Instruction
If the equation is presented as: dy/dx = q(x)/n(y), refer to the category of 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 parts. 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 as ydy = xdx and integrate. You should get y^2 = x^2 + c.
to linear equations attribute the equations "first". An unknown function with its derivatives is included in 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.
Keep in mind that many differential equations are second-order equations (containing second derivatives). For example, this is the equation of simple harmonic motion, written as a general one: md 2x / dt 2 \u003d -kx. Such equations have, in , partial 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 conditions of the problem, then you are given additional conditions due to which you can find a solution. Read the problem carefully to find these conditions. If a variables x and y are distance, speed, weight - feel free to set the limit x≥0 and y≥0. It is quite possible that x or y is hiding the number of , apples, etc. – then the values can only be . If x is the age of the son, 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
Tasks for differential and integral calculus are important elements of consolidating the theory of mathematical analysis, a section of higher mathematics studied in universities. differential the equation is solved by the integration method.
Instruction
The differential calculus investigates properties. Conversely, the integration of a function allows, according to the given properties, i.e. derivatives or differentials of a function to find it itself. This is the solution of the differential equation.
Any is a ratio between an unknown value and known data. In the case of a differential equation, the role of the unknown is played by the function, and the role of the known quantities is played by its derivatives. In addition, the ratio 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 there are several independent variables in the relation and partial derivatives (differentials) of functions with respect to these variables, then the equation is called a differential equation with partial derivatives and has the form: x∂z/∂y - ∂z/∂x = 0, where z(x, y) is the desired function.
So, in order to learn how to solve differential equations, you need to be able to find antiderivatives, i.e. solve the problem of inverse differentiation. For example: Solve the first order equation y’ = -y/x.
SolutionReplace y' with dy/dx: dy/dx = -y/x.
Bring 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| = - log |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 particular value of C, the solution will be unique. Such a solution is a particular solution of a differential equation.
Solution of most equations of higher degrees does not have a clear formula, like finding the roots of a square equations. However, there are several reduction methods that allow you to transform a higher degree equation to a more visual form.
Instruction
The most common method for solving equations of higher degrees is expansion. This approach is a combination of the selection of integer roots, divisors of the free term, and the 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 member of this polynomial is -3, therefore, its integer divisors can be ±1 and ±3. Substitute them one by one into the equation and find out if you get the identity: 1: 1 + 1 + 2 - 1 - 3 = 0.
Second root x = -1. Divide by the expression (x + 1). Write the resulting equation (x - 1) (x + 1) (x² + x + 3) = 0. The degree has dropped 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 square it. This approach is used when all the 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 enter into the reaction are called initial, and those that are formed as a result of this process are called products. It happens that in the course of a chemical reaction, the elements that make up the starting materials change their oxidation state. That is, they can accept other people's electrons and give their own. In both cases, their charge changes. Such reactions are called redox reactions.
1. The first order differential equation has the form
If this equation can be solved with respect to ta, it can be written as
In this case, we say that the differential equation is solved with respect to the derivative. For such an equation, the following theorem is valid, which is called the theorem on the existence and uniqueness of a solution to a differential equation. Theorem. If in the equation
function and its partial derivative with respect to y are continuous in some domain D on a plane containing some point , then there is a unique solution to this equation
satisfying the condition at
This theorem will be proved in § 27 Ch. XVI.
The geometric meaning of the theorem is that there exists and, moreover, a unique function whose graph passes through the point
It follows from the theorem just stated that the equation has an infinite number of different solutions (for example, a solution whose graph passes through a point, another solution whose graph passes through a point, etc., if only these points lie in the region
The condition that when the function y must be equal to a given number is called the initial condition. It is often written as
Definition 1. A general solution of a first-order differential equation is a function
which depends on one arbitrary constant C and satisfies the following conditions:
a) it satisfies the differential equation for any particular value of the constant C;
b) whatever the initial condition for, you can find such a value that the function satisfies the given initial condition. It is assumed that the values belong to the region of variation of the variables x and y, in which the conditions of the theorem of existence and uniqueness of the solution are satisfied.
2. In the process of searching for a general solution of a differential equation, we often come to a relation of the form
not permitted with respect to Resolving this relation with respect to y, we obtain the general solution. However, it is not always possible to express y from relation (2) in elementary functions; in such cases the general solution is left implicit. An equality of the form implicitly specifying a general solution is called a general integral of a differential equation.
Definition 2. A particular solution is any function that is obtained from a general solution if a certain value is given in the last arbitrary constant C. The ratio is called in this case the partial integral of the equation.
Example 1. For a first order equation
the general solution will be a family of functions; this can be checked by a simple substitution in the equation.
