Instructions

If the equation is presented in the form: dy/dx = q(x)/n(y), classify them as differential equations with separable variables. They can be solved by writing the condition in differentials as follows: n(y)dy = q(x)dx. Then integrate both sides. In some cases, the solution is written in the form of integrals taken from known functions. For example, in the case of dy/dx = x/y, we get q(x) = x, n(y) = y. Write it in the form ydy = xdx and integrate. It should be y^2 = x^2 + c.

To linear equations relate the equations to “first”. An unknown function with its derivatives enters such an equation only to the first degree. Linear has the form dy/dx + f(x) = j(x), where f(x) and g(x) are functions depending on x. The solution is written using integrals taken from known functions.

Please note that many differential equations- these are second-order equations (containing second derivatives). For example, this is the equation of simple harmonic motion, written in the general form: md 2x/dt 2 = –kx. Such equations have, in , particular solutions. The equation of simple harmonic motion is an example of something quite important: linear differential equations that have a constant coefficient.

If there is only one linear equation in the problem conditions, then you have been given additional conditions, thanks to which a solution can be found. Read the problem carefully to find these conditions. If variables x and y indicate distance, speed, weight - feel free to set the limit x≥0 and y≥0. It is quite possible that x or y hides the number of apples, etc. – then the values can only be . If x is the son’s age, it is clear that he cannot be older than his father, so indicate this in the conditions of the problem.

Sources:

how to solve an equation with one variable

Differential and integral calculus problems are important elements consolidation of the theory mathematical analysis, section higher mathematics, studied in universities. Differential the equation solved by the integration method.

Instructions

Differential calculus explores the properties of . And vice versa, integrating a function allows for given properties, i.e. derivatives or differentials of a function to find it itself. This is the solution to the differential equation.

Anything is a relationship between an unknown quantity and known data. In the case of a differential equation, the role of the unknown is played by a function, and the role of known quantities is played by its derivatives. In addition, the relation may contain an independent variable: F(x, y(x), y'(x), y''(x),…, y^n(x)) = 0, where x is an unknown variable, y (x) is the function to be determined, the order of the equation is the maximum order of the derivative (n).

Such an equation is called an ordinary differential equation. If the relationship contains several independent variables and partial derivatives (differentials) of the function with respect to these variables, then the equation is called a partial differential equation and has the form: x∂z/∂y - ∂z/∂x = 0, where z(x, y) is the required function.

So, in order to learn how to solve differential equations, you need to be able to find antiderivatives, i.e. solve the problem inverse to differentiation. For example: Solve the first order equation y’ = -y/x.

SolutionReplace y’ with dy/dx: dy/dx = -y/x.

Reduce the equation to a form convenient for integration. To do this, multiply both sides by dx and divide by y:dy/y = -dx/x.

Integrate: ∫dy/y = - ∫dx/x + Сln |y| = - ln |x| + C.

This solution is called the general differential equation. C is a constant whose set of values determines the set of solutions to the equation. For any specific value of C, the solution will be unique. This solution is a partial solution of the differential equation.

Solving most higher-order equations degrees does not have a clear formula for finding square roots equations. However, there are several reduction methods that allow you to transform a higher degree equation into a more visual form.

Instructions

The most common method for solving higher degree equations is expansion. This approach is a combination of selecting integer roots, divisors of the free term, and subsequent division of the general polynomial into the form (x – x0).

For example, solve the equation x^4 + x³ + 2 x² – x – 3 = 0. Solution: The free term of this polynomial is -3, therefore, its integer divisors can be the numbers ±1 and ±3. Substitute them one by one into the equation and find out whether you get the identity: 1: 1 + 1 + 2 – 1 – 3 = 0.

Second root x = -1. Divide by the expression (x + 1). Write down the resulting equation (x - 1)·(x + 1)·(x² + x + 3) = 0. The degree has been reduced to the second, therefore, the equation can have two more roots. To find them, solve the quadratic equation: x² + x + 3 = 0D = 1 – 12 = -11

The discriminant is a negative value, which means that the equation no longer has real roots. Find the complex roots of the equation: x = (-2 + i·√11)/2 and x = (-2 – i·√11)/2.

