Analytic Methods for Solving Higher Order Ordinary Differential Equations

Abstract

In this work, an analytic approach for solving higher order ordinary differential equations (ODEs) is developed. The techniques offer analytic flexibility in many research areas such as physics, engineering, and applied sciences and are effective for solving complex ODEs.

1. Introduction

The solutions of differential equations (DEs) are of much interest to engineers, physicists, mathematicians and researchers since many physical phenomena are modeled by using differential equations. For example, in physics, Legendre DE [1], which is a self-adjoint ODE, arises in the solutions of Hydrogen atom wave functions and angular momentum in single-particle quantum mechanics. Their solutions form the polar angle part of the spherical harmonics basis for the multi pole expansion, which is used in both electromagnetic and gravitational statics. In engineering, for example, many difficult problems in the field of static and dynamic mechanics can be solved by computing the solutions self-adjoint Bessel equations (see [2]).

In general, solving higher order DEs is complex and numerical methods are usually needed to solve these equations with initial or boundary conditions. For instance, Khawaja et al. (2018) [3] used iterative power series of $sech(x)$ to solve nonlinear ordinary differential equations (ODEs). Vitoriano (2016) [4] also used the finite element method to solve partial differential equations (PDEs). Mariani and Tweneboah (2016); Mariani et al. (2016, 2018) and Habtemicael and SenGupta (2014) used Itô’s calculus to solve stochastic differential equations (SDEs) (see [5,6,7,8]). The above studies suggest that different numerical techniques are needed when solving complex DEs.

Leighton and Nehari (1958) [9] studied the oscillation properties of the solutions of fourth order self-adjoint DEs i.e.,

$${(r(x)y^{″})}^{″}+{(q(x)y^{\prime })}^{\prime }+p(x)y=0,$$

(1)

where $r(x)>0$ and $p(x)>0$. In particular, the authors studied the case where the functions $r(x)$ and $p(x)$ do not change sign in $(0,\infty )$. Barret [10] investigated a special case of Label (1), where $q(x)=0$ and both $r(x)$ and $p(x)$ are positive and continuous on $(0,\infty )$. The author extended this with the object of paralleling the self-adjoint DE of the second order with positive and continuous coefficients. In the paper by [11], the authors used change of variables to find the analytic solution to some self-adjoint DEs of the Fourth Order. In general, the solutions to self-adjoint equations are complex, hence we most often tend to solve them using numerical methods, which contain approximation errors. The solution technique proposed in this paper produces analytic solutions unlike numerical methods that have approximation errors. In this work, we extend the concept of self-adjoint equations to solve higher order differential equations including odd orders, arguing that this work may serve as a reference for solving other higher order self-adjoint type ODEs. According to the authors’ knowledge, no comprehensive work was dedicated to solving odd order self-adjoint type ODEs. Indeed, at present, general research in self-adjoint type ODEs are concerned with solving even orders ODEs (see [2,9,11], etc.). In addition, little work has been carried out to study integrating factor type techniques for solving higher order ODEs and previous works have only studied the integrating factor technique for solving first order ODEs [12]. An integrating factor is a function by which an ordinary differential equation can be multiplied in order to make it integrable.

In this paper, we provide two new analytic methods to solve some particular classes of higher order differential equations (ODEs), without the need of solving them numerically. The paper is outlined as follows: In Section 2, we will briefly review the background of self-adjoint operators and present some known results. In Section 3, we present the technique for solving higher order ODEs in the self-adjoint form. This section also discusses the conditions for a higher order ODEs to be in the self-adjoint form. Examples and applications are also presented. In Section 4, we discuss the integrating factor theory for solving higher order ODEs. Conclusions are contained in Section 5.

2. Second Order ODEs in the Self-Adjoint Form

This section presents background on the self-adjoint form and techniques for solving second order ODEs. We begin our discussion with the necessary conditions for a second order ODE to be in the self-adjoint form.