Let us find a particular solution that satisfies the following initial condition: for Substituting these values into the formula, we obtain or Therefore, the required particular solution will be the function
From the geometric point of view, the general integral is a family of curves on the coordinate plane, depending on one arbitrary constant C (or, as they say, on one parameter C).
These curves are called integral curves of the given differential equation. The partial integral corresponds to one curve of this family passing through some given point of the plane.
So, in the last example, the general integral is geometrically represented by a family of hyperbolas, and the partial integral, defined by the indicated initial condition, is represented by one of these hyperbolas passing through the point. 251 shows family curves corresponding to some parameter values: etc.
To make the reasoning more clear, we will henceforth call the solution of an equation not only a function that satisfies the equation, but also the corresponding integral curve. In this connection, we will speak, for example, of a solution passing through the point .
Comment. The equation does not have a solution passing through a point lying on the axis of Fig. 251), since the right side of the equation for is not defined and, therefore, is not continuous.
Solving or, as is often said, integrating a differential equation means:
a) find its general solution or general integral (if the initial conditions are not given), or
b) find that particular solution of the equation that satisfies the given initial conditions (if any).
3. Let us give a geometric interpretation of the first-order differential equation.
Let a differential equation be given that is solved with respect to the derivative:
and let be the general solution of this equation. This general solution defines a family of integral curves in the plane
Equation (D) for each point M with coordinates x and y determines the value of the derivative, i.e., the slope of the tangent to the integral curve passing through this point. Thus, the differential equation (D) gives a set of directions, or, as they say, determines the field of directions on the plane
Therefore, from a geometric point of view, the problem of integrating a differential equation is to find curves whose tangent direction coincides with the direction of the field at the corresponding points.
For the differential equation (1), the locus of points at which the relation holds is called the isocline of the given differential equation.
For different values of k, we obtain different isoclines. The equation of the isocline corresponding to the value of k will obviously be: By constructing a family of isoclines, one can approximately construct a family of integral curves. It is said that, knowing the isoclines, one can qualitatively determine the location of the integral curves on the plane.
In some problems of physics, a direct connection between the quantities describing the process cannot be established. But there is a possibility to obtain an equality containing the derivatives of the functions under study. This is how differential equations arise and the need to solve them in order to find an unknown function.
This article is intended for those who are faced with the problem of solving a differential equation in which the unknown function is a function of one variable. The theory is built in such a way that with a zero understanding of differential equations, you can do your job.
Each type of differential equations is associated with a solution method with detailed explanations and solutions of typical examples and problems. You just have to determine the type of differential equation for your problem, find a similar analyzed example and carry out similar actions.
To successfully solve differential equations, you will also need the ability to find sets of antiderivatives (indefinite integrals) of various functions. If necessary, we recommend that you refer to the section.
First, we consider the types of ordinary differential equations of the first order that can be solved with respect to the derivative, then we move on to second-order ODEs, then we dwell on higher-order equations and finish with systems of differential equations.
Recall that if y is a function of the argument x .
First order differential equations.
The simplest differential equations of the first order of the form .
Let us write down several examples of such DE 
.
Differential Equations 
can be resolved with respect to the derivative by dividing both sides of the equality by f(x) . In this case, we arrive at the equation , which will be equivalent to the original one for f(x) ≠ 0 . Examples of such ODEs are .
If there are values of the argument x for which the functions f(x) and g(x) simultaneously vanish, then additional solutions appear. Additional solutions to the equation 
given x are any functions defined for those argument values. Examples of such differential equations are .
Second order differential equations.
Second Order Linear Homogeneous Differential Equations with Constant Coefficients.
LODE with constant coefficients is a very common type of differential equations. Their solution is not particularly difficult. First, the roots of the characteristic equation are found 
. For different p and q, three cases are possible: the roots of the characteristic equation can be real and different, real and coinciding 
or complex conjugate. Depending on the values of the roots of the characteristic equation, the general solution of the differential equation is written as 
, or 
, or respectively.
For example, consider a second-order linear homogeneous differential equation with constant coefficients. The roots of his characteristic equation are k 1 = -3 and k 2 = 0. The roots are real and different, therefore, the general solution to the LDE with constant coefficients is
Linear Nonhomogeneous Second Order Differential Equations with Constant Coefficients.
The general solution of the second-order LIDE with constant coefficients y is sought as the sum of the general solution of the corresponding LODE 
and a particular solution of the original inhomogeneous equation, that is, . The previous paragraph is devoted to finding a general solution to a homogeneous differential equation with constant coefficients. And a particular solution is determined either by the method of indefinite coefficients for a certain form of the function f (x) , standing on the right side of the original equation, or by the method of variation of arbitrary constants.
