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 a 2nd order differential equation (1.1) consists of finding a particular solution to the equation that satisfies the initial conditions: for: , . It should be noted that the graphs of solutions to a 2nd order equation can intersect, unlike the graphs of solutions to a 1st order equation. However, the solution to the Cauchy problem for second-order equations (1.1) under fairly broad assumptions for the functions included in the equation is unique, i.e. any two solutions with a common initial condition coincide at the intersection of the definition intervals.
It is not always possible to obtain a general solution or solve the Cauchy problem for a 2nd order differential equation analytically. However, in some cases it is possible to lower the order of the equation by introducing various substitutions. Let's look at these cases.
1. Equations that do not explicitly contain an independent variable.
Let the 2nd order differential equation have the form: , i.e. there is clearly no independent variable in equation (1.1). 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 is not explicitly contained has been reduced to a 1st order equation for a function. Integrating this equation, we obtain the 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, depending 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 a as a new independent variable, and - as. Then the equation takes the following form for the function: .
This is a differential equation with separable variables: . Where does it follow, 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 obtain that. Consequently, the partial integral of the equation that satisfies the initial conditions has the form: .
2. Equations that do not explicitly contain the desired function.
Let the 2nd order differential equation have the form: , i.e. the equation clearly does not include the desired function. In this case, a statement is introduced. Then the 2nd order equation for the function turns into a 1st order equation for the function. Having integrated it, we obtain a 1st order differential equation for the function: . Solving the last equation, we obtain the general integral of the given differential equation, depending on two arbitrary constants: .
Therefore, there is a natural desire to reduce an equation of order higher than the first to an equation of lower order. In some cases this can be done. Let's look at them.
1. Equations of the form y (n) =f(x) are solved by sequential 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 (that is, not explicitly containing an unknown function and some of its derivatives), the order is reduced using changing the 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 obtain the equation y (n-k) = φ(x,C 1 ,C 2 ,…, C n - k) considered in the case of type 1.
Example 1. Solve the equation x 2 y"" = (y") 2. 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 in the form , from where . 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=-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 replacing 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, for p≠0, we have. Integrating, we get 
or, which is the same thing, . Then or. Integrating the last equality, we finally obtain 
When separating variables, we could lose the solution y=C, which is obtained for p=0, or, what is the same, for y"=0, but it is contained in the one obtained above.
4. Sometimes it is possible to notice a feature that allows you to lower the order of the equation in ways different from those discussed above. Let's show this with examples.
Examples.
1. If both sides of the equation yy"""=y′y″ are divided by yy″, we obtain an equation that 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 discussed earlier.
2. Similarly, for the equation yy″=y′(y′+1) we have, or (ln(y"+1))" = (lny)". From the last relation it follows 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 equations that can be reduced in order possible using a special service
One of the methods for integrating higher order DEs is the order reduction method. The essence of the method is that, by replacing a variable (substitution), this DE is reduced to an equation of lower order.
Let us consider three types of equations that allow a reduction in order.
I. Let the equation be given
The order can be lowered by introducing a new function p(x), setting y " =p(x). Then y "" =p " (x) and we obtain a first order DE: p " =ƒ(x). Having solved it, i.e. That is, having found the function p = p (x), we solve the equation y " = p (x). Let us obtain a general solution to the given equation (3.6).
In practice, they act differently: the order is reduced directly by sequential integration of the equation.
Because 
equation (3.6) can be written in the form dy " =ƒ(x) dx. Then, integrating the equation y "" =ƒ(x), we obtain: y " = or y " =j1 (x) + с 1. Further, integrating the resulting equation in x, we find: - the general solution of this equation. If the equation is given 
then, having integrated it successively n times, we find the general solution of the equation:
Example 3.1. Solve the equation 
Solution: Consistently integrating this equation four times, we obtain
Let the equation be given
Let us denote y " =р, where р=р(х) is a new unknown function. Then y "" =p " and equation (3.7) takes the form p " =ƒ(х;р). Let р=j(х;с 1) is the general solution of the resulting first-order DE. Replacing the function p with y ", we obtain the DE: y " = 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 have the form
A special case of equation (3.7) is the equation
which also does not explicitly contain the desired function, then its order can be lowered 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
Using the replacement 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 assume y"=p, where 
Then 
This is a separable equation: 
Integrating, we get Returning to the original variable, we get y"=c 1 x,
- general solution of the equation.
III. Consider the equation
which does not explicitly contain the independent variable x.
To reduce 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 =p(y(x)):
i.e. 
Now equation (3.10) will be written in the form 
Let p=j(y;c 1) be the general solution of this first order DE. Replacing the function p(y) with y", we obtain y"=j(y;c 1) - DE with separable variables. Integrating it, we find the general integral of equation (3.10): 
A special case of equation (3.10) is the differential equation
This equation can be solved using a similar substitution: y " =p(y),
We do the same when solving the equation F(y; y " ; y";...; y (n)) = 0. Its order can be lowered by one by setting y"=p, where p=p(y). Using the rule of differentiation of a complex function, we find Then we find
p=uv=((-1+y)e -y +e -y +c 1) e+y, or p=c 1 ey+y. Replacing p with y ", we get: y"=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 this