Consider the second order ODE:

$$p_0(x)u^{″}(x)+p_1(x)u^{\prime }(x)+p_2(x)u(x)=0.$$

(2)

Suppose we have an operator $\mathcal{L}$ defined as

$$\mathcal{L}=p_0(x)\frac{d^2}{d{x}^2}+p_1(x)\frac{d}{dx}+p_2(x).$$

(3)

Using the substitution $p_0^{\prime }=p_1$, we can rewrite Label (2) as

$$\overline{\mathcal{L}}u=\frac{d^2}{d{x}^2}(p_{0}u)-\frac{d}{dx}(p_{1}u)+p_{2}u=0.$$

(4)

For a given operator $\mathcal{L}$, there exists a corresponding operator $\overline{\mathcal{L}}$, known as the adjoint operator associated with $\mathcal{L}$. If the condition

$$p_0^{\prime }=p_1$$

(5)

is satisfied, we see that $\mathcal{L}=\overline{\mathcal{L}}$ and we say that the operator $\mathcal{L}$ is self-adjoint.

Definition 1.

A second order ODE (2) is said to be in the self-adjoint form if and only if:

$$\mathcal{L}u=\overline{\mathcal{L}}u=\frac{d}{dx}(p_0{u}^{\prime })+p_{2}u=0,$$

(6)

where $p_0(x)>0$ on $(a,b)$, $p_0^{\prime }(x)$, and $p_2(x)$ are continuous functions on $[a,b]$ and condition (5) is satisfied.

A classical example of a second order self-adjoint ODE is the Legendre’s equation:

$$(1-x^2)y^{″}-2x{y}^{\prime }+n(n+1)y=0.$$

Definition 2.

A second-order ordinary differential equation is called a Sturm–Liouville Equation if it is of the form:

$$\frac{d}{dx}(p(x)u^{\prime })+[\lambda w(x)-q(x)]u=0,$$

(7)

where λ is a constant and $w(x)$ is a known function called either the density or weighting function.

The constant $\lambda$ is called an eigenvalue and each eigenvalue has a corresponding $u_{\lambda }(x)$ eigenfunction. The solutions of (7) satisfy important mathematical properties under appropriate boundary conditions. See [1,2,13,14,15,16,17] and the references therein for more background on the self-adjoint approach.

A second order differential equation can always be presented in a self-adjoint formulation by using the Sturm–Liouville theory. Refer to [11] for details of transforming a non self-adjoint differential operator into the self-adjoint form.

Next, we state the following results in [11].

Theorem 1.

If a second order self-adjoint ODE

$$\frac{d}{dx}(p_0(x)u^{\prime })+p_2(x)u=0$$

(8)

verifies the condition

$$p_2(x)=\frac{p_0^{″}}{2}-\frac{p_0^{\prime 2}}{4p_0},$$

(9)

then the solution to Label (8) is

$$u(x)=\frac{1}{\sqrt{p_0(x)}}\left(C_{1}x+C_0\right),$$

(10)

where $C_1$ and $C_2$ are arbitrary constants.

We now present an example.

Example 1.

Consider the third order ODE

$$\frac{d}{dx}\left[\frac{x}{2}{(y^{\prime }+\frac{1}{2x}y)}^{\prime }\right]-\frac{1}{8x}(y^{\prime }+\frac{1}{2x}y)=0.$$

(11)

To obtain a solution for this equation, we let $v=y^{\prime }+\frac{1}{2x}y$. Thus, (11) reduces to the form:

$$\frac{d}{dx}\left[\frac{x}{2}{v}^{\prime }\right]-\frac{1}{8x}v=0,$$

(12)

which is a second order ODE for v. Next, we obtain a solution for v using the second order self-adjoint technique. From (12), we observe that $p_0(x)=\frac{x}{2}$ and $p_2(x)=-\frac{1}{8x}$. We observe that (9) is satisfied:

$$\begin{array}{c}\hfill p_2(x)=\frac{p_0^{″}}{2}-\frac{p_0^{\prime 2}}{4p_0}=0-\frac{\frac{1}{4}}{4\frac{x}{2}}=-\frac{1}{8x}.\end{array}$$

Therefore, the solution of (12) is

$$\begin{array}{cc}\hfill v(x)& =C_3\sqrt{x}+\frac{C_4}{\sqrt{x}}\hfill \end{array}$$

and then:

$$\begin{array}{cc}\hfill y^{\prime }+\frac{1}{2x}y& =C_3\sqrt{x}+\frac{C_4}{\sqrt{x}}.\hfill \end{array}$$

(13)

Equation (13) is a linear differential equation that can easily be solved.

