Definition. A differential equation of the form
𝐹(𝑥, 𝑦, 𝑦ˊ, 𝑦ˊˊ) = 0
(1)
is called the second order differential equation.
In some cases a second-order differential equation (1) can be solved by reducing it to a firstorder equation, using an appropriate variable substitution. In the following examples we consider such
particular types of equations.
Example 1. Solve the equation 𝑥 2 𝑦ˊˊ = 1.
Solution. When the equation (1) does not contain 𝑦 and 𝑦ˊ, and can be written as 𝑦ˊˊ = 𝑓(𝑥), we
use the substitution 𝑦ˊ = 𝑧. This implies
𝑦ˊˊ = 𝑧ˊ,
and the original equation turns into an equation of the first order:
𝑧ˊ = 𝑓(𝑥).
1
The given equation, 𝑥 2 𝑦ˊˊ = 1, can be rewritten as 𝑦ˊˊ = 𝑥 2 . By the substitution 𝑦ˊ = 𝑧 we arrive
at the equation of the first order
𝑧ˊ =
1
.
𝑥2
Integrating implies
1
𝑧 = − 𝑥 + 𝐶1.
Thus, we arrive at the equation
1
𝑥
𝑦ˊ = − + 𝐶1,
which has the general solution 𝑦 = − ln 𝑥 + 𝐶1 𝑥 + 𝐶2 .
Example 2. Solve the equation
𝑦ˊˊ +
𝑦ˊ
𝑥
= 𝑥.
Solution. When the equation (1) does not contain 𝑦, the substitution 𝑦ˊ = 𝑧 can be used to
reduce the given second-order differential equation to one of the first-order. Thus, the given equation,
𝑦ˊ
𝑦ˊˊ + 𝑥 = 𝑥, becomes
𝑧
𝑧ˊ + 𝑥 = 𝑥.
This is a linear equation of the first order. Put 𝑧(𝑥) = 𝑢(𝑥)𝑣(𝑥), where 𝑢(𝑥) and 𝑣(𝑥) are to be
determined. On substituting into the equation, we obtain
𝑢ˊ𝑣 + 𝑢𝑣ˊ +
𝑢𝑣
𝑥
𝑣
= 𝑥 or 𝑢ˊ𝑣 + 𝑢(𝑣ˊ + 𝑥) = 𝑥.
So, the functions 𝑢(𝑥) and 𝑣(𝑥) can be found from the following equations:
𝑣
1. 𝑣ˊ + 𝑥 =0, and 2. 𝑢𝑣ˊ = 𝑥.
1
𝑥3
3
The first equation provides 𝑣(𝑥) = 𝑥 and, hence, the second equation has the solution 𝑢(𝑥) =
𝑥3
1
+ 𝐶1. Thus, 𝑧 = 𝑢(𝑥)𝑣(𝑥) = ( 3 + 𝐶1 ) 𝑥 =
𝑦ˊ =
𝑥2
3
+
𝑥2
3
+
𝐶1
.
𝑥
Since 𝑧 = 𝑦ˊ, the next equation to be solved is
𝐶1
.
𝑥
On integrating both sides of the equation we get the general solution to the original equation:
𝑥3
9
𝑦=
+ 𝐶1 ln 𝑥 + 𝐶2 .
Example 3. Solve the equation
2𝑦𝑦ˊˊ + 𝑦ˊ2 = 0 for 𝑦 > 0.
Solution. When the equation (1) does not contain 𝑥, it can be reduced to a first-order equation
using the substitution 𝑦ˊ = 𝑝, where 𝑝 is considered a function of 𝑦. The latter implies:
𝑑𝑝
𝑑𝑝 𝑑𝑦
𝑑𝑝
𝑦ˊˊ = 𝑑𝑥 = 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑝.
Using these, the given equation turns to
𝑑𝑝
𝑑𝑝
2𝑦𝑝 𝑑𝑦 + 𝑝2 = 0 or 𝑝 (2𝑦 𝑑𝑦 + 𝑝) = 0.
