Uniform Du 1 th order. How to solve a homogeneous differential equation
To solve the homogeneous differential equation of the 1st order, use the 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, rational fraction On the left side we declare on simple factors:
(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:
For example, a function 
- homogeneous function of the first measurement, since
- homogeneous function of the third dimension, since
- uniform function of zero measurement, since
. 
.
Definition 2. First order differential equation y." = f.(x., y.) is called homogeneous if the function f.(x., y.) There is a homogeneous zero measurement function relative to x. and y., or, as they say f.(x., y.) - a homogeneous function of the degree of zero.
It can be represented as
which allows you to determine a homogeneous equation as such a differential, which can be converted to mind (3.3).
Replacement 
it gives a homogeneous equation to the equation with separating variables. Indeed after substitution y \u003d.xz.receive 
,
Sharing variables and integrating, we find:
,
Example 1. Hold the equation.
Δ Consider y \u003d.zX.,
We substitute these expressions y.
and dY.in this equation:
or 
We share variables: 
and integrate: 
,
Replacing z.on the 
, get 
.
Example 2. Find a general solution of the equation.
Δ in this equation P.
(x.,y.)
=x. 2 -2y. 2 ,Q.(x.,y.)
=2xY.- The homogeneous functions of the second measurement, therefore, this equation is homogeneous. It can be represented as 
and to solve the same way as presented above. But we use another form of recording. Put y. =
zX.From! dY. =
zDX
+
xDZ.. Substituting these expressions into the original equation, we will have
dX+2 zxdz. = 0 .
Separate variables, counting 
.
We integrate soil this equation
From! 
i.e 
. Returning to the former function 
we find a general solution 
Example 3.
.
Find a general solution equation 
.
Δ Transformation chain: 
,y. =
zX.,
,
,
,
,
,
,
,
,
,
.
Lecture 8.
4. Linear differential equations of the first order The first-order linear differential equation is
Here is a free member, also called the right-hand part of the equation. In this form we will consider linear equation further.
If a 
0, then equation (4.1a) is called linear inhomogeneous. If 
0, the equation takes the form
|
|
and called linear homogeneous.
The name of equation (4.1a) is explained by the fact that an unknown function y.
and its derivative 
linearly enter it, i.e. in the first degree.
In a linear homogeneous equation, variables are separated. Rewriting it in the form 
from 
and integrating, we get: 
,those.
|
|
When dividing on 
we lose a decision 
. However, it can be included in the family foundation found (4.3), if we assume that FROMmay receive and value 0.
There are several methods for solving equation (4.1a). According to bernoulli methodThe solution is searched as a product of two functions from h.:
|
|
One of these functions can be selected arbitrarily, as only a piece uV. must satisfy the initial equation, the other is determined on the basis of equation (4.1a).
Differentiating both parts of equality (4.4), we find 
.
Substituting the resulting expression derivative 
as well as value w.
in equation (4.1a), we get 
, or
those. As a function v.take the solution of a homogeneous linear equation (4.6):
|
|
(Here C.it is necessary to write, otherwise it will not be common, but a private solution).
Thus, we see that as a result of the substitution used (4.4), equation (4.1a) is reduced to two equations with separating variables (4.6) and (4.7).
Substituting 
and v.(x) in formula (4.4), we finally get
,
|
|
.Example 1.
Find a general solution equation 
Put 
, then 
. Substituting expressions 
and 
in the original equation, we get 
or 
(*)
Equate zero the coefficient when 
:
Separating variables in the resulting equation, we have 
(arbitrary constant C.
we do not write), hence v.=
x.. Found value v.we substitute to equation (*):
,
,
.
Hence, 
general solution of the initial equation.
Note that the equation (*) could be recorded in an equivalent form:
.
Arbitrarily choosing a function u., but not v.we could believe 
. This path of solutions differs from the substitute only considered v.on the u.(and, therefore, u.on the v.) so the final value w.it turns out the same.
Based on the above, we obtain the algorithm for solving a linear differential equation of the first order.
Note further that sometimes the first order equation becomes linear if w.consider an independent variable, and x.- dependent, i.e. change the role x. and y.. This can be done provided that x.and dXpart linously is included in the equation.
Example 2.
.
Solve equation 
.
By type, this equation is not linear relative to the function w..