3. Higher Order ODEs in the Self-Adjoint Form

In this section, we develop a technique that gives conditions for higher order ODEs to be in a self-adjoint form, providing an analytic solution.

We begin with a third order self-adjoint ODE.

3.1. Third Order ODEs in the Self-Adjoint Form

A third order linear differential operator is defined as

$$\mathcal{L}=r_0(x)\frac{d^3}{d{x}^3}+r_1(x)\frac{d^2}{d{x}^2}+r_2(x)\frac{d}{dx}+r_3(x),a<x<b,$$

(14)

where $r_0(x)>0$ on $(a,b)$, and $r_1(x),r_2(x),r_3(x)$ are continuous functions on $[a,b]$. If the operator in (14) acts on $y(x)$, and taking into account that

$${(r_o{y}^{\prime })}^{″}=r_0{y}^{‴}+2r_0^{\prime }{y}^{″}+r_0^{″}{y}^{\prime },$$

(15)

we have:

$$\mathcal{L}y={(r_o{y}^{\prime })}^{″}+(r_1-2r_0^{\prime })y^{″}+(r_2-r_0^{″})y^{\prime }+r_3(x)y(x).$$

(16)

To define the self-adjoint operator, we now assume the following conditions:

$$r_1(x)=2r_0^{\prime }(x),q(x)=r_2(x)-r_0^{″}(x),p(x)=r_3(x)-q^{\prime }(x),$$

(17)

where $r_1(x)$, $q(x)$, and $p(x)$ are continuous functions on $[a,b]$. Using these conditions, we define an adjoint operator $\overline{\mathcal{L}}$ corresponding to $\mathcal{L}$ in (16) as follows:

$$\overline{\mathcal{L}}y={(r_o{y}^{\prime })}^{″}+q(x)y^{\prime }+r_3(x)y(x)=0.$$

(18)

The second term of (18) can be written as:

$$q(x)y^{\prime }={(qy)}^{\prime }-q^{\prime }y.$$

(19)

Thus, (18) can be rewritten as:

$$\begin{array}{cc}\hfill \overline{\mathcal{L}}y& ={(r_o{y}^{\prime })}^{″}+{(q(x)y)}^{\prime }+p(x)y(x)=0.\hfill \end{array}$$

(20)

Therefore, we see that $\mathcal{L}=\overline{\mathcal{L}}$ based on the conditions (17), and we say that the operator $\mathcal{L}$ is self-adjoint.

Definition 3.

A third order ODE is said to be in the self-adjoint form if and only if:

$$\mathcal{L}y=\overline{\mathcal{L}}y={(r_o{y}^{\prime })}^{″}+{(q(x)y)}^{\prime }+p(x)y=0,$$

(21)

where $r_0(x)>0$ and conditions (17) are satisfied.

We present a theorem for solving higher order self-adjoint type ODEs.

Theorem 2.

If a third order self-adjoint ODE

$${(r(x)y^{\prime })}^{″}+{(q(x)y)}^{\prime }+p(x)y=0$$

(22)

verifies the conditions $q=r^{″}-\frac{2}{3r}{(r^{\prime })}^2$ and $p=-\frac{r^{‴}}{3}+\frac{2}{3r}{r}^{\prime }{r}^{″}-\frac{10}{27r^2}{(r^{\prime })}^3,$ then the solution to (22) is

$$y(x)=\frac{1}{\sqrt[3]{r^2(x)}}\left(C_1{x}^2+C_{2}x+C_3\right),$$

where $r(x)>0$, $p(x)$, $q(x)$ are continuous differentiable functions, and $C_1$, $C_2$, $C_3$ are arbitrary constants.

Proof. 