Another method for solving a higher degree equation is to change variables to make it quadratic. This approach is used when all powers of the equation are even, for example: x^4 – 13 x² + 36 = 0

Now find the roots of the original equation: x1 = √9 = ±3; x2 = √4 = ±2.

Tip 10: How to Determine Redox Equations

A chemical reaction is a process of transformation of substances that occurs with a change in their composition. Those substances that react are called initial substances, and those that are formed as a result of this process are called products. It happens that during chemical reaction the elements that make up the starting substances change their oxidation state. That is, they can accept someone else's electrons and give away their own. In both cases, their charge changes. Such reactions are called redox reactions.

Lecture notes on

differential equations

Differential equations

Introduction

When studying certain phenomena, a situation often arises when the process cannot be described using the equation y=f(x) or F(x;y)=0. In addition to the variable x and the unknown function, the derivative of this function enters the equation.

Definition: The equation connecting the variable x, the unknown function y(x) and its derivatives is called differential equation. IN general view the differential equation looks like this:

F(x;y(x); ;;...;y (n))=0

Definition: The order of a differential equation is the order of the highest derivative included in it.

–differential equation of 1st order

–differential equation of 3rd order

Definition: The solution to a differential equation is a function that, when substituted into the equation, turns it into an identity.

1st order differential equations

Definition: Equation of the form =f(x;y) or F(x;y; )=0is called a 1st order differential equation.

Definition: The general solution of a 1st order differential equation is the function y=γ(x;c), where (c –const), which, when substituted into the equation, turns it into an identity. Geometrically, on the plane, the general solution corresponds to a family of integral curves depending on the parameter c.

Definition: The integral curve passing through a point in the plane with coordinates (x 0 ;y 0) corresponds to a particular solution of the differential equation satisfying the initial condition:

Theorem on the existence of uniqueness of a solution to a 1st order differential equation

Given a 1st order differential equation
and the function f(x;y) is continuous along with partial derivatives in some region D of the XOY plane, then through the point M 0 (x 0 ;y 0) D passes through the only curve corresponding to a particular solution of the differential equation corresponding to the initial condition y(x 0)=y 0

One integral curve passes through a point in the plane with given coordinates.

If it is not possible to obtain a general solution to a 1st order differential equation in explicit form, i.e.
, then it can be obtained implicitly:

F(x; y; c) =0 – implicit form

The general solution in this form is called general integral differential equation.

In relation to the 1st order differential equation, 2 problems are posed:

1) Find the general solution (general integral)

2) Find a particular solution (partial integral) that satisfies the given initial condition. This problem is called the Cauchy problem for a differential equation.

Differential equations with separable variables

Equations of the form:
is called a differential equation with separable variables.

Let's substitute

multiply by dx

let's separate the variables

divide by

Note: it is necessary to consider the special case when

variables are separated

let's integrate both sides of the equation

- common decision

A differential equation with separable variables can be written as:

An isolated case
!

Let's integrate both sides of the equation:

2)
beginning conditions:

Homogeneous differential equations of 1st order

Definition: Function
is called homogeneous of order n if

Example: - homogeneous function of ordern=2

Definition: A homogeneous function of order 0 is called homogeneous.

Definition: Differential equation
is called homogeneous if
- homogeneous function, i.e.

Thus, the homogeneous differential equation can be written as:

Using replacement , where t is a function of the variable x, the homogeneous differential equation is reduced to an equation with separable variables.

- substitute into the equation

Variables separated, let's integrate both sides of the equation

Let's make the reverse substitution by substituting , we obtain a general solution in implicit form.

A homogeneous differential equation can be written in differential form.

M(x;y)dx+N(x;y)dy=0, where M(x;y) and N(x;y) are homogeneous functions of the same order.

Divide by dx and express

Ordinary differential equation is an equation that relates an independent variable, an unknown function of this variable and its derivatives (or differentials) of various orders.

The order of the differential equation is called the order of the highest derivative contained in it.

In addition to ordinary ones, partial differential equations are also studied. These are equations relating independent variables, an unknown function of these variables and its partial derivatives with respect to the same variables. But we will only consider ordinary differential equations and therefore, for the sake of brevity, we will omit the word “ordinary”.