As examples of second-order LIDEs with constant coefficients, we present 
To understand the theory and get acquainted with the detailed solutions of examples, we offer you on the page linear inhomogeneous differential equations of the second order with constant coefficients.
Linear Homogeneous Differential Equations (LODEs) 
and second-order linear inhomogeneous differential equations (LNDEs).
A special case of differential equations of this type are LODE and LODE with constant coefficients.
The general solution of the LODE on a certain interval is represented by a linear combination of two linearly independent particular solutions y 1 and y 2 of this equation, that is, 
.
The main difficulty lies precisely in finding linearly independent partial solutions of this type of differential equation. Usually, particular solutions are chosen from the following systems of linearly independent functions: 
However, particular solutions are not always presented in this form.
An example of a LODU is 
.
The general solution of the LIDE is sought in the form , where is the general solution of the corresponding LODE, and is a particular solution of the original differential equation. We just talked about finding, but it can be determined using the method of variation of arbitrary constants.
An example of an LNDE is 
.
Higher order differential equations.
Differential equations admitting order reduction.
Order of differential equation 
, which does not contain the desired function and its derivatives up to k-1 order, can be reduced to n-k by replacing .
In this case , and the original differential equation reduces to . After finding its solution p(x), it remains to return to the replacement and determine the unknown function y .
For example, the differential equation 
after the replacement becomes a separable equation , and its order is reduced from the third to the first.
A first order equation of the form a 1 (x) y "+ a 0 (x) y \u003d b (x) is called a linear differential equation. If b (x) ≡ 0 then the equation is called homogeneous, otherwise - heterogeneous. For a linear differential equation, the existence and uniqueness theorem has a more concrete form.
Service assignment. An online calculator can be used to check the solution homogeneous and non-homogeneous linear differential equations like y"+y=b(x) .
To obtain a solution, the original expression must be reduced to the form: a 1 (x)y" + a 0 (x)y = b(x) . For example, for y"-exp(x)=2*y it will be y"-2 *y=exp(x) .Theorem. Let a 1 (x) , a 0 (x) , b(x) be continuous on the interval [α,β], a 1 ≠0 for ∀x∈[α,β]. Then for any point (x 0 , y 0), x 0 ∈[α,β], there is a unique solution to the equation that satisfies the condition y(x 0) = y 0 and is defined on the entire interval [α,β].
Consider a homogeneous linear differential equation a 1 (x)y"+a 0 (x)y=0 .
Separating the variables, we get , or, integrating both parts, 
The last relation, taking into account the notation exp(x) = e x , is written in the form 
Let us now try to find a solution to the equation in the indicated form, in which the function C(x) is substituted instead of the constant C, that is, in the form 
Substituting this solution into the original solution, after the necessary transformations, we obtain 
Integrating the latter, we have 
where C 1 is some new constant. Substituting the resulting expression for C(x), we finally obtain the solution of the original linear equation 
.
Example. Solve the equation y" + 2y = 4x. Consider the corresponding homogeneous equation y" + 2y = 0. Solving it, we get y = Ce -2 x. We are now looking for a solution to the original equation in the form y = C(x)e -2 x . Substituting y and y" = C"(x)e -2 x - 2C(x)e -2 x into the original equation, we have C"(x) = 4xe 2 x, whence C(x) = 2xe 2 x - e 2 x + C 1 and y(x) = (2xe 2 x - e 2 x + C 1)e -2 x = 2x - 1 + C 1 e -2 x is the general solution to the original equation.In this solution, y 1 ( x) = 2x-1 - movement of the object under the action of force b(x) = 4x, y 2 (x) = C 1 e -2 x - proper movement of the object.
Example #2. Find the general solution of the first order differential equation y"+3 y tan(3x)=2 cos(3x)/sin 2 2x.
This is an inhomogeneous equation. Let's make a change of variables: y=u v, y" = u"v + uv".
3u v tg(3x)+u v"+u" v = 2cos(3x)/sin 2 2x or u(3v tg(3x)+v") + u" v= 2cos(3x)/sin 2 2x
The solution consists of two steps:
1. u(3vtg(3x)+v") = 0
2. u "v \u003d 2cos (3x) / sin 2 2x
1. Equate u=0, find solution for 3v tg(3x)+v" = 0
Represent in the form: v" = -3v tg(3x)
Integrating, we get:
ln(v) = ln(cos(3x))
v = cos(3x)
2. Knowing v, Find u from the condition: u "v \u003d 2cos (3x) / sin 2 2x
u" cos(3x) = 2cos(3x)/sin 2 2x
u" = 2/sin 2 2x
Integrating, we get:
From the condition y=u v, we get:
y = u v = (C-cos(2x)/sin(2x)) cos(3x) or y = C cos(3x)-cos(2x) ctg(3x)
First order differential equations solved with respect to the derivative
How to solve first order differential equations
Let we have a first-order differential equation solved with respect to the derivative:
.