The third order self-adjoint ODE

$${(r(x)y^{\prime })}^{″}+{(q(x)y)}^{\prime }+p(x)y=0$$

can be written as:

$$r{y}^{‴}+2r^{\prime }{y}^{″}+(r^{″}+q)y^{\prime }+(q^{\prime }+p)y=0,$$

which implies that

$$y^{‴}+2\frac{r^{\prime }}{r}{y}^{″}+\frac{r^{″}+q}{r}{y}^{\prime }+\frac{q^{\prime }+p}{r}y=0.$$

(23)

Considering the change of variable for $y=u(x)v(x)$ in (23), where $u(x)$ and $v(x)$ are continuous and differentiable functions, we obtain

$$\begin{array}{c}{u}^{‴}v+\left(3v^{\prime }+\frac{2r^{\prime }}{r}v\right)u^{″}+\left(3v^{″}+4\frac{r^{\prime }}{r}{v}^{\prime }+\frac{r^{″}+q}{r}v\right)u^{\prime }\hfill \\ +\left(v^{‴}+2\frac{r^{\prime }}{r}{v}^{″}+\frac{r^{″}+q}{r}{v}^{\prime }+\frac{q^{\prime }+p}{r}v\right)u=0.\hfill \end{array}$$

(24)

Assuming that the coefficient of $u^{″}$ is zero, we can solve for $v(x)$ as follows:

$$3v^{\prime }+\frac{2r^{\prime }}{r}v=0.$$

(25)

Then, without loss of generality, the corresponding solution to (25) is

$$v(x)=e^{-\frac{2}{3}\int \frac{r^{\prime }}{r}dr}=r^{-\frac{2}{3}}=\frac{1}{\sqrt[3]{r^2}}.$$

(26)

Next, we also assume that the coefficients of $u^{\prime }$ and u are both zero i.e.,

$$3v^{″}+4\frac{r^{\prime }}{r}{v}^{\prime }+\frac{r^{″}+q}{r}v=0,$$

(27)

$$\mathrm{and},v^{‴}+2\frac{r^{\prime }}{r}{v}^{″}+\frac{r^{″}+q}{r}{v}^{\prime }+\frac{q^{\prime }+p}{r}v=0.$$

(28)

By replacing (26) into (27), we obtain:

$$r{r}^{″}-\frac{2}{3}{(r^{\prime })}^2-qr=0.$$

(29)

Similarly, replacing (26) into (28):

$$\frac{2}{3}{r}^{‴}-\frac{2}{3r}{r}^{\prime }{r}^{″}+\frac{8}{27r^2}{(r^{\prime })}^3-q^{\prime }=0$$

(30)

Thus, from (29) and (30), we obtain the corresponding conditions:

$$q=r^{″}-\frac{2}{3r}{(r^{\prime })}^2$$

(31)

and

$$p=-\frac{r^{‴}}{3}+\frac{2}{3r}{r}^{\prime }{r}^{″}-\frac{10}{27r^2}{(r^{\prime })}^3.$$

(32)

From Equations (25), (27), (28), (31) and (32), we obtain

$$v{u}^{‴}=0,\mathrm{then}{u}^{‴}=0$$

and

$$u(x)=C_1{x}^2+C_{2}x+C_3,$$

where $C_1$, $C_2$, and $C_3$ are arbitrary constants. Hence, the solution of the third order self-adjoint differential Equation (22) is as follows:

$$y(x)=u(x)v(x)$$

and we conclude that:

$$y(x)=\frac{1}{\sqrt[3]{r^2(x)}}\left(C_1{x}^2+C_{2}x+C_3\right).$$

☐

3.2. Fourth Order ODEs in the Self-Adjoint Form

A fourth order linear differential operator $\mathcal{L}$ acting on a function y is given by

$$\mathcal{L}\left[y\right]=r_0(x)\frac{d^{4}y}{d{x}^4}+r_1(x)\frac{d^{3}y}{d{x}^3}+r_2(x)\frac{d^{2}y}{d{x}^2}+r_3(x)\frac{dy}{dx}+p(x)y(x),$$

(33)

where $r_0(x)>0$ on $(a,b)$, and $r_1(x),r_2(x),r_3(x),p(x)$ are continuous functions on $[a,b]$. Then, using the same approach discussed in the previous subsection, we state the following definition:

Definition 4.