From 𝑝 = 0, that implies 𝑦ˊ = 0, we obtain the solution 𝑦 = 𝑐𝑜𝑛𝑠𝑡. Alternatively, we have the
equation
𝑑𝑝
2𝑦 𝑑𝑦 + 𝑝 = 0,
which, after separating variables and integrating, provides the solution
𝑝=
𝐶1
.
√𝑦
Thus, the next equation to be solved is
𝑦ˊ =
𝐶1
.
√𝑦
Its general solution can be easily determined:
3
𝑦 = √(𝐶1 𝑥 + 𝐶2 )2, where 𝐶1 and 𝐶2 are arbitrary constants.
3
Note that 𝑦 = 𝑐𝑜𝑛𝑠𝑡 is included in the latter expression. Thus, 𝑦 = √(𝐶1 𝑥 + 𝐶2 )2 is the general
solution to the original equation.
Consider a nonhomogeneous linear equation
𝑦ˊˊ + 𝑎1 𝑦ˊ + 𝑎2 𝑦 = 𝑓(𝑥).
(D)
To find the general solution to the equation (D), we have, according to Theorem 3, to determine
the general solution to the corresponding homogeneous equation (C):
𝑦ˊˊ + 𝑎1 𝑦ˊ + 𝑎2 𝑦 = 0,
(C)
and one particular solution to equation (D). Sometimes the particular solution can be determined from
the right-hand side of the equation. In the following examples we consider some of such cases.
Example 4. Solve the equation
𝑦ˊˊ − 2𝑦ˊ + 𝑦 = 1 + 𝑥.
Solution. When the right-hand side of the linear equation (D) has the form 𝑓(𝑥) = 𝑃(𝑥)𝑒 𝑡𝑥 ,
where 𝑃(𝑥) is a polynomial and 𝑡 is a real number, then the given equation possesses a particular
solution 𝑦𝑝 = 𝑥 𝑘 𝑄(𝑥)𝑒 𝑡𝑥 , where 𝑄(𝑥) is a polynomial of the same degree as 𝑃(𝑥), and 𝑘 takes the
following values:
𝑘={
0 𝑖𝑓 𝑡 𝑖𝑠 𝑛𝑜𝑡 𝑎 𝑟𝑜𝑜𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
.
𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡 (1 𝑜𝑟2) 𝑖𝑓 𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑟𝑜𝑜𝑡
To start, we consider the homogeneous equation
𝑦ˊˊ − 2𝑦ˊ + 𝑦 = 0.
Its characteristic equation, 𝑟 2 − 2𝑟 + 1 = 0, has a double root 𝑟 = 1. Hence, the general
solution to the homogeneous equation is
𝑦𝑔 = 𝑒 𝑥 (𝐶1 + 𝐶2 𝑥),
where 𝐶1 and 𝐶2 are arbitrary constants. To find a particular solution to the original equation, we
examine the right-hand side of the equation, the function 𝑓(𝑥) = 1 + 𝑥. This function has the form
𝑃(𝑥)𝑒 𝑡𝑥 , where 𝑃(𝑥) = 1 + 𝑥 (a polynomial of the first degree) and 𝑡 = 0. Hence, the particular
solution to the original equation will have the form
𝑦𝑝 = 𝑥 𝑘 𝑄(𝑥)𝑒 𝑡𝑥 ,
with 𝑘 = 0, since 0 is not a characteristic root, and 𝑄(𝑥) = 𝐴𝑥 + 𝐵. Thus,
𝑦𝑝 = 𝐴𝑥 + 𝐵.
where 𝐴 and 𝐵 are constants to be determined.
Substituting 𝑦𝑝 , 𝑦𝑝 ˊ, and 𝑦𝑝 ˊˊ into initial equation yields 𝐴 = 1 and 𝐵 = 3. So, the particular
solution is 𝑦𝑝 = 𝑥 + 3, and, the general solution to the given equation is
𝑦𝐺 (𝑥) = 𝑦𝑔 (𝑥) + 𝑦𝑝 (𝑥) = 𝑒 𝑥 (𝐶1 + 𝐶2 𝑥) + (𝑥 + 3).
Example 5. Solve the equation
𝑦ˊˊ + 4𝑦ˊ + 13𝑦 = 5 sin 2𝑥.