Dividing this equation by , at , we get an equation of the form:
,
where .
Next, we look to see if these equations belong to one of the types listed below. If not, then we rewrite the equation in the form of differentials. To do this, we write and multiply the equation by . We get the equation in the form of differentials:
.
If this equation is not an equation in total differentials, then we consider that in this equation is an independent variable, and is a function of . Let's divide the equation by:
.
Next, we look to see if this equation belongs to one of the types listed below, considering that and have been swapped.
If a type is not found for this equation, then we look to see if it is possible to simplify the equation by a simple substitution. For example, if the equation is:
,
then we notice that . Then we make a substitution. After that, the equation will take a simpler form:
.
If this does not help, then we try to find an integrating factor.
Separable Variable Equations
;
.
Divide by and integrate. When we get:
.
Equations that reduce to equations with separable variables
Homogeneous equations
We solve by substitution:
,
where is a function of . Then
;
.
Separate variables and integrate.
Equations Reducing to Homogeneous
We introduce variables and :
;
.
The constants and are chosen so that the free terms vanish:
;
.
As a result, we obtain a homogeneous equation in variables and .
Generalized homogeneous equations
We make a substitution. We obtain a homogeneous equation in variables and .
Linear differential equations
There are three methods for solving linear equations.
2) Bernoulli method.
We are looking for a solution in the form of a product of two functions and from a variable:
.
;
.
We can choose one of these functions arbitrarily. Therefore, as we choose any non-zero solution of the equation:
.
3) The method of variation of the constant (Lagrange).
Here we first solve the homogeneous equation:
The general solution of the homogeneous equation has the form:
,
where is a constant. Next, we replace the constant with a function depending on the variable :
.
Substitute in the original equation. As a result, we obtain an equation from which we determine .
Bernoulli's equations
By substitution, the Bernoulli equation is reduced to a linear equation.
This equation can also be solved by the Bernoulli method. That is, we are looking for a solution in the form of a product of two functions depending on the variable :
.
We substitute into the original equation:
;
.
As we choose any non-zero solution of the equation:
.
Having determined , we obtain an equation with separable variables for .
Riccati equations
It is not resolved in a general way. Substitution
the Riccati equation is reduced to the form:
,
where is a constant; ; .
Next, substitution:
it looks like:
,
where .
Properties of the Riccati equation and some special cases of its solution are presented on the page
Riccati differential equation >>>
Jacobi equations
Solved by substitution:
.
Equations in Total Differentials
Given that
.
When this condition is met, the expression on the left side of the equality is the differential of some function:
.
Then
.
From here we obtain the integral of the differential equation:
.
To find the function , the most convenient way is the method of successive selection of the differential. For this, formulas are used:
;
;
;
.
Integrating factor
If the first-order differential equation is not reduced to any of the listed types, then you can try to find an integrating factor. An integrating factor is such a function, when multiplied by it, the differential equation becomes an equation in total differentials. A first order differential equation has an infinite number of integrating factors. However, there are no general methods for finding the integrating factor.
Equations not solved for the derivative y"
Equations admitting a solution with respect to the derivative y"
First you need to try to solve the equation with respect to the derivative. If possible, then the equation can be reduced to one of the types listed above.
Equations Allowing Factorization
If you can factorize the equation:
,
then the problem is reduced to the sequential solution of simpler equations:
;
;
;
. We believe . Then
or .
Next, we integrate the equation:
;
.
As a result, we obtain the expression of the second variable through the parameter.
More general equations:
or
are also solved in a parametric form. To do this, you need to choose a function so that from the original equation you can express or through the parameter .
To express the second variable in terms of the parameter , we integrate the equation:
;
.
Equations resolved with respect to y
Clairaut's equations
This equation has a general solution
Lagrange equations
We are looking for a solution in a parametric form. We assume , where is a parameter.
Equations leading to the Bernoulli equation
These equations are reduced to the Bernoulli equation if we look for their solutions in a parametric form by introducing a parameter and making a substitution .
References:
V.V. Stepanov, Course of Differential Equations, LKI, 2015.
N.M. Gunther, R.O. Kuzmin, Collection of problems in higher mathematics, Lan, 2003.