Examples of differential equations:

(1) ;

(3) ;

(4) ;

Equation (1) is fourth order, equation (2) is third order, equations (3) and (4) are second order, equation (5) is first order.

Differential equation n th order does not necessarily have to contain an explicit function, all its derivatives from the first to n-th order and independent variable. It may not explicitly contain derivatives of certain orders, a function, or an independent variable.

For example, in equation (1) there are clearly no third- and second-order derivatives, as well as a function; in equation (2) - the second-order derivative and the function; in equation (4) - the independent variable; in equation (5) - functions. Only equation (3) contains explicitly all the derivatives, the function and the independent variable.

Solving a differential equation every function is called y = f(x), when substituted into the equation it turns into an identity.

The process of finding a solution to a differential equation is called its integration.

Example 1. Find the solution to the differential equation.

Solution. Let's write this equation in the form . The solution is to find the function from its derivative. The original function, as is known from integral calculus, is an antiderivative for, i.e.

That's what it is solution to this differential equation . Changing in it C, we will obtain different solutions. We found out that there is an infinite number of solutions to a first order differential equation.

General solution of the differential equation n th order is its solution, expressed explicitly with respect to the unknown function and containing n independent arbitrary constants, i.e.

The solution to the differential equation in Example 1 is general.

Partial solution of the differential equation a solution in which arbitrary constants are given specific numerical values is called.

Example 2. Find the general solution of the differential equation and a particular solution for .

Solution. Let's integrate both sides of the equation a number of times equal to the order of the differential equation.

As a result, we received a general solution -

of a given third order differential equation.

Now let's find a particular solution under the specified conditions. To do this, substitute their values instead of arbitrary coefficients and get

If, in addition to the differential equation, the initial condition is given in the form , then such a problem is called Cauchy problem . Substitute the values and into the general solution of the equation and find the value of an arbitrary constant C, and then a particular solution of the equation for the found value C. This is the solution to the Cauchy problem.

Example 3. Solve the Cauchy problem for the differential equation from Example 1 subject to .

Solution. Let us substitute the values from the initial condition into the general solution y = 3, x= 1. We get

We write down the solution to the Cauchy problem for this first-order differential equation:

Solving differential equations, even the simplest ones, requires good integration and derivative skills, including complex functions. This can be seen in the following example.

Example 4. Find the general solution to the differential equation.

Solution. The equation is written in such a form that you can immediately integrate both sides.

We apply the method of integration by change of variable (substitution). Let it be then.

Required to take dx and now - attention - we do this according to the rules of differentiation of a complex function, since x and there is complex function("apple" - extraction square root or, what is the same thing - raising to the power “one-half”, and “minced meat” is the very expression under the root):

We find the integral:

Returning to the variable x, we get:

This is the general solution to this first degree differential equation.

Not only skills from previous sections of higher mathematics will be required in solving differential equations, but also skills from elementary, that is, school mathematics. As already mentioned, in a differential equation of any order there may not be an independent variable, that is, a variable x. Knowledge about proportions from school that has not been forgotten (however, depending on who) from school will help solve this problem. This is the next example.

Either have already been solved with respect to the derivative, or they can be solved with respect to the derivative .

General solution of differential equations of the type on the interval X, which is given, can be found by taking the integral of both sides of this equality.

We get .

If you look at the properties indefinite integral, then we find the desired general solution:

y = F(x) + C,

Where F(x)- one of antiderivative functions f(x) in between X, A WITH- arbitrary constant.

Please note that in most problems the interval X do not indicate. This means that a solution must be found for everyone. x, for which and the desired function y, and the original equation make sense.

If you need to calculate a particular solution to a differential equation that satisfies the initial condition y(x 0) = y 0, then after calculating the general integral y = F(x) + C, it is still necessary to determine the value of the constant C = C 0, using the initial condition. That is, a constant C = C 0 determined from the equation F(x 0) + C = y 0, and the desired partial solution of the differential equation will take the form:

y = F(x) + C 0.

Let's look at an example:

Let's find a general solution to the differential equation and check the correctness of the result. Let us find a particular solution to this equation that would satisfy the initial condition.