A fourth order ODE is said to be in the self-adjoint form if and only if:

$$\mathcal{L}y=\overline{\mathcal{L}}y={(r_o{y}^{″})}^{″}+{(q(x)y^{\prime })}^{\prime }+p(x)y(x)=0$$

(34)

and the conditions

$$r_1(x)=2r_0^{\prime }(x),q(x)=r_2(x)-r_0^{″}(x),q^{\prime }(x)=r_3(x)$$

(35)

are satisfied and where $r_0(x)$, $r_1(x)$, $q(x)$, and $p(x)$ are continuous functions on $[a,b]$.

An example of a fourth order self adjoint ODE is given below:

$${(1+x)}^4{y}^{(4)}+8{(1+x)}^3{y}^{‴}+20{(1+x)}^2{y}^{″}+16(1+x)y^{\prime }+12(1+x)y=0.$$

(36)

3.3. Examples and Applications

In this subsection, we present applications and examples of self-adjoint ordinary differential equations.

Example 2.

Consider the equation

$${\left(x^2{y}^{\prime }\right)}^{″}+{\left(-\frac{2}{3}y\right)}^{\prime }+\left(-\frac{8}{27x}\right)y=0.$$

(37)

In this example, $r(x)=x^2$. Now, we check the conditions of Theorem 2 to solve the ODE (37):

$$q=r^{″}-\frac{2}{3r}{(r^{\prime })}^2=-\frac{2}{3},$$

$$and p=-\frac{r^{‴}}{3}+\frac{2}{3r}{r}^{\prime }{r}^{″}-\frac{10}{27r^2}{(r^{\prime })}^3=-\frac{8}{27x}.$$

We observe that the conditions are satisfied, hence we can obtain the analytic solution from Theorem 2 as:

$$\begin{array}{cc}\hfill y(x)& =\frac{1}{\sqrt[3]{x^4}}\left(C_1{x}^2+C_{2}x+C_3\right)\hfill \\ \hfill & =x^{-\frac{4}{3}}\left(C_1{x}^2+C_{2}x+C_3\right).\hfill \end{array}$$

(38)

Example 3.

Consider a third order ODE,

$${\left(e^x{y}^{\prime }\right)}^{″}+{\left(\frac{1}{3}{e}^{x}y\right)}^{\prime }+\left(-\frac{1}{27}{e}^x\right)y=0.$$

(39)

We have $r(x)=e^x$. Now, we check the conditions of Theorem 2 to solve the ODE (39):

$$q=r^{″}-\frac{2}{3r}{(r^{\prime })}^2=\frac{e^x}{3},$$

$$and,p=-\frac{r^{‴}}{3}+\frac{2}{3r}{r}^{\prime }{r}^{″}-\frac{10}{27r^2}{(r^{\prime })}^3=-\frac{1}{27}{e}^x.$$

Since the conditions of Theorem 2 are satisfied, we can obtain the analytic solution of our ODE as

$$\begin{array}{cc}\hfill y(x)& =\frac{1}{\sqrt[3]{e^{2x}}}\left(C_1{x}^2+C_{2}x+C_3\right)\hfill \\ \hfill & =e^{-\frac{2x}{3}}\left(C_1{x}^2+C_{2}x+C_3\right).\hfill \end{array}$$

(40)

4. Integrating Factor Approach for Solving Higher Order ODEs

In this section, we present an integrating factor methodology for solving ODEs. The aim is to develop a technique for solving higher order differential equations. To proceed, we present the definition of a Ricatti ODE of the first order.

Definition 5.

A Ricatti ODE of the first order is any equation of the form

$$y^{\prime }+P(x)+Q(x)y=R(x)y^n,$$

(41)

where $P(x),Q(x)$ and $R(x)$ are continuous functions and $n>1$ is a real number.

4.1. Integrating Factor Technique for Second Order ODEs

In order to solve a second order ODE, we propose a first order Ricatti ODE for a variable b. In fact, this Ricatti equation is obtained by using the coefficients of the second order ODE and the particular solution for this equation helps to obtain the integrating factor of the second order ODE. Using integrating type techniques, we obtain a particular solution of the second order differential equation. We proceed with the methodology as follows.

