Homogeneous dy of the 1st order. How to solve a homogeneous differential equation
To solve a homogeneous differential equation of the 1st order, use the substitution u = y / x, that is, u is a new unknown function depending on x. Hence y = ux. We find the derivative y ’using the product differentiation rule: y’ = (ux) ’= u’x + x’u = u’x + u (since x’ = 1). For another notation: dy = udx + xdu After substitution, we simplify the equation and arrive at an equation with separable variables.
Examples of solving homogeneous differential equations of the 1st order.
1) Solve the equation
We check that this equation is homogeneous (see How to determine homogeneous equation). After making sure, we make the replacement u = y / x, whence y = ux, y ’= (ux)’ = u’x + x’u = u’x + u. Substitute: u'x + u = u (1 + ln (ux) -lnx). Since the logarithm of the product is equal to the sum of the logarithms, ln (ux) = lnu + lnx. From here
u'x + u = u (1 + lnu + lnx-lnx). After reducing similar terms: u'x + u = u (1 + lnu). Now expand the brackets
u'x + u = u + u lnu. Both parts contain u, hence u'x = u lnu. Since u is a function of x, u ’= du / dx. Substitute,
We got an equation with separable variables. We separate the variables, for which we multiply both sides by dx and divide by x u lnu, provided that the product x u lnu ≠ 0
We integrate:
On the left is a tabular integral. On the right, we make the change t = lnu, whence dt = (lnu) ’du = du / u
ln│t│ = ln│x│ + C. But we have already discussed that in such equations instead of C it is more convenient to take ln│C│. Then
ln│t│ = ln│x│ + ln│C│. By the property of logarithms: ln│t│ = ln│Сx│. Hence t = Cx. (by condition, x> 0). It's time to do the reverse replacement: lnu = Cx. And one more reverse replacement:
By the property of logarithms:
This is the general integral of the equation.
We recall the condition product x u lnu ≠ 0 (and hence x ≠ 0, u ≠ 0, lnu ≠ 0, whence u ≠ 1). But x ≠ 0 from the condition, it remains u ≠ 1, whence x ≠ y. Obviously, y = x (x> 0) are included in common decision.
2) Find the partial integral of the equation y ’= x / y + y / x satisfying the initial conditions y (1) = 2.
First, we check that this equation is homogeneous (although the presence of the terms y / x and x / y already indirectly indicates this). Then we make the change u = y / x, whence y = ux, y ’= (ux)’ = u’x + x’u = u’x + u. We substitute the obtained expressions into the equation:
u'x + u = 1 / u + u. Simplifying:
u'x = 1 / u. Since u is a function of x, u ’= du / dx:
We got an equation with separable variables. To separate the variables, we multiply both sides by dx and u and divide by x (x ≠ 0 by hypothesis, hence u ≠ 0 too, so there is no loss of solutions in this case).
We integrate:
and since both parts contain tabular integrals, we immediately obtain
We carry out the reverse replacement:
This is the general integral of the equation. We use the initial condition y (1) = 2, that is, we substitute y = 2, x = 1 into the resulting solution:
3) Find the general integral of the homogeneous equation:
(x²-y²) dy-2xydx = 0.
Change u = y / x, whence y = ux, dy = xdu + udx. Substitute:
(x²- (ux) ²) (xdu + udx) -2ux²dx = 0. Move x² out of parentheses and divide both sides by it (assuming x ≠ 0):
x² (1-u²) (xdu + udx) -2ux²dx = 0
(1-u²) (xdu + udx) -2udx = 0. Expand the brackets and simplify:
xdu-u²xdu + udx-u³dx-2udx = 0,
xdu-u²xdu-u³dx-udx = 0. We group the terms with du and dx:
(x-u²x) du- (u³ + u) dx = 0. We take the common factors out of the brackets:
x (1-u²) du-u (u² + 1) dx = 0. Separating variables:
x (1-u²) du = u (u² + 1) dx. To do this, we divide both sides of the equation by xu (u² + 1) ≠ 0 (respectively, we add the requirements x ≠ 0 (already noted), u ≠ 0):
We integrate:
On the right side of the equation is a tabular integral, rational fraction on the left side we decompose into prime factors:
(or in the second integral one could replace t = 1 + u², dt = 2udu instead of bringing under the differential sign - who likes which method better). We get:
By the properties of logarithms:
Reverse replacement
We recall the condition u ≠ 0. Hence y ≠ 0. At С = 0, y = 0, which means that there is no loss of solutions, and y = 0 is included in the general integral.