Solution:

After we integrate the given differential equation, we get:

Let's take this integral using the method of integration by parts:

That., is a general solution to the differential equation.

To make sure the result is correct, let's do a check. To do this, we substitute the solution we found into the given equation:

That is, when the original equation turns into an identity:

therefore, the general solution of the differential equation was determined correctly.

The solution we found is a general solution to the differential equation for every real value of the argument x.

It remains to calculate a particular solution to the ODE that would satisfy the initial condition. In other words, it is necessary to calculate the value of the constant WITH, at which the equality will be true:

Then, substituting C = 2 into the general solution of the ODE, we obtain a particular solution of the differential equation that satisfies the initial condition:

Ordinary differential equation can be solved for the derivative by dividing the 2 sides of the equation by f(x). This transformation will be equivalent if f(x) does not turn to zero under any circumstances x from the integration interval of the differential equation X.

There are likely situations when, for some values of the argument x ∈ X functions f(x) And g(x) simultaneously become zero. For similar values x the general solution of a differential equation is any function y, which is defined in them, because .

If for some argument values x ∈ X the condition is satisfied, which means that in this case the ODE has no solutions.

For everyone else x from the interval X the general solution of the differential equation is determined from the transformed equation.

Let's look at examples:

Example 1.

Let's find a general solution to the ODE: .

Solution.

From the properties of the basic elementary functions it is clear that the function natural logarithm is defined for non-negative argument values, so the scope of the expression is ln(x+3) there is an interval x > -3 . This means that the given differential equation makes sense for x > -3 . For these argument values, the expression x+3 does not vanish, so you can solve the ODE for the derivative by dividing the 2 parts by x + 3.

We get .

Next, we integrate the resulting differential equation, solved with respect to the derivative: . To take this integral, we use the method of subsuming it under the differential sign.

The content of the article

DIFFERENTIAL EQUATIONS. Many physical laws that govern certain phenomena are written in the form mathematical equation, expressing a certain relationship between some quantities. Often we are talking about the relationship between quantities that change over time, for example, engine efficiency, measured by the distance that a car can travel on one liter of fuel, depends on the speed of the car. The corresponding equation contains one or more functions and their derivatives and is called a differential equation. (The rate of change of distance over time is determined by speed; therefore, speed is a derivative of distance; similarly, acceleration is a derivative of speed, since acceleration determines the rate of change of speed with time.) Great importance, which differential equations have for mathematics and especially for its applications, are explained by the fact that the study of many physical and technical problems comes down to solving such equations. Differential equations also play a significant role in other sciences, such as biology, economics and electrical engineering; in fact, they arise wherever there is a need for a quantitative (numerical) description of phenomena (since the world changes over time and conditions change from one place to another).

Examples.

The following examples provide a better understanding of how various problems are formulated in the language of differential equations.

1) The law of decay of some radioactive substances is that the decay rate is proportional to the available amount of this substance. If x– the amount of substance at a certain point in time t, then this law can be written as follows:

Where dx/dt is the decay rate, and k– some positive constant characterizing a given substance. (The minus sign on the right side indicates that x decreases over time; a plus sign, always implied when the sign is not explicitly stated, would mean that x increases over time.)

2) The container initially contains 10 kg of salt dissolved in 100 m 3 of water. If pure water pours into the container at a speed of 1 m 3 per minute and mixes evenly with the solution, and the resulting solution flows out of the container at the same speed, then how much salt will be in the container at any subsequent point in time? If x– amount of salt (in kg) in the container at a time t, then at any time t 1 m 3 of solution in the container contains x/100 kg salt; therefore the amount of salt decreases at a rate x/100 kg/min, or

3) Let there be masses on the body m suspended from the end of the spring, a restoring force acts proportional to the amount of tension in the spring. Let x– the amount of deviation of the body from the equilibrium position. Then, according to Newton's second law, which states that acceleration (the second derivative of x by time, designated d 2 x/dt 2) proportional to force:

The right side has a minus sign because the restoring force reduces the stretch of the spring.

4) The law of body cooling states that the amount of heat in a body decreases in proportion to the difference in body temperature and environment. If a cup of coffee heated to a temperature of 90°C is in a room where the temperature is 20°C, then

Where T– coffee temperature at time t.