Consider a general second order ODE,

$$y^{″}+P(x)y^{\prime }+Q(x)y=R(x).$$

(42)

We can express the second order ODE as:

$${(u(y^{\prime }+by))}^{\prime }=uR(x),$$

(43)

where u is an integrating factor. This equation can be rewritten as:

$$y^{\prime }(u^{\prime }+ub)+y(u^{\prime }b+u{b}^{\prime })+u{y}^{″}=uR(x).$$

(44)

From Labels (42) and (44), we can compare the coefficients as follows:

$$u^{\prime }+ub=uP(x),$$

(45)

$$u^{\prime }b+u{b}^{\prime }=uQ(x).$$

(46)

We then simplify the above two equations and obtain the Ricatti equation for b:

$$b^{\prime }=Q(x)-bP(x)+b^2.$$

(47)

If we know the homogeneous solution $y_1$ of (42), then $-\frac{y_1^{\prime }}{y_1}=b$ is a solution of the Ricatti equation $b^{\prime }=b^2+Q-bP$ and if u in (45) is given by $\frac{u^{\prime }}{u}=P-b$, we obtain that u is an integrating factor of (42) and this allows us to obtain a particular solution. We also observe that, from (45) and (46), it is possible to obtain the second order differential equation for the integrating factor, which is $u^{″}-u^{\prime }P+u(Q-P^{\prime })=0$. As $ub=uP-u^{\prime }$, it implies that ${(ub)}^{\prime }=uQ=u^{\prime }P+u{P}^{\prime }-u^{″}$. Therefore, $u^{″}-u^{\prime }P+u(Q-P^{\prime })=0$.

Next, we present an example.

Example 4.

Consider a second order ODE,

$$y^{″}-x{y}^{\prime }+y=x^k,$$

(48)

where k is any even number.

In particular, we consider the case where $k=2$ i.e.,

$$y^{″}-x{y}^{\prime }+y=x^2.$$

(49)

Comparing this equation with (42), we observe that $P=-x$, $Q=1$, and $R=x^2$.

The Ricatti equation for b is given by:

$$b^{\prime }=Q-bP+b^2.$$

(50)

Knowing that $y_1=x$ is a solution of the homogeneous equation associated with (49) i.e., $y^{″}-x{y}^{\prime }+y=0$, as $b=-\frac{y_1^{\prime }}{y_1}=-\frac{1}{x}$, then $b=-\frac{1}{x}$ is a particular solution of (50). We now compare between Equations (42) and (44) and obtain:

$$\frac{u^{\prime }}{u}=P-b.$$

Then,

$$\frac{u^{\prime }}{u}=-x+\frac{1}{x}.$$

(51)

Integrating both sides of (51), we have

$$\ln u=-\frac{x^2}{2}+lnx+C.$$

Without loss of generality, $u=e^{-x^2/2}x$.

Now plugging $u,b$ and R into (43) and integrating both sides, we obtain

$$x{e}^{-x^2/2}\left(y^{\prime }-\frac{1}{x}y\right)=\int x^3{e}^{-x^2/2}dx=-e^{-\frac{x^2}{2}}(x^2+2)+C.$$

This implies that

$$y^{\prime }-\frac{1}{x}y=-\left(x+\frac{2}{x}\right)+\frac{C{e}^{x^2/2}}{x}.$$

The solution to this last equation can be found by using first degree order integrating factor theory.

4.2. Integrating Factor Technique for Third Order ODEs

We begin this subsection with a theorem.

Theorem 3.

Given

$$y^{‴}+P(x)y^{″}+Q(x)y^{\prime }+R(x)y=f(x),$$

(52)

if we know a solution to the associated integrating factor equation

$$u^{‴}-P{u}^{″}+(Q-2P^{\prime })u^{\prime }+(Q^{\prime }-P^{″}-R)u=0$$

(53)

or, alternatively, a solution to

$$y^{‴}+2P{y}^{″}+(P^{\prime }+P^2+Q)y^{\prime }+(Q^{\prime }-R+QP)y=0,$$

(54)

then we can find a particular solution to (52).

Proof. 

