Higher-order equations that allow lowering the order. Methods for lowering the order of an equation

Therefore, there is a natural desire to reduce an equation of order higher than the first to an equation of a lower order. In some cases, this can be done. Let's consider them.

1. Equations of the form y (n) =f(x) are solved by successive integration n times
, ,… .
Example. Solve the equation xy""=1 . We can write, therefore, y"=ln|x| + C 1 and, integrating again, we finally get y=∫ln|x| + C 1 x + C 2

2. In equations of the form F(x,y (k) ,y (k +1) ,..,y (n))=0 (i.e., not explicitly containing an unknown function and some of its derivatives), the order is reduced using change of variable y (k) = z(x). Then y (k +1) =z"(x),…,y (n) = z (n - k) (x) and we get the equation F(x,z,z",..,z (n - k)) of order n-k. Its solution is the function z = φ(x,C 1 ,C 2 ,…,C n) or, remembering what z is, we get the equation y (n- k) = φ(x,C 1 ,C 2 ,…, C n - k) of the type considered in case 1.
Example 1 . Solve the equation x 2 y "" = (y") 2. We make the replacement y "=z (x) . Then y""=z"(x) . Substituting into the original equation, we get x 2 z"=z 2. Separating the variables, we get . Integrating, we have , or, which is the same, . The last relation is written as , whence . Integrating, we finally get
Example 2 . Solve the equation x 3 y"" +x 2 y"=1. We make a change of variables: y"=z; y""=z"
x 3 z "+x 2 z=1. We make a change of variables: z=u/x; z"=(u"x-u)/x 2
x 3 (u "x-u) / x 2 + x 2 u / x = 1 or u" x 2 -xu + xu = 1 or u "x ^ 2 = 1. From: u" = 1 / x 2 or du / dx=1/x 2 or u = int(dx/x 2) = -1/x+c 1
Since z=u/x, then z = -1/x 2 +c 1 /x. Since y "=z, then dy / dx \u003d -1 / x 2 + c 1 / x
y = int(c 1 dx/x-dx/x 2) =c 1 ln(x) + 1/x + c 2 . Answer: y = c 1 ln(x) + 1/x + c 2

3. The next equation that can be reduced in order is an equation of the form F(y,y",y"",…,y (n))=0 , which does not explicitly contain an independent variable. The order of the equation is reduced by changing the variable y" =p(y) , where p is the new desired function depending on y. Then
= and so on. By induction, we have y (n) =φ(p,p",..,p (n-1)). Substituting into the original equation, we lower its order by one.

Example. Solve the equation (y") 2 +2yy""=0. We make the standard replacement y"=p(y) , then y″=p′·p . Substituting into the equation, we get Separating the variables, at p≠0, we have Integrating, we get or, which is the same, . Then or . Integrating the last equality, we finally obtain When separating variables, we could lose the solution y=C, which is obtained at p=0, or, what is the same, at y"=0, but it is contained in the one obtained above.

4. Sometimes it is possible to notice a feature that makes it possible to lower the order of the equation in ways different from those considered above. Let's show this with examples.

Examples.
1. If both parts of the equation yy"""=y′y″ are divided by yy″, then we get the equation , which can be rewritten as (lny″)′=(lny)′. From the last relation it follows that lny″=lny +lnC , or, what is the same, y″=Cy... The result is an equation an order of magnitude lower and of the type considered earlier.
2. Similarly, for the equation yy″=y′(y′+1) we have , or (ln(y"+1))" = (lny)" . It follows from the last relation that ln(y"+1) = lny + lnC 1 , or y"=C 1 y-1. Separating the variables and integrating, we get, ln(C 1 y-1) = C 1 x+C 2
Decide lower-order equations possible with the help of a special service

one of the methods for integrating higher order differential equations is the order reduction method. The essence of the method is that with the help of a change of variable (substitution), this DE is reduced to an equation, the order of which is lower.

Let us consider three types of equations that can be reduced in order.

I. Let the equation

The order can be lowered by introducing a new function p(x), setting y"=p(x). Then y""=p"(x) and we get the first-order DE: p"=ƒ(x). Solving it, i.e. i.e. having found the function p \u003d p (x), we solve the equation y "\u003d p (x). Let us obtain the general solution of the given equation (3.6).

In practice, they act differently: the order is lowered directly by successive integration of the equation.