Solution. When the right-hand side of the equation (D) is the function
𝑓(𝑥) = 𝑎 cos 𝑞𝑥 + 𝑏 sin 𝑞𝑥,
then the particular solution to equation (D) has the form
𝐴 cos 𝑞𝑥 + 𝐵 sin 𝑞𝑥 𝑖𝑓 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 ± 𝑞𝑖 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑟𝑜𝑜𝑡𝑠
𝑦𝑝 (𝑥) = {
.
𝑥(𝐴 cos 𝑞𝑥 + 𝐵 sin 𝑞𝑥) 𝑖𝑓 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 ± 𝑞𝑖 𝑎𝑟𝑒 𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑟𝑜𝑜𝑡𝑠
The characteristic equation,
𝑟 2 + 4𝑟 + 13 = 0,
has complex conjugate roots 𝑟1 = −2 + 3𝑖 and 𝑟2 = −2 − 3𝑖. Hence, the general solution to the
corresponding homogeneous equation is
𝑦𝑔 = 𝑒 −2𝑥 (𝐶1 cos 3𝑥 + 𝐶2 sin 3𝑥).
Since the numbers ±2𝑖 are not the characteristic roots, the particular solution to the original
equation will be written as
𝑦𝑝 = 𝐴 cos 2𝑥 + 𝐵 sin 2𝑥,
where 𝐴 and 𝐵 are constants to be determined.
8
9
Substituting 𝑦𝑝 , 𝑦𝑝 ˊ, and 𝑦𝑝 ˊˊ into initial equation gives 𝐴 = − 29 and 𝐵 = 29. Therefore, the
particular solution is
𝑦𝑝 = −
8
29
cos 2𝑥 +
9
29
sin 2𝑥,
and the general solution to the initial equation is
8
9
𝑦𝐺 = 𝑒 −2𝑥 (𝐶1 cos 3𝑥 + 𝐶2 sin 3𝑥) − 29 cos 2𝑥 + 29 sin 2𝑥.
Example 6. Solve the equation
𝑦ˊˊ + 𝑦 = 4𝑥 sin 𝑥.
Solution. When the right-hand side of the equation (D) has the form
𝑓(𝑥) = 𝑒 𝑝𝑥 (𝑃1 (𝑥) cos 𝑞𝑥 + 𝑃2 (𝑥) sin 𝑞𝑥),
where 𝑃1 (𝑥) and 𝑃2 (𝑥) are polynomials, then the particular solution is
𝑦𝑝 =
𝑒 𝑝𝑥 (𝑄1 (𝑥) cos 𝑞𝑥 + 𝑄2 (𝑥) sin 𝑞𝑥) 𝑖𝑓 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑝 ± 𝑞𝑖 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑟𝑜𝑜𝑡𝑠
.
{ 𝑝𝑥
𝑥[𝑒 (𝑄1 (𝑥) cos 𝑞𝑥 + 𝑄2 (𝑥) sin 𝑞𝑥)] 𝑖𝑓 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑝 ± 𝑞𝑖 𝑎𝑟𝑒 𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑟𝑜𝑜𝑡𝑠
The characteristic equation,
𝑟2 + 1 = 0
has complex conjugate roots ±𝑖. Therefore, the general solution to the homogeneous equation is
𝑦𝑔 = 𝐶1 cos 𝑥 + 𝐶2 sin 𝑥,
while the particular solution will be of the form
𝑦𝑝 = 𝑥[((𝐴𝑥 + 𝐵) cos 𝑥 + (𝐶𝑥 + 𝐷) sin 𝑥)],
where 𝐴, 𝐵, 𝐶, and 𝐷 are constants to be determined.
Substituting 𝑦𝑝 , 𝑦𝑝 ˊ, and 𝑦𝑝 ˊˊ into initial equation gives 𝐴 = −1 and 𝐵 = 0, 𝐶 = 0, and 𝐷 = 1.
Thus, the particular solution is 𝑦𝑝 = 𝑥(−𝑥 cos 𝑥 + sin 𝑥), and the general solution to the given equation
is
𝑦𝐺 = 𝐶1 cos 𝑥 + 𝐶2 sin 𝑥 + 𝑥(−𝑥 cos 𝑥 + sin 𝑥).