Comment
You can get the solution in a different form if you leave the term with x on the left:
The geometric meaning of the integral curve in this case is a family of circles with centers on the Oy axis and passing through the origin.
Self-test tasks:
1) (x² + y²) dx-xydy = 0
1) We check that the equation is homogeneous, after which we make the change u = y / x, whence y = ux, dy = xdu + udx. Substitute in the condition: (x² + x²u²) dx-x²u (xdu + udx) = 0. Dividing both sides of the equation by x² ≠ 0, we get: (1 + u²) dx-u (xdu + udx) = 0. Hence dx + u²dx-xudu-u²dx = 0. Simplifying, we have: dx-xudu = 0. Hence xudu = dx, udu = dx / x. We integrate both parts:
For example, the function 
is a homogeneous function of the first measurement, since
is a homogeneous function of the third dimension, since
- homogeneous function of zero measurement, since
, i.e. 
.
Definition 2. First order differential equation y" = f(x, y) is called homogeneous if the function f(x, y) is a homogeneous function of zero dimension with respect to x and y, or, as they say, f(x, y) Is a homogeneous function of degree zero.
It can be represented as
which allows us to define a homogeneous equation as a differential one that can be transformed to the form (3.3).
Replacement 
leads a homogeneous equation to an equation with separable variables. Indeed, after the substitution y =xz get 
,
Separating the variables and integrating, we find:
,
Example 1: Solve an equation.
Δ We put y =zx,
Substitute these expressions y
and dy v given equation:
or 
Separating variables: 
and integrate: 
,
Replacing z on 
, we get 
.
Example 2. Find the general solution to the equation.
Δ In this equation P
(x,y)
=x 2 -2y 2 ,Q(x,y)
=2xy- homogeneous functions of the second dimension, therefore, this equation is homogeneous. It can be represented as 
and solve in the same way as above. But we use a different form of notation. We put y =
zx, where dy =
zdx
+
xdz... Substituting these expressions into the original equation, we will have
dx+2 zxdz = 0 .
Separate variables by counting 
.
We integrate this equation term by term
, where 
that is 
... Returning to the old function 
find a general solution 
Example 3
.
Find the general solution to the equation 
.
Δ Transformation chain: 
,y =
zx,
,
,
,
,
,
,
,
,
,
.
Lecture 8.
4. Linear differential equations of the first order The linear differential equation of the first order has the form
Here is the free term, also called the right side of the equation. In this form, we will consider linear equation further.
If 
0, then equation (4.1a) is called linear inhomogeneous. If 
0, then the equation takes the form
|
|
and is called linear homogeneous.
The name of equation (4.1a) is explained by the fact that the unknown function y
and its derivative 
enter it linearly, i.e. in the first degree.
In a linear homogeneous equation, the variables are separated. Rewriting it as 
where 
and integrating, we get: 
,those.
|
|
When divided by 
we lose the solution 
... However, it can be included in the found family of solutions (4.3) if we assume that WITH can also take the value 0.
There are several methods for solving equation (4.1a). According to Bernoulli method, the solution is sought in the form of a product of two functions of NS:
|
|
One of these functions can be chosen arbitrarily, since only the product uv must satisfy the original equation, the other is determined based on equation (4.1a).
Differentiating both sides of equality (4.4), we find 
.
Substituting the resulting expression for the derivative 
and also the value at
into equation (4.1a), we obtain 
, or
those. as a function v we take the solution of the homogeneous linear equation (4.6):
|
|
(Here C be sure to write, otherwise you will get not a general, but a particular solution).
Thus, we see that as a result of the substitution (4.4) used, equation (4.1a) is reduced to two equations with separable variables (4.6) and (4.7).
Substituting 
and v(x) into formula (4.4), we finally obtain
,
|
|
.Example 1.
Find the general solution to the equation 
Put 
, then 
... Substituting expressions 
and 
into the original equation, we get 
or 
(*)
Let us equate to zero the coefficient at 
:
Separating the variables in the resulting equation, we have 
(arbitrary constant C
do not write), hence v=
x... Found value v we substitute in the equation (*):
,
,
.
Hence, 
general solution of the original equation.