5) The Foreign Minister of the State of Blefuscu claims that the arms program adopted by Lilliput forces his country to increase military spending as much as possible. The Minister of Foreign Affairs of Lilliput makes similar statements. The resulting situation (in its simplest interpretation) can be accurately described by two differential equations. Let x And y- expenses for armament of Lilliput and Blefuscu. Assuming that Lilliput increases its expenditures on armaments at a rate proportional to the rate of increase in expenditures on armaments of Blefuscu, and vice versa, we obtain:

where the members are ax And - by describe the military expenditures of each country, k And l are positive constants. (This problem was first formulated in this way in 1939 by L. Richardson.)

After the problem is written in the language of differential equations, you should try to solve them, i.e. find the quantities whose rates of change are included in the equations. Sometimes solutions are found in the form of explicit formulas, but more often they can only be presented in approximate form or qualitative information can be obtained about them. It can often be difficult to determine whether a solution even exists, let alone find one. An important section of the theory of differential equations consists of the so-called “existence theorems”, in which the existence of a solution for one or another type of differential equation is proved.

The original mathematical formulation of a physical problem usually contains simplifying assumptions; the criterion of their reasonableness can be the degree of consistency of the mathematical solution with the available observations.

Solutions of differential equations.

Differential equation, for example dy/dx = x/y, is satisfied not by a number, but by a function, in this particular case such that its graph at any point, for example at a point with coordinates (2,3), has a tangent with slope, equal to the ratio of coordinates (in our example 2/3). This is easy to verify if you build big number points and from each set aside a short segment with a corresponding slope. The solution will be a function whose graph touches each of its points to the corresponding segment. If there are enough points and segments, then we can approximately outline the course of the solution curves (three such curves are shown in Fig. 1). There is exactly one solution curve passing through each point with y No. 0. Each individual solution is called a partial solution of a differential equation; if it is possible to find a formula containing all the particular solutions (with the possible exception of a few special ones), then they say that a general solution has been obtained. A particular solution represents one function, while a general solution represents a whole family of them. Solving a differential equation means finding either its particular or general solution. In the example we are considering, the general solution has the form y 2 – x 2 = c, Where c– any number; a particular solution passing through the point (1,1) has the form y = x and it turns out when c= 0; a particular solution passing through point (2,1) has the form y 2 – x 2 = 3. The condition requiring that the solution curve pass, for example, through the point (2,1), is called the initial condition (since it specifies the starting point on the solution curve).

It can be shown that in example (1) the general solution has the form x = ce –kt, Where c– a constant that can be determined, for example, by indicating the amount of substance at t= 0. The equation from example (2) is a special case of the equation from example (1), corresponding k= 1/100. Initial condition x= 10 at t= 0 gives a particular solution x = 10e –t/100 . The equation from example (4) has a general solution T = 70 + ce –kt and private solution 70 + 130 – kt; to determine the value k, additional data is needed.

Differential equation dy/dx = x/y is called a first-order equation, since it contains the first derivative (the order of a differential equation is usually considered to be the order of the highest derivative included in it). For most (though not all) differential equations of the first kind that arise in practice, only one solution curve passes through each point.

There are several important types of first-order differential equations that can be solved in the form of formulas containing only elementary functions - powers, exponents, logarithms, sines and cosines, etc. Such equations include the following.

Equations with separable variables.

Equations of the form dy/dx = f(x)/g(y) can be solved by writing it in differentials g(y)dy = f(x)dx and integrating both parts. In the worst case, the solution can be represented in the form of integrals of known functions. For example, in the case of the equation dy/dx = x/y we have f(x) = x, g(y) = y. By writing it in the form ydy = xdx and integrating, we get y 2 = x 2 + c. Equations with separable variables include equations from examples (1), (2), (4) (they can be solved in the manner described above).

Equations in total differentials.