As equation (3.6) can be written as dy "=ƒ(х) dx. Then, integrating the equation y""=ƒ(х), we obtain: y"=or y"=j1(x)+с 1. Further, integrating the resulting equation for x, we find: - the general solution of this equation.If the equation is given then, integrating it successively n times, we find the general solution of the equation:

Example 3.1. solve the equation

Solution: By successively integrating this equation four times, we obtain

Let the equation

Let us denote y "=p, where p=p(x) is a new unknown function. Then y" "=p" and equation (3.7) takes the form p "=ƒ(x; p). Let p=j(x; c 1) - the general solution of the obtained first-order DE. Replacing the function p by y ", we obtain the DE: y" \u003d j (x; c 1). It has the form (3.6). To find y, it is enough to integrate the last equation. The general solution of the equation ( 3.7) will look like

A special case of equation (3.7) is the equation

which also does not contain the explicitly desired function, then its order can be reduced by k units by setting y (k) = p(x). Then y (k + 1) = p "; ...; y (n) = p (n-k) and equation (3.9) takes the form F(x; p; p ";...; p (n-κ) )=0. A special case of equation (3.9) is the equation

By replacing y (n-1) =p(x), y (n) =p ", this equation is reduced to a first-order DE.

Example 3.2. solve the equation

Solution: We put y "=p, where Then This is a separable equation: Integrating, we get Returning to the original variable, we get y "=c 1 x,

is the general solution of the equation.

III. Consider the equation

which does not contain an explicitly independent variable x.

To lower the order of the equation, we introduce a new function p = p (y), depending on the variable y, setting y "= p. We differentiate this equality with respect to x, taking into account that p \u003d p (y (x)):

i.e. Now equation (3.10) can be written in the form

Let р=j(y;с 1) be the general solution of this first-order DE. Replacing the function p(y) by y", we obtain y"=j(y;c 1) - a DE with separable variables. Integrating it, we find the general integral of equation (3.10):

A special case of equation (3.10) is the DE

Such an equation is solved using a similar substitution: y " \u003d p (y),

We do the same when solving the equation F (y; y "; y" "; ...; y (n)) \u003d 0. Its order can be reduced by one by setting y "= p, where p \u003d p (y). By the rule of differentiation of a complex function, we find Then we find

p \u003d uv \u003d ((-1 + y) e -y + e -y + c 1) e + y, or p \u003d c 1 ey + y. Replacing p with y ", we get: y" \u003d c 1 -e y + y. Substituting y"=2 and y=2 into this equality, we find with 1:

2=c 1 e 2 +2, c 1 =0.

We have y"=y. Hence y=c 2 e x. We find c 2 from the initial conditions: 2=c 2 e°, c 2 =2. Thus, y=2e x is a particular solution of the given

The 2nd order differential equation has the form:

The general solution of the equation is a family of functions depending on two arbitrary constants and: (or - the general integral of a 2nd order differential equation). The Cauchy problem for the 2nd order differential equation (1.1) is to find a particular solution of the equation that satisfies the initial conditions: for: , . It should be noted that the graphs of solutions to the 2nd order equation can intersect, in contrast to the graphs of solutions to the 1st order equation. However, the solution of the Cauchy problem for the second-order equations (1.1) under rather broad assumptions for the functions entering the equation is unique, i.e. any two solutions with a common initial condition coincide at the intersection of the definition intervals.

Obtaining a general solution or solving the Cauchy problem for a 2nd order differential equation is not always possible analytically. However, in some cases it is possible to lower the order of the equation by introducing various substitutions. Let's analyze these cases.

1. Equations that do not contain an explicitly independent variable.

Let the 2nd order differential equation have the form: , i.e. Equation (1.1) does not explicitly contain the independent variable. This allows us to take it as a new argument, and take the 1st order derivative as a new function. Then.

Thus, a 2nd order equation for a function that does not explicitly contain has been reduced to a 1st order equation for a function. Integrating this equation, we obtain a general integral or, and this is a 1st order differential equation for the function. Solving it, we obtain the general integral of the original differential equation, which depends on two arbitrary constants: .

Example 1. Solve a differential equation for given initial conditions: , .

Since there is no explicit argument in the original equation, we will take for a new independent variable, and - for. Then the equation takes the following form for the function: .

This is a differential equation with separable variables: . Whence follows, i.e. .

Since for and, then substituting the initial conditions into the last equality, we obtain that and, which is equivalent. As a result, for the function we have an equation with separable variables, solving which we obtain. Using the initial conditions, we get that. Therefore, the partial integral of the equation that satisfies the initial conditions has the form: .

2. Equations that do not contain the explicitly desired function.

Let the 2nd order differential equation have the form: , i.e. the desired function is clearly not included in the equation. In this case, the setting is entered. Then the 2nd order equation for the function goes into the 1st order equation for the function. After integrating it, we obtain a differential equation of the 1st order for the function: . Solving the last equation, we obtain the general integral of the given differential equation, which depends on two arbitrary constants: .