Note that the equation (*) could be written in an equivalent form:
.
Arbitrarily choosing a function u, but not v, we could believe 
... This solution path differs from the one considered only by replacing v on u(and therefore u on v), so that the final value at turns out to be the same.
Based on the above, we obtain an algorithm for solving a first-order linear differential equation.
Note further that sometimes a first-order equation becomes linear if at be considered an independent variable, and x- dependent, i.e. change roles x and y... This can be done provided that x and dx enter the equation linearly.
Example 2
.
Solve the equation 
.
In appearance, this equation is not linear with respect to the function at.
However, if we consider x as a function of at, then, considering that 
, it can be reduced to the form
|
|
(4.1 b) |
Replacing 
on 
, we get 
or 
... Dividing both sides of the last equation by the product ydy, let's bring it to the form
, or 
.
(**)
Here P (y) =, 
... This is a linear equation with respect to x... We believe 
,
... Substituting these expressions in (**), we obtain
or 
.
We choose v so that 
,
, where 
;
... Further, we have 
,
,
.
Because 
, then we arrive at the general solution of this equation in the form
.
Note that in equation (4.1a) P(x) and Q (x) can enter not only in the form of functions of x but also constants: P= a,Q= b... Linear Equation
|
|
can also be solved using the substitution y = uv and separating variables:
;
.
From here 
;
;
; where 
... Freeing ourselves from the logarithm, we obtain the general solution of the equation
(here 
).
At b= 0 we arrive at the solution of the equation
|
|
|
|
(see the exponential growth equation (2.4) for 
).
First, we integrate the corresponding homogeneous equation (4.2). As indicated above, its solution has the form (4.3). We will consider the factor WITH in (4.3) as a function of NS, i.e. essentially doing the variable change
whence, integrating, we 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 general solution of the corresponding homogeneous equation (4.3) and the particular solution of the inhomogeneous equation determined by the second term in (4.14) (and in ( 4.9)).
When solving specific equations, one should repeat the above calculations, and not use the cumbersome formula (4.14).
We apply the Lagrange method to the equation considered in example 1 :
.
We integrate the corresponding homogeneous equation 
.
Separating the variables, we get 
and further 
... Solving an expression by a formula y
=
Cx... We seek the solution of the original equation in the form y
=
C(x)x... Substituting this expression into the given equation, we get 
;
;
,
... The general solution of the original equation has the form
.
In conclusion, we note that the Bernoulli equation is reduced to a linear equation
|
|
,
(
)which can be written as
|
|
.Replacement 
it is reduced to a linear equation:
,
,
.
Bernoulli's equations are also solved by the above methods.
Example 3
.
Find the general solution to the equation 
.
Chain of transformations: 
,
,,
,
,
,
,
,
,
,
,
,
,
,
I think we should start with the history of such a glorious mathematical tool as differential equations... Like all differential and integral calculus, these equations were invented by Newton in the late 17th century. He considered this discovery of his to be so important that he even encrypted a message that today can be translated something like this: "All laws of nature are described by differential equations." This may seem like an exaggeration, but it is. Any law of physics, chemistry, biology can be described by these equations.
The mathematicians Euler and Lagrange made an enormous contribution to the development and creation of the theory of differential equations. Already in the 18th century, they discovered and developed what is now being studied in the senior years of universities.
A new milestone in the study of differential equations began thanks to Henri Poincaré. He created the "qualitative theory of differential equations", which, in combination with the theory of functions of a complex variable, made a significant contribution to the foundation 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 detail the whole essence of this very useful mathematical apparatus, which is actually not as complicated as the name suggests. In order to start talking about differential equations of the first order, you should first get acquainted with the basic concepts that are inherently associated with this definition. And we'll start with the differential.
Differential
Many people know this concept from school. However, let's dwell on it in more detail. Imagine a graph of a function. We can enlarge it to such an extent that any segment of it takes the form of a 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 infinitesimal. It is called the differential and denoted by the signs dy (differential from y) and dx (differential from x). It is very important to understand that the differential is not a finite value, and this is its meaning and main function.
Now it is necessary to consider next item, which will be useful to us when explaining the concept of a differential equation. This is a derivative.