If the differential equation has the form dy/dx = M(x,y)/N(x,y), Where M And N are two given functions, then it can be represented as M(x,y)dx – N(x,y)dy= 0. If the left side is the differential of some function F(x,y), then the differential equation can be written as dF(x,y) = 0, which is equivalent to the equation F(x,y) = const. Thus, the solution curves of the equation are the “lines of constant levels” of the function, or the locus of points that satisfy the equations F(x,y) = c. The equation ydy = xdx(Fig. 1) - with separable variables, and the same - in total differentials: to make sure of the latter, we write it in the form ydy – xdx= 0, i.e. d(y 2 – x 2) = 0. Function F(x,y) in this case is equal to (1/2)( y 2 – x 2); Some of its constant level lines are shown in Fig. 1.

Linear equations.

Linear equations are equations of “first degree” - the unknown function and its derivatives appear in such equations only to the first degree. Thus, the first order linear differential equation has the form dy/dx + p(x) = q(x), Where p(x) And q(x) – functions that depend only on x. Its solution can always be written using integrals of known functions. Many other types of first-order differential equations are solved using special techniques.

Higher order equations.

Many differential equations that physicists encounter are second-order equations (i.e., equations containing second derivatives). Such, for example, is the equation of simple harmonic motion from example (3), md 2 x/dt 2 = –kx. Generally speaking, we can expect that a second-order equation has partial solutions that satisfy two conditions; for example, one might require that the solution curve pass through a given point in a given direction. In cases where the differential equation contains a certain parameter (a number whose value depends on the circumstances), solutions of the required type exist only for certain values of this parameter. For example, consider the equation md 2 x/dt 2 = –kx and we will demand that y(0) = y(1) = 0. Function yє 0 is obviously a solution, but if it is an integer multiple p, i.e. k = m 2 n 2 p 2, where n is an integer, but in reality only in this case, there are other solutions, namely: y= sin npx. The parameter values for which the equation has special solutions are called characteristic or eigenvalues; they play an important role in many tasks.

The equation of simple harmonic motion is an example of an important class of equations, namely linear differential equations with constant coefficients. A more general example (also of second order) is the equation

Where a And b– given constants, f(x) is a given function. Such equations can be solved different ways, for example, using the integral Laplace transform. The same can be said about linear equations of higher orders with constant coefficients. They also play an important role linear equations with variable odds.

Nonlinear differential equations.

Equations containing unknown functions and their derivatives to powers higher than the first or in some more complex manner are called nonlinear. IN last years they are attracting more and more attention. The fact is that physical equations are usually linear only to a first approximation; Further and more accurate research, as a rule, requires the use of nonlinear equations. In addition, many problems are nonlinear in nature. Since solutions to nonlinear equations are often very complex and difficult to represent using simple formulas, a significant part modern theory devoted to qualitative analysis their behavior, i.e. the development of methods that make it possible, without solving the equation, to say something significant about the nature of the solutions as a whole: for example, that they are all limited, or have a periodic nature, or depend in a certain way on the coefficients.

Approximate solutions to differential equations can be found numerically, but this requires a lot of time. With the advent of high-speed computers, this time was greatly reduced, which opened up new possibilities for the numerical solution of many problems that were previously intractable to such a solution.

Existence theorems.

An existence theorem is a theorem that states that, under certain conditions, a given differential equation has a solution. There are differential equations that have no solutions or have more of them than expected. The purpose of an existence theorem is to convince us that a given equation actually has a solution, and most often to assure us that it has exactly one solution of the required type. For example, the equation we have already encountered dy/dx = –2y has exactly one solution passing through each point of the plane ( x,y), and since we have already found one such solution, we have thereby completely solved this equation. On the other hand, the equation ( dy/dx) 2 = 1 – y 2 has many solutions. Among them are straight y = 1, y= –1 and curves y= sin( x + c). The solution may consist of several segments of these straight lines and curves, passing into each other at points of contact (Fig. 2).

Partial differential equations.

An ordinary differential equation is a statement about the derivative of an unknown function of one variable. A partial differential equation contains a function of two or more variables and derivatives of that function with respect to at least two different variables.

In physics, examples of such equations are Laplace's equation

X, y) inside the circle if the values u specified at each point of the bounding circle. Since problems with more than one variable in physics are the rule rather than the exception, it is easy to imagine how vast the subject of the theory of partial differential equations is.