However, if we consider x.as a function OT w., then considering that 
it can be given to mind
|
|
(4.1 b.) |
Replace 
on the 
, get 
or 
. Sharing both parts of the last equation for the work yDY., give it to mind
, or 
.
(**)
Here p (y) \u003d, 
. This linear equation is relative x.. Believe 
,
. Substituting these expressions in (**), we get
or 
.
Choose Vatus to 
,
From! 
;
. Next, have 
,
,
.
Because 
, we come to a general solution of this equation in the form
.
Note that equation (4.1a) P.(x.) I. Q. (x.) may include not only in the form of functions from x., but also a constant: P.= a.,Q.= b.. Linear equation
|
|
can be solved using the substitution Y \u003d uV. and variables separation:
;
.
From here 
;
;
; Where 
. Free from logarithm, we get a general solution of the equation
(here 
).
For b.= 0 We arrive in solving the equation
|
|
|
|
(see equation of indicative growth (2.4) when 
).
First, we integrate the corresponding homogeneous equation (4.2). As indicated above, its solution has the form (4.3). We will consider a factory FROMin (4.3) the function from h.. Essentially we replace the variable
where, integrating, find
Note that according to (4.14) (see also (4.9)), the general solution of the inhomogeneous linear equation is equal to the sum of the overall solution of the corresponding homogeneous equation (4.3) and the private solution of the inhomogeneous equation determined by the second term included in (4.14) (and in ( 4.9)).
When solving specific equations, the above calculations should be repeated, and not to use a bulky formula (4.14).
Apply the Lagrange method to the equation discussed in example 1. :
.
We integrate the corresponding homogeneous equation 
.
Separating variables, get 
and further 
. Solving formula expression y.
=
CX.. The solution of the original equation is looking for in the form of y.
=
C.(x.)x.. Substituting this expression in the specified equation, we get 
;
;
,
. The general solution of the source equation has the form
.
In conclusion, we note that the Bernoulli equation is given to the linear equation
|
|
,
(
)which can be written as
|
|
.Replacement 
it is provided to a linear equation:
,
,
.
Bernoulli equations are also solved above methods.
Example 3.
.
Find general solutions equation 
.
Transformation chain: 
,
,,
,
,
,
,
,
,
,
,
,
,
,
I think we should start with the history of such a glorious mathematical instrument as differential equations. Like all differential and integral calculations, these equations were invented by Newton at the end of the 17th century. He considered it that his discovery is so important that even encrypted the message, which today can be translated as follows: "All laws of nature are described by differential equations." This may seem exaggeration, but everything is so. Any law of physics, chemistry, biology can be described by these equations.
A huge contribution to the development and creation of the theory of differential equations made mathematics Euler and Lagrange. Already in the 18th century, they opened and developed what is now studying at the senior courses of universities.
The new milestone in the study of differential equations began thanks to Henri Poincare. He created a "high-quality theory of differential equations", which, in combination with the theory of complex variable functions, has made a significant contribution to the basis of topology - the science of space and its properties.
What are differential equations?
Many are afraid of one phrase. However, in this article we will present in detail the whole essence of this very useful mathematical apparatus, which is actually not as folded as it seems from the name. In order to begin talking about the differential equations of the first order, you should first get acquainted with the basic concepts that are inherently associated with this definition. And we will start with differential.
Differential
Many know this concept since school. However, still focus on it in more detail. Imagine a graph of the function. We can increase it to such an extent that any segment will take the kind of straight line. On it, we take two points that are infinitely close to each other. The difference between their coordinates (x or y) will be infinitely low. It is called differential and denote DY signs (differential from Y) and DX (differential from X). It is very important to understand that the differential is not the ultimate magnitude, and this is its meaning and the main function.
And now it is necessary to consider next elementwhich we will come in handy with the explanation of the concept of a differential equation. This is a derivative.
Derivative
We all have probably heard at school and this is a concept. It is said that the derivative is the rate of growth or decrease of the function. However, much of this definition becomes incomprehensible. Let's try to explain the derivative through differentials. Let's return to an infinitely small segment of a function with two points that are at the minimum distance from each other. But even for this distance the function has time to change on some size. And to describe this change and invented a derivative, which otherwise can be written as the ratio of differentials: F (x) "\u003d DF / DX.
