eye is needed. From a technical point of view, it is more advantageous to switch to the "bar" designation of the derivative, for this we divide all members of the equation for:

And here we have made a "dangerous" transformation! The zero differential corresponds to the family of direct, parallel axles. Are they rooted by our do? Substitute and in the original equation:

This equality is valid if, that is, when dividing on we risked to lose the decision, and we lost it - Since it no longer satisfy The resulting equation .

It should be noted that if we initially An equation was given The root of speech would not go. But we have it, and we are in time "caught."

We continue the decision standard replacement :
:

After the substitution, the equation is easily simplified:

We share variables:

And here again stop: when dividing on we risks, lose two functions. Since, these are functions:

Obviously, the first function is a solution to the equation . We check the second - we substitute it and its derivative:

- Received faithful equalityIt means that the function is also solving the differential equation.

And when dividing on we, these solutions risk losing. However, they can enter the overall integral. But may not fit

Take this note and integrate both parts:

The integral of the left side is standardly solved using allocation of full squarebut in the diffusers it is much more convenient to use method of uncertain coefficients:

Using the method of indefinite coefficients, decompose the integrated function in the amount elementary fractions:


In this way:

We find integrals:

"Since we painted some logarithms, then the constant also pushed under the logarithm."

Before replacement again we simplify everything you can simplify:

Reset chains:

And reverse replacement:

Now I remember "Losses": the solution went into a common integral when, but "flew past the cash register", because It turned out in the denominator. Therefore, in response, it is honored with a separate phrase, and yes - do not forget about the lost solution, which, by the way, was also below.

Answer: General integral: . More solutions:

It is not so difficult to express a general solution:
But this is already Ponte.

Comfortable, however, for verification. Find a derivative:

and substitute In the left part of the equation:

- as a result right part Equations that needed to check.

The following diffuse is independently:

Example 6.

Solve differential equation

Complete solution and answer at the end of the lesson. Try at the same time for training and here to express a general solution.

In the final part of the lesson, consider a couple more of characteristic tasks on the topic:

Example 7.

Solve differential equation

Decision: We go like a dear way. This equation is homogeneous, we will replace:

With "XOM" everything is fine here, but what about square triple? Since he is indecomposable for multipliers:, we definitely do not lose decisions. It would always be so! We allocate the full square on the left side and integrate:

Simplify nothing to do here, and therefore the reverse replacement:

Answer: General integral:

Example 8.

Solve differential equation

This is an example for an independent solution.

so:

For non-uniform transformations, always check (at least orally), do you lose some decisions! What are these transformations? As a rule, the reduction on something or division into something. So, for example, when dividing on, it is necessary to check whether the functions of the solutions of the differential equation are. At the same time, when dividing on the need for such an inspection already disappears - due to the fact that this divider does not turn to zero.

Here is one more dangerous situation:

Here, getting rid of, should be checked if the solution is not a solution. Often as such a multiplier encounters "X", "Igrek", and cutting on them, we lose the functions that may be solutions.

On the other hand, if something is initially in the denominator, there is no reason for such anxiety. So, in a homogeneous equation, you can not worry about the function, as it is "declared" in the denominator.

Listed subtleties do not lose relevance, even if the task requires only a particular solution. There is a small, but the chance that we will lose exactly the required private solution. truth cauchy task in practical tasks With homogeneous equations, it is rarely requested. Nevertheless, such examples are in the article Equations reduced to homogeneouswhich I recommend to study "hot pixels" to consolidate your solution skills.

There are more complex homogeneous equations. The complexity consists not in replacing variable or simplifications, but in sufficiently difficult or rare integrals that arise as a result of the separation of variables. I have examples of solutions of such homogeneous equations - terrible integrals and terrible answers. But we will not be about them, because at the nearest lessons (see below) I still have time to torture you, I want to see you fresh and optimistic!

Successful promotion!

Solutions and answers:

Example 2: Decision: Check the equation for homogeneity, for this in the original equation instead Substitute, A. instead Substitute:

As a result, the initial equation was obtained, which means that this is homogeneous.

Currently, only 4 hours are provided for the basic level of mathematics to study mathematics in high-grade classes (2 hours of algebra, 2 hours of geometry). In rural small-scale schools, they try to increase the number of hours at the expense of the school component. But if the class is humanitarian, then the school component is added to the study of the objects of the humanitarian direction. In a small village, it is often not necessary to choose a schoolboy, he studies in the class; What is in school. To become a lawyer, a historian or journalist (there are such cases) is not going to, but wants to become an engineer or economist, therefore the exam in mathematics should pass on high balls. Under such circumstances, a teacher of mathematics have to find their way out of the current situation, besides, according to the Kolmogorov textbook, the study of the topic "homogeneous equations" is not provided. In past years, for the introduction of this topic and consolidation, I needed two dual lessons. Unfortunately, the test of educational supervision has banned dual lessons at school, so the number of exercises had to cut up to 45 minutes, and, accordingly, the level of difficulty of exercises to lower to the average. I bring to your attention a plan-abstract lesson on this topic in the 10th grade with a basic level of learning mathematics in a rural little complete school.