We solve (52) by formulating

$${\left[u(y^{″}+b{y}^{\prime }+ay)\right]}^{\prime }=uf(x),$$

(55)

where u satisfies (53). Multiplying through (52) by u, we obtain

$$u(y^{‴}+P(x)y^{″}+Q(x)y^{\prime }+R(x)y)=uf(x).$$

(56)

From (55) and (56), we get

$$u^{\prime }(y^{″}+b{y}^{\prime }+ay)+u{(y^{″}+b{y}^{\prime }+ay)}^{\prime }=uf(x).$$

(57)

Expanding, we get

$$y^{‴}u+y^{″}(u^{\prime }+ub)+y^{\prime }(u^{\prime }b+u{b}^{\prime }+ua)+y(u^{\prime }a+u{a}^{\prime })=uf(x).$$

(58)

From (55)–(58), we obtain the following three conditions:

(i)

$uP=u^{\prime }+ub,$

(ii)

$uQ=u^{\prime }b+u{b}^{\prime }+ua,$

(iii)

$uR=u^{\prime }a+u{a}^{\prime }.$

From (i), we have that:

$$\frac{u^{\prime }}{u}=P-b$$

(59)

and from (ii):

$$\frac{u^{\prime }}{u}=\frac{Q-b^{\prime }-a}{b}.$$

(60)

Thus, from (59) and (60), $P-b=\frac{Q-b^{\prime }-a}{b}$; then, we have that $(P-b)b=Q-b^{\prime }-a$ and, solving for a, we obtain

$$a=Q-b^{\prime }-b(P-b).$$

(61)

Finally from (iii), we have $u^{\prime }a=u(R-a^{\prime })$ and so

$$\frac{u^{\prime }}{u}=\frac{R-a^{\prime }}{a}.$$

(62)

Combining with (59), we obtain

$$\frac{u^{\prime }}{u}=\frac{R-a^{\prime }}{a}=P-b.$$

(63)

Thus, $a(P-b)=R-a^{\prime }$. Replacing in (63) a and $a^{\prime }$, we get

$$\left[Q-b^{\prime }-b(P-b)\right](P-b)=R-{\left[Q-b^{\prime }-b(P-b)\right]}^{\prime }.$$

(64)

From (64), we obtain

$$b^{″}=Q^{\prime }-R+QP+b(-Q-P^{\prime }-P^2)+3b{b}^{\prime }-2b^{\prime }P+2P{b}^2-b^3).$$

(65)

Using the change of variable $b=-\frac{y^{\prime }}{y}$, we have that

$$b^{\prime }=-\frac{y^{″}}{y}+\frac{y^{\prime 2}}{y^2}.$$

(66)

Then, replacing $b=-\frac{y^{\prime }}{y}$ into (66), we conclude that $b^{\prime }=-\frac{y^{″}}{y}+b^2$, and

$$b^{″}=-\frac{y^{‴}}{y}+\frac{y^{\prime }}{y}\frac{y^{″}}{y}+2b{b}^{\prime }.$$

(67)

Since $b^{\prime }=-\frac{y^{″}}{y}+b^2$, replacing $\frac{y^{\prime }}{y}=-b$ into (67), we have

$$b^{″}=-\frac{y^{‴}}{y}-b\frac{y^{″}}{y}+2b{b}^{\prime }$$

and, using $b^{\prime }=-\frac{y^{″}}{y}+b^2$, we get $\frac{y^{″}}{y}=b^2-b^{\prime }$; therefore,

$$b^{″}=-\frac{y^{‴}}{y}-b^3+3b{b}^{\prime }.$$

(68)

Considering the homogeneous differential equation associated with (52) i.e.,

$$y^{‴}+P(x)y^{″}+Q(x)y^{\prime }+R(x)y=0,$$

we obtain

$$\frac{y^{‴}}{y}=-P(x)\frac{y^{″}}{y}-Q(x)\frac{y^{\prime }}{y}-R(x).$$

(69)