Derivative
We all probably heard this concept at school. The derivative is said to be the rate at which a function rises or falls. However, from this definition, much becomes incomprehensible. Let's try to explain the derivative in terms of differentials. Let's go back to the infinitesimal segment of a function with two points that are at a minimum distance from each other. But even for this distance, the function has time to change by some amount. And to describe this change and came up with a derivative, which can otherwise be written as the ratio of differentials: f (x) "= df / dx.
Now it is worth considering the basic properties of the derivative. There are only three of them:
- The derivative of the sum or difference can be represented as the sum or difference of the derivatives: (a + b) "= a" + b "and (a-b)" = a "-b".
- The second property is related to multiplication. The derivative of a product is the sum of the products of one function by the derivative of another: (a * b) "= a" * b + a * b ".
- The derivative of the difference can be written as the following equality: (a / b) "= (a" * b-a * b ") / b 2.
All these properties will be useful to us for finding solutions to first-order differential equations.
There are also partial derivatives. Let's say we have a function z that depends on the variables x and y. To calculate the partial derivative of this function, say, with respect to x, we need to take the variable y as a constant and just differentiate.
Integral
Another important concept is the integral. In fact, this is the exact opposite of a derivative. Integrals are of several types, but to solve the simplest differential equations we need the most trivial
So, let's say we have some dependence of f on x. We take the integral from it and get the function F (x) (often called the antiderivative), the derivative of which is equal to the original function. Thus, F (x) "= f (x). It also follows that the integral of the derivative is equal to the original function.
When solving differential equations, it is very important to understand the meaning and function of the integral, since you will very often have to take them to find a solution.
Equations are different depending on their nature. In the next section, we will look at the types of first-order differential equations, and then we will learn how to solve them.
Classes of differential equations
"Diffures" are divided according to the order of the derivatives involved in them. Thus, there is the first, second, third and more order. They can also be divided into several classes: ordinary and partial derivatives.
In this article, we will look at first order ordinary differential equations. We will also discuss examples and how to solve them in the following sections. We will consider only ODEs, because these are the most common types of equations. Ordinary are divided into subspecies: with separable variables, homogeneous and heterogeneous. Next, you will learn how they differ from each other and learn how to solve them.
In addition, these equations can be combined so that after we get a system of first order differential equations. We will also consider such systems and learn how to solve.
Why are we only considering the first order? Because you need to start simple, and it is simply impossible to describe everything related to differential equations in one article.
Separable Equations
These are perhaps the simplest first-order differential equations. These include examples that can be written like this: y "= f (x) * f (y). To solve this equation, we need a formula for representing the derivative as a ratio of differentials: y" = dy / dx. Using it, we get the following equation: dy / dx = f (x) * f (y). Now we can turn to the method for solving standard examples: we will divide the variables in parts, that is, we will transfer everything from the y variable to the part where dy is located, and do the same with the x variable. We get an equation of the form: dy / f (y) = f (x) dx, which is solved by taking integrals from both parts. Do not forget about the constant, which must be set after taking the integral.
The solution to any "diffusion" is a function of the dependence of x on y (in our case) or, if there is a numerical condition, then the answer is in the form of a number. Let's analyze on specific example the whole course of the solution:
We transfer variables in different directions:
Now we take the integrals. All of them can be found in a special table of integrals. And we get:
ln (y) = -2 * cos (x) + C
If required, we can express "game" as a function of "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 (n / 2) = e. Then we simply substitute the value of these variables into the solution and find the value of the constant. In our example, it is equal to 1.
Homogeneous differential equations of the first order
Now let's move on to the more difficult part. Homogeneous differential equations of the first order can be written in general view so: y "= z (x, y). It should be noted that the right-hand function of two variables is homogeneous, and it cannot be divided into two dependences: z on x and z on y. It is quite simple to check whether the equation is homogeneous or not : we make the replacement x = k * x and y = k * y. Now we cancel all k. If all these letters are canceled, then the equation is homogeneous and we can safely start solving it. Looking ahead, let's say: the principle of solving these examples is also very simple ...
We need to make a replacement: y = t (x) * x, where t is a function that also depends on x. Then we can express the derivative: y "= t" (x) * x + t. Substituting all of this into our original equation and simplifying it, we get an example with separable variables t and x. We solve it and get the dependence t (x). When we get it, we simply substitute y = t (x) * x in our previous replacement. Then we get the dependence of y on x.
To make it clearer, let's look at an example: x * y "= y-x * e y / x.