Now it is worth considering the main properties of the derivative. There are only three of them:
- The derivative of the amount or difference can be represented as a sum or difference of derivatives: (a + b) "\u003d a" + b "and (a-b)" \u003d a "-b".
- The second property is associated with multiplication. The derivative of the work is the amount of the works of one function on the different derivative: (A * b) "\u003d A" * B + A * B ".
- The difference derivative can be written in the form of the following equality: (A / B) "\u003d (A" * B-A * B ") / b 2.
All these properties will be useful to us for finding solutions of first-order differential equations.
There are also private derivatives. Suppose we have a z function, which depends on the variables x and y. To calculate the private derivative of this function, let's say, by x, we need to take the variable y for the permanent and simply to indifferentiate.
Integral
Another important concept is an integral. In fact, it is the direct opposite of the derivative. Integrals are several species, but to solve the simplest differential equations, we will need the most trivial
So, let's say we have some dependence F from x. We take the integral from it and obtain the function f (x) (often called the primitive), the derivative of which is equal to the original function. Thus, f (x) "\u003d f (x). Hence it also follows that the integral from the derivative is equal to the original function.
When solving differential equations, it is very important to understand the meaning and the function of the integral, as it will have to take them very often to find a solution.
Equations are different depending on their nature. In the following section, we consider the types of differential equations of the first order, and then learn to decide them.
CLASS OF DIFFERENTIAL EQUATIONS
"Diffuras" are divided in order of derivatives participating in them. Thus, the first, second, third and more order. They can also be divided into several classes: ordinary and private derivatives.
In this article, we will consider ordinary differential equations of the first order. Examples and ways to solve them will also discuss in the following sections. We will consider only ODU, because these are the most common types of equations. Ordinary are divided into subspecies: with separating variables, homogeneous and inhomogeneous. Next, you will learn what they differ from each other, and learn them to decide.
In addition, these equations can be combined so that after we have a first-order differential equation system. Such systems we will also consider and learn to decide.
Why do we only consider the first order? Because you need to start with a simple, but describe everything connected with differential equations, in one article is simply impossible.
Equations with separating variables
This is perhaps the most simple differential equations of the first order. These include examples that can be written as follows: y "\u003d f (x) * f (y). To solve this equation, we will need a formula for the representation of the derivative as a ratio of differentials: y" \u003d dy / dx. With the help of it we get such an equation: dy / dx \u003d f (x) * f (y). Now we can refer to the method of solving standard examples: we divide variables in parts, i.e., we move everything from the variable Y to the part where the DY is located, and we will also make a variable x. We obtain the equation of the form: DY / F (Y) \u003d F (X) DX, which is solved by taking the integrals from both parts. Do not forget about the constant that you need to put after taking the integral.
The solution of any "Diffur" is a function of dependence X from Y (in our case) or, if a numerical condition is present, then the answer in the form of a number. Let's wonder specific example All decision:
We carry variables in different directions:
Now we take the integrals. All of them can be found in the special integral table. And we get:
ln (y) \u003d -2 * cos (x) + c
If required, we can express "igrek" as a function from "X". Now we can say that our differential equation is solved if the condition is not specified. A condition can be specified, for example, y (p / 2) \u003d e. Then we simply substitute the value of these variables into the solution and find the value of constant. In our example it is equal to 1.
Uniform first-order differential equations
Now go to a more complex part. Uniform differential equations of the first order can be recorded in general So: y "\u003d z (x, y). It should be noted that the right function from two variables is homogeneous, and it cannot be divided into two dependences: z from x and z from y. Check whether the equation is uniform or not, just enough : We make the replacement x \u003d k * x and y \u003d k * y. Now we reduce all k. If all these letters decreased, it means the equation is homogeneous and can be safely started to solve it. Running forward, say: the principle of solving these examples is also very simple .
We need to make a replacement: y \u003d t (x) * x, where T is a certain function that also depends on x. Then we can express the derivative: y "\u003d t" (x) * x + t. Substituting all this in our original equation and simplifying it, we obtain an example with separating variables T and X. We solve it and obtain the dependence T (x). When we got it, we simply substitute in our previous replacement y \u003d t (x) * x. Then we obtain the dependence Y from x.