Type of lesson: Traditional.

purpose: Learn to solve typical homogeneous equations.

Tasks:

Cognitive:

Developing:

Educational:

• Education of hard work through patient performance of tasks, sense of partnership through work in pairs and groups.

During the classes

I. Organizational stage (3 min.)

II. Check the knowledge necessary for the assimilation of the new material (10 min.)

To identify the main difficulties with further analysis of the tasks performed. The guys are performed by choosing 3 options. The tasks differentiated by the degree of complexity and by the level of the preparedness of the guys, followed by an explanation at the board.

Level 1. Decide equations:

1. 3 (x + 4) \u003d 12,
2. 2 (x-15) \u003d 2x-30
3. 5 (2) \u003d - 3x-2 (x + 5)
4. x 2 -10x + 21 \u003d 0 Replies: 7; 3

2 level. Decide the simplest trigonometric equations and a biquette equation:

answers:

b) x 4 -13x 3 + 36 \u003d 0 Answers: -2; 2; -3; 3.

3 level. Solving equations by replacement of variables:

b) x 6 -9x 3 + 8 \u003d 0 Replies:

III. Message themes, setting goals and tasks.

Subject: Uniform equations

purpose: Learn to solve typical homogeneous equations

Tasks:

Cognitive:

• get acquainted with homogeneous equations, learn to solve the most common types of such equations.

Developing:

• Development of analytical thinking.
• The development of mathematical skills: to learn how to identify the main features for which homogeneous equations differ from other equations, be able to establish the similarity of homogeneous equations in their various manifestations.

IV. Mastering new knowledge (15 min.)

1. The lecture moment.

Definition 1. (Write to the notebook). The equation of the form p (x; y) \u003d 0 is called homogeneous if P (x; y) is a homogeneous polynomial.

The polynomial of two variables x and y is called homogeneous if the degree of each of its member is equal to the same number to.

Definition 2.(Just familiarizing). View equations

called a homogeneous equation of degree n relative to U (x) and V (x). Objects both parts of the equation on (V (x)) n, using the replacement to obtain the equation

What makes it possible to simplify the original equation. The case V (x) \u003d 0 must be considered separately, since it is impossible to divide on 0.

2. Examples of homogeneous equations:

Explain: Why are they homogeneous, bring our examples of such equations.

3. Task on the definition of homogeneous equations:

Among specified equations Determine homogeneous equations and explain your choice:

After they explained their choice on one example show a method for solving a homogeneous equation:

4. Solve yourself:

Answer:

b) 2sin x - 3 cos x \u003d 0

We divide both parts of the equation on COS X, we obtain 2 TG x -3 \u003d 0, TG x \u003d ⅔, x \u003d arctg⅔ +

5. Show Solution Example from brochure "P.V. Stockings. Equations and inequalities in the school course of mathematics. Moscow Pedagogical University "First September" 2006 p.22 ". As one of the possible examples of the level S.

V.. Decide to consolidate the textbook Bashmakov

page 183 No. 59 (1.5) or by the textbook edited by Kolmogorov: St81 №169 (A, B)

answers:

VI. Verification, independent work (7 min.)

1 option	Option 2
Solve equations:
a) sin 2 x-5sinxcosx + 6cos 2 x \u003d 0	a) 3sin 2 x + 2sin x cos x-2cos 2 x \u003d 0
b) COS 2 -3SIN 2 \u003d 0	b)

Answers to tasks:

1 Option a) Answer: arctg2 + πn, N € Z; b) answer: ± π / 2 + 3πn, N € Z; in)

2 Option a) Answer: arctg (-1 ± 31/2) + πn, N € Z; b) answer: -Arctg3 + πn, 0.25π + πk ,; c) (-5; -2); (5; 2)

VII. Homework

№169 in Kolmogorov, No. 59 in Shmakov.

In addition, solve the system of equations:

Answer: arctg (-1 ± √3) + πN,

References:

1. P.V. Stockings. Equations and inequalities in the school course of mathematics. - M.: Pedagogical University "First September", 2006. p. 22
2. A. Merzlyak, V. Polonsky, E. Rabinovich, M. Yakir. Trigonometry. - M.: "AST-PRESS", 1998, p. 389
3. Algebra for grade 8 edited by N.Ya. Vilenkin. - M.: "Enlightenment", 1997.
4. Algebra for grade 9 edited by N.Ya. Vilenkin. Moscow "Enlightenment", 2001.
5. M.I. Shoes. Algebra and start analysis. For 10-11 classes - M.: Enlightenment 1993
6. Kolmogorov, Abramov, Dudnitsyn. Algebra and start analysis. For 10-11 classes. - M.: "Enlightenment", 1990.
7. A.G. Mordkovich. Algebra and start analysis. Part 1 Tutorial 10-11 classes. - M.: Mnemozina, 2004.