When checking and replacing, everything is reduced. This means that the equation is really homogeneous. Now we make another replacement, which we talked about: y = t (x) * x and y "= t" (x) * x + t (x). After simplification, we get the following equation: t "(x) * x = -et. Solve the resulting example with separated variables and get: e -t = ln (C * x). We only need to replace t with y / x (after all, if y = t * x, then t = y / x), and we get the answer: e -y / x = ln (x * C).
Linear differential equations of the first order
It's time to consider another broad topic. We will analyze first order inhomogeneous differential equations. How are they different from the previous two? Let's figure it out. Linear differential equations of the first order in general form can be written as follows: y "+ g (x) * y = z (x). It is worth clarifying that z (x) and g (x) can be constant values.
And now an example: y "- y * x = x 2.
There are two ways to solve this, and we will deal with 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 side to zero and solve the resulting equation, which after transferring the parts will take the form:
ln | y | = x 2/2 + C;
y = e x2 / 2 * y C = C 1 * e x2 / 2.
Now we need to replace the constant C 1 with the function v (x), which we have to find.
Let's replace the derivative:
y "= v" * e x2 / 2 -x * v * e x2 / 2.
And we substitute these expressions into the original equation:
v "* e x2 / 2 - x * v * e x2 / 2 + x * v * e x2 / 2 = x 2.
You can see that two terms are canceled on the left. If in some example this did not happen, then you did something wrong. Let's continue:
v "* e x2 / 2 = x 2.
Now we solve the usual equation in which we need to separate the variables:
dv / dx = x 2 / e x2 / 2;
dv = x 2 * e - x2 / 2 dx.
To extract the integral, we have to apply integration by parts here. However, this is not the subject of our article. If you're interested, you can learn how to do these things yourself. It is not difficult, and with enough skill and attention, it does not take a lot of time.
Let's turn to the second solution inhomogeneous equations: Bernoulli method. Which approach is faster and easier is up to you.
So, when solving the equation by this method, we need to make a substitution: y = k * n. Here k and n are some x-dependent functions. Then the derivative will look like this: y "= k" * n + k * n "Substitute both substitutions in the equation:
k "* n + k * n" + x * k * n = x 2.
We group:
k "* n + k * (n" + x * n) = x 2.
Now we need to equate to zero what is in parentheses. Now, if you combine the two resulting equations, you get a system of first-order differential equations that needs to be solved:
We solve the first equality as an ordinary equation. To do this, you need to separate the variables:
We take the integral and get: ln (n) = x 2/2. Then, if we express n:
Now we substitute the resulting equality into the second equation of the system:
k "* e x2 / 2 = x 2.
And converting, we get the same equality as in the first method:
dk = x 2 / e x2 / 2.
We will also not analyze further actions. It should be said that at first the solution of first-order differential equations causes significant difficulties. However, as you delve deeper into the topic, it starts to get better and better.
Where are differential equations used?
Differential equations are very actively used in physics, since almost all basic laws are written 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 basic laws are deduced with their help. In biology, differential equations are used to model the behavior of systems, such as a predator-prey. They can also be used to create breeding models for, say, a microbial colony.
How do differential equations help in life?
The answer to this question is simple: nothing. If you are not a scientist or engineer, then they are unlikely to be useful to you. However, for general development, it does not hurt to know what a differential equation is and how it is solved. And then the question of a son or daughter "what is a differential equation?" will not confuse you. 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 the question "how to solve a first-order differential equation?" you can always give an answer. Agree, it's always nice when you understand what people are even afraid to understand.
The main problems when studying
The main problem in understanding this topic is poor skill in integrating and differentiating functions. If you are not good at taking derivatives and integrals, then it is probably worth learning more, mastering different methods of integration and differentiation, and only then start studying the material that was described in the article.
Some people are surprised when they find out that dx can be carried over, because earlier (in school) it was stated that the fraction dy / dx is indivisible. Here you need to read the literature on the derivative and understand that it is the ratio of infinitesimal quantities that can be manipulated when solving equations.
Many people do not immediately realize that solving first-order differential equations is often a function or non-trivial integral, and this delusion gives them a lot of trouble.
What else can you study for a better understanding?
It is best to start your further immersion in the world of differential calculus with specialized textbooks, for example, in mathematical analysis for students of non-mathematical specialties. Then you can move on to more specialized literature.