To be clearer, we will analyze an example: x * y "\u003d y-x * e y / x.
When checking with a replacement, everything is reduced. So the equation is really homogeneous. Now we make another replacement that we said: y \u003d t (x) * x and y "\u003d t" (x) * x + t (x). After simplification, we obtain the following equation: T "(x) * x \u003d -et. We solve the resulting example with separated variables and get: E -T \u003d ln (C * x). We can only replace T to Y / X (after all if y \u003d T * x, then T \u003d Y / X), and we get the answer: E -y / x \u003d ln (x * c).
Linear differential equations of first order
It's time to consider another extensive topic. We will analyze inhomogeneous differential equations of the first order. What do they differ from the previous two? Let's figure out. Linear differential equations of the first order in a general form can be written by such an equality: y "+ g (x) * y \u003d z (x). It is necessary to clarify that Z (x) and G (x) may be permanent values.
And now an example: y "- y * x \u003d x 2.
There are two ways to solve, and we will analyze both in order. The first is the method of variation of arbitrary constants.
In order to solve the equation in this way, you must first equate right part To zero and solve the resulting equation, which, after carrying parts, takes the form:
ln | y | \u003d x 2/2 + c;
y \u003d E x2 / 2 * y C \u003d C 1 * E x2 / 2.
Now you need to replace the C 1 constant to the V (X) function, which we have to find.
We will replace the derivative:
y "\u003d V" * E x2 / 2 -x * V * E x2 / 2.
And we will substitute these expressions to the original equation:
v "* E x2 / 2 - x * v * e x2 / 2 + x * v * e x2 / 2 \u003d x 2.
It can be seen that two terms are reduced in the left side. If this did not happen in some example, then you didn't do something wrong. Let us continue:
v "* E x2 / 2 \u003d x 2.
Now we solve the usual equation in which the variables should be divided:
dv / dx \u003d x 2 / e x2 / 2;
dV \u003d X 2 * E - X2 / 2 DX.
To remove the integral, we will have to apply integration in parts. However, this is not the topic of our article. If you are interested, you can learn how to perform such actions. It is not difficult, and with sufficient skill and attentiveness does not take much time.
Turn to the second way to solve inhomogeneous equations: Method Bernoulli. What approach is faster and easier to solve only you.
So, when solving the equation, we need to be replaced by this method: y \u003d k * n. Here k and n are some of the X function dependent. Then the derivative will look like this: y "\u003d k" * n + k * n ". We substitute both replacements to the equation:
k "* n + k * n" + x * k * n \u003d x 2.
We group:
k "* n + k * (n" + x * n) \u003d x 2.
Now it is necessary to equate to zero what is in brackets. Now, if you combine the two of the resulting equations, the system of differential equations of the first order is obtained, which must be solved:
The first equality is solved as the usual equation. To do this, share variables:
We take the integral and get: ln (n) \u003d x 2/2. Then, if you express n:
Now we substitute the resulting equality in the second equation of the system:
k "* E x2 / 2 \u003d x 2.
And converting, we get the same equality as in the first method:
dk \u003d x 2 / e x2 / 2.
We will also not disassemble further actions. It is worth saying that at first the solution of the differential equations of the first order causes significant difficulties. However, with a deeper immersion in the topic, it starts to get better and better.
Where are differential equations?
The differential equations are very active in physics, since almost all major laws are recorded in differential form, and the formulas that we see are the solution of these equations. In chemistry, they are used for the same reason: the main laws are derived from them. In biology, differential equations are used to model the behavior of systems, such as a predator - the victim. They can also be used to create models of breeding, say, colony of microorganisms.
How will differential equations help in life?
The answer to this question is simple: no way. If you are not a scientist or engineer, then they are unlikely to use you. However, for general development, it will not hurt to know what the differential equation is and how it is solved. And then the question of the son or daughter "What is a differential equation?" Will not put you in a dead end. Well, if you are a scientist or engineer, then you yourself understand the importance of this topic in any science. But the most important thing is that now on the question "how to solve the first order differential equation?" You will always be able to answer. Agree, it is always nice when you understand what people are even afraid to figure out.
Major problems when studying
The main problem in understanding this topic is a bad skill of integration and differentiation of functions. If you do not take the derivatives and integrals badly, it is probably worth learning, mastering different methods of integration and differentiation, and only then begin to study the material that was described in the article.