It is worth saying that, in addition to differential equations, there are also integral equations, so you will always have something to strive for and what to study.
Conclusion
We hope that after reading this article, you have an idea of what differential equations are and how to solve them correctly.
In any case, mathematics will in some way be useful to us in life. It develops logic and attention, without which every person is like no hands.
Stop! Let's all the same try to understand this cumbersome formula.
In the first place should be the first variable to the degree with a certain coefficient. In our case it is
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 is in place. Coefficient.
We have it.
The first variable is in power, and the second variable is squared, with a coefficient. This is the last term in the equation.
As you can see, our equation fits the definition of a formula.
Let's look at the second (verbal) part of the definition.
We have two unknowns and. It converges here.
Consider all the terms. In them, the sum of the degrees of the unknowns must be the same.
The sum of the degrees is equal.
The sum of the degrees is equal to (for and for).
The sum of the degrees is equal.
As you can see, it all fits together !!!
Now let's practice defining homogeneous equations.
Determine which of the equations are homogeneous:
Homogeneous equations - equations numbered:
Let us consider the equation separately.
If we divide each term by expanding each term, we get
And this equation completely falls under the definition of homogeneous equations.
How to solve homogeneous equations?
Example 2.
Divide the equation by.
By condition, y cannot be equal to us. Therefore, we can safely divide by
By replacing, we get a simple quadratic equation:
Since this is a reduced quadratic equation, we use Vieta's theorem:
Having made the reverse substitution, we get the answer
Answer:
Example 3.
Divide the equation by (by condition).
Answer:
Example 4.
Find if.
Here you need not divide, but multiply. Let's multiply the whole equation by:
Let's make the replacement and solve the quadratic equation:
Having made the reverse replacement, we get the answer:
Answer:
Solving homogeneous trigonometric equations.
Solving homogeneous trigonometric equations is no different from the solutions described above. Only here, among other things, you need to know a little trigonometry. And be able to decide trigonometric equations(for this you can read the section).
Let's consider such equations by examples.
Example 5.
Solve the equation.
We see a typical homogeneous equation: and are unknowns, and the sum of their powers in each term is equal.
Such homogeneous equations are not difficult to solve, but before dividing the equations into, consider the case when
In this case, the equation will take the form:, then. But sine and cosine cannot be equal at the same time, because according to the basic trigonometric identity. Therefore, we can safely divide into it:
Since the equation is reduced, then by Vieta's theorem:
Answer:
Example 6.
Solve the equation.
As in the example, you need to divide the equation by. Consider the case when:
But sine and cosine cannot be equal at the same time, because according to the basic trigonometric identity. That's why.
Let's make a substitution and solve a quadratic equation:
Let's make the reverse replacement and find and:
Answer:
Solving homogeneous exponential equations.
Homogeneous equations are solved in the same way as those discussed above. If you forgot how to decide exponential equations- see the corresponding section ()!
Let's look at a few examples.
Example 7.
Solve the equation
Let's imagine how:
We see a typical homogeneous equation, with two variables and a sum of degrees. Divide the equation into:
As you can see, making the substitution, we get the reduced quadratic equation (in this case, there is no need to be afraid of division by zero - it is always strictly greater than zero):
By Vieta's theorem:
Answer: .
Example 8.
Solve the equation
Let's imagine how:
Divide the equation into:
Let's make the replacement and solve the quadratic equation:
The root does not satisfy the condition. Let's make a reverse replacement and find:
Answer:
HOMOGENEOUS EQUATIONS. AVERAGE LEVEL
First, using one problem as an example, let me remind you what are homogeneous equations and what is the solution of homogeneous equations.
Solve the problem:
Find if.
Here you can notice a curious thing: if you divide each term by, we get:
That is, now there are no separate and, - now the variable in the equation is the desired value. And this is an ordinary quadratic equation that can be easily solved using Vieta's theorem: the product of the roots is equal, and the sum is the numbers and.
Answer:
Equations of the form
called homogeneous. That is, this is an equation with two unknowns, each term of which has the same sum of the powers of these unknowns. For example, in the example above, this amount is. The solution of homogeneous equations is carried out by dividing by one of the unknowns to this degree:
And the subsequent replacement of variables:. Thus, we obtain an equation of degree with one unknown:
Most often we will come across equations of the second degree (that is, quadratic), and we are able to solve them:
Note that dividing (and multiplying) the entire equation by a variable is only possible 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 by. In cases where it is not so obvious, it is necessary to separately check the case when this variable is equal to zero. For example:
Solve the equation.
Solution:
We see here a typical homogeneous equation: and are unknowns, and the sum of their powers in each term is equal.
But, before dividing by and getting a quadratic equation for, we must consider the case when. In this case, the equation will take the form:, hence,. But sine and cosine cannot be equal to zero at the same time, because according to the main trigonometric identity:. Therefore, we can safely divide into it:
Hope this solution is completely clear? If not, read the section. If it's not clear where it came from, you need to return even earlier - to the section.
Decide for yourself:
- Find if.
- Find if.
- Solve the equation.
Here I will briefly write directly the solution of homogeneous equations:
Solutions:
Answer: .
And here we must not divide, but multiply:
Answer:
If you have not done trigonometric equations yet, you can skip this example.
Since here we need to divide by, let's make sure first that it does not is zero:
This is impossible.
Answer: .
HOMOGENEOUS EQUATIONS. BRIEFLY ABOUT THE MAIN
The solution of all homogeneous equations is reduced to dividing by one of the unknowns in power and further by changing the variables.
Algorithm:
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Homogeneous differential equation of the first order
is an equation of the form
, where f is a function.
How to define a homogeneous differential equation
In order to determine whether a first-order differential equation is homogeneous, it is necessary to introduce a constant t and replace y with ty and x with tx: y → ty, x → tx. If t is canceled, then this homogeneous differential equation... The derivative y ′ does not change under this transformation.
.
Example
Determine if a given equation is homogeneous
Solution
We make the replacement y → ty, x → tx.
Divide by t 2
.
.
The equation does not contain t. Hence, this is a homogeneous equation.
Method for solving a homogeneous differential equation
A homogeneous first-order differential equation is reduced to an equation with separable variables using the substitution y = ux. Let's show it. Consider the equation:
(i)
We make the substitution:
y = ux,
where u is a function of x. Differentiate by x:
y ′ =
Substitute in the original equation (i).
,
,
(ii) .
Separating variables. Multiply by dx and divide by x (f (u) - u).
For f (u) - u ≠ 0 and x ≠ 0
we get:
We integrate:
Thus, we have obtained the general integral of the equation (i) in quadratures:
We replace the constant of integration C by ln C, then
We omit the modulus sign, since the required sign is determined by the choice of the sign of the constant C. Then the general integral will take the form:
Next, consider the case f (u) - u = 0.
If this equation has roots, then they are a solution to the equation (ii)... Since the equation (ii) does not coincide with the original equation, then you should make sure that the additional solutions satisfy the original equation (i).
Whenever we, in the process of transformations, divide any equation by some function, which we denote as g (x, y), then further transformations are valid for g (x, y) ≠ 0... Therefore, the case g (x, y) = 0.
An example of solving a homogeneous first-order differential equation
Solve the equation
Solution
Let us check whether the given equation is homogeneous. We make the replacement y → ty, x → tx. Moreover, y ′ → y ′.
,
,
.
Reduce by t.
The constant t has decreased. Therefore, the equation is homogeneous.
We make the substitution y = ux, where u is a function of x.
y ′ = (ux) ′ = u ′ x + u (x) ′ = u ′ x + u
Substitute in the original equation.
,
,
,
.
For x ≥ 0
, | x | = x. For x ≤ 0
, | x | = - x. We write | x | = x implying that the upper sign refers to values x ≥ 0
, and the lower one - to the values x ≤ 0
.
,
Multiply by dx and divide by.
For u 2 - 1 ≠ 0
we have:
We integrate:
Integrals are tabular,
.
Let's apply the formula:
(a + b) (a - b) = a 2 - b 2.
We put a = u,.
.
We take both sides modulo and logarithm,
.
From here
.
Thus, we have:
,
.
We omit the modulus sign, since the required sign is provided by choosing the sign of the constant C.
Multiply by x and substitute ux = y.
,
.
Squaring.
,
,
.
Now consider the case u 2 - 1 = 0
.
The roots of this equation
.
It is easy to verify that the functions y = x satisfy the original equation.
Answer
,
,
.
References:
N.M. Gunther, R.O. Kuzmin, Collection of problems on higher mathematics, "Doe", 2003.