Some people are surprised when they find out that the DX can be transferred, because previously (at school) argued that the fraction of DY / DX is indivisible. Here you need to read the literature on the derivative and understand that it is the attitude of infinitely small values, which can be manipulated when solving equations.
Many do not immediately realize that the solution of the first-order differential equations is often a function or an unbearable integral, and this delusion delivers them a lot of trouble.
What else can be learned for a better understanding?
It is best to start a further immersion in the world of differential calculus from specialized textbooks, for example, according to mathematical analysis for students of non-imaging specialties. Then you can move to more specialized literature.
It is worth saying that, besides differential, there are still integral equations, so you will always seek what to strive for and what to study.
Conclusion
We hope that after reading this article, you have an idea of \u200b\u200bwhat differential equations are and how to solve them correctly.
In any case, mathematics in any way come in handy in life. It develops logic and attention, without which every person is without hand.
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 on:
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:
- Find if.
- Find if.
- 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:
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Fill a hand by solving tasks on this topic.
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Uniform Differential Equation of First Order
- this is the equation of type
where F is a function.
How to determine a homogeneous differential equation
In order to determine whether the differential equation of the first order is uniform, it is necessary to introduce a constant T and replace Y on TY and X on TX: Y → TY, X → TX. If t decreases, then this uniform differential equation. The y derive y 'is not changing with such a transformation.
.
Example
Determine whether this equation is homogeneous
Decision
We make the replacement Y → TY, X → TX.
We divide on T. 2
.
.
The equation does not contain t. Consequently, this is a homogeneous equation.
Method of solving a homogeneous differential equation
The homogeneous differential equation of the first order is provided to the equation with separating variables by substitution Y \u003d UX. Show it. Consider the equation:
(i)
Making a substitution:
y \u003d UX,
where u is a function from x. Differentiate by x:
y '\u003d.
Substitute to the original equation (i).
,
,
(ii) .
We share variables. Multiply on DX and divide on x (F (U) - U).
With F. (U) - U ≠ 0 and x ≠ 0
We get:
We integrate:
So we got a common integral equation (i) in quadratures:
Replace the permanent integration C on lN C., then
We lower the sign of the module, since the desired sign is determined by the selection of a constant sign. Then the common integral will take the form:
Next should consider the case of F (U) - U \u003d 0.
If this equation has a root, then they are a solution to the equation (ii). Since equation (ii) does not coincide with the initial equation, then you should make sure that additional solutions satisfy the source equation (i).
Whenever we, in the process of transformations, divide any equation to some function that we denote as G (x, y), then further transformations are valid for g (x, y) ≠ 0. Therefore, one should consider the case of G (x, y) \u003d 0.
An example of solving a homogeneous differential equation of first order
Solve equation
Decision
Check whether this equation is homogeneous. We make the replacement Y → TY, X → TX. In this case, Y '→ Y'.
,
,
.
Reducing on t.
Permanent t decreased. Therefore, the equation is homogeneous.
We make the substitution Y \u003d UX, where u is a function from x.
y '\u003d. (UX) '\u003d u' x + u (x) '\u003d u' x + u
Substitute to the original equation.
,
,
,
.
At x ≥ 0
, | x | \u003d x. At x ≤ 0
, | x | \u003d - x. We write | x | \u003d x implying that the upper sign refers to the values \u200b\u200bx ≥ 0
, and the bottom - to the values \u200b\u200bx ≤ 0
.
,
We multiply on DX and divide on.
Under U. 2 - 1 ≠ 0
We have:
We integrate:
Table integrals,
.
Apply the formula:
(a + b) (a - b) \u003d a 2 - b 2.
Put a \u003d u ,.
.
Take both parts by module and logarithm
.
From here
.
Thus, we have:
,
.
Lower the module sign, since the desired sign is provided by the selection of a constant sign.
We multiply on x and substitute ux \u003d y.
,
.
We are erected into a square.
,
,
.
Now consider the case, u 2 - 1 = 0
.
Roots of this equation
.
It is easy to make sure that the functions y \u003d x satisfy the source equation.
Answer
,
,
.
References:
N.M. Gunter, R.O. Kuzmin, Collection of tasks higher Mathematics, "Lan", 2003.