Replacing $\frac{y^{″}}{y}=b^2-b^{\prime }$ and $\frac{y^{\prime }}{y}=-b$ in (69):

$$\frac{y^{‴}}{y}=-P(x)\left[b^2-b^{\prime }\right]+Q(x)b-R(x)$$

and, replacing the last expression into (68), we get

$$b^{″}=R(x)+b^{2}P(x)+3b{b}^{\prime }-b^3-bQ(x)-b^{\prime }P(x).$$

(70)

Therefore, (70) represents the Ricatti equation associated with (52). On the other hand, if we replace (68) into (65) and use the relations $b=-\frac{y^{\prime }}{y}$ and $b^{\prime }=-\frac{y^{″}}{y}+b^2$, we get:

$$y^{‴}+2P{y}^{″}+(P^{\prime }+P^2+Q)y^{\prime }+(Q^{\prime }-R+QP)y=0,$$

(71)

which is the corresponding homogeneous equation associated with (65).

We recall from conditions (i) and (ii) that $ub=uP-u^{\prime }$ and ${(ub)}^{\prime }+ua=uQ$, respectively. Thus, from (i), ${(ub)}^{\prime }=u^{\prime }P+u{P}^{\prime }-u^{″}$ and replacing this into (ii), we have:

$$ua=uQ-u^{\prime }P-u{P}^{\prime }+u^{″}.$$

(72)

From condition (iii) ${(ua)}^{\prime }=uR$, so, taking the derivative of (72), we obtain the integrating factor equation

$$u^{‴}-u^{″}P+u^{\prime }(Q-2P^{\prime })+u(Q^{\prime }-P^{″}-R)=0.$$

(73)

Thus, if we know one solution of (54), by using the change of variable $b=-\frac{y^{\prime }}{y}$, we have one solution of (70) which means we have a solution for b, consequently for a from (i), (ii) and (iii)—hence the proof. ☐

Remark 1.

If we have a solution for the integrating factor equation (73), we can find a particular solution for

$$y^{‴}+P(x)y^{″}+Q(x)y^{\prime }+R(x)y=f(x)$$

and, conversely, if we know a particular solution for (52), we can obtain a solution for the integrating factor equation (73). In fact, knowing one solution for

$$y^{‴}+P(x)y^{″}+Q(x)y^{\prime }+R(x)y=0,$$

we can find a solution for

$$u^{‴}-P{u}^{″}+u^{\prime }(Q-2P^{\prime })+u(Q^{\prime }-P^{″}-R)=0.$$

Example 5.

Consider the third order ODE

$$u^{‴}+x^2{u}^{″}+6x{u}^{\prime }+6u=0.$$

(74)

One solution for (74) is $u_1=x^2{e}^{-x^3/3}$ and another solution is $u_2=x^2{e}^{-x^3/3}\int \frac{e^{x^3/3}}{x^3}dx$.

Comparing (74) to the integrating factor equation (73), we observe that $-P=x^2$, $Q-2P^{\prime }=6x$ and $Q^{\prime }-P^{″}-R=6$. Solving for $P,P^{\prime },P^{″},Q,Q^{\prime }$ and R, the corresponding equation for (52) is

$$y^{‴}-x^2{y}^{″}+2x{y}^{\prime }-2y=f(x).$$

Since $\frac{u^{\prime }}{u}=P-b$ from (59), we can solve for b using $u=e^{-x^3/3}{x}^2$. Thus, $\frac{u^{\prime }}{u}=\frac{2}{x}-x^2$ and as $\frac{u^{\prime }}{u}=P-b=-x^2-b$ solving for b we get $b=-\frac{2}{x}$. Now, from (61), since $a=Q-b^{\prime }-b(P-b)$, replacing $b,P,Q$ and $b^{\prime }$, we get $a=\frac{2}{x^2}$. Thus, replacing $u,b$ and a in (55) and integrating both sides we get:

$$e^{-x^3/3}{x}^2\left(y^{″}-\frac{2}{x}{y}^{\prime }+\frac{2}{x^2}y\right)=\int e^{-x^3/3}{x}^{2}f(x)dx.$$

Suppose $f(x)=c$, then

$$e^{-x^3/3}{x}^2\left(y^{″}-\frac{2}{x}{y}^{\prime }+\frac{2}{x^2}y\right)=-c{e}^{-x^3/3}+k$$

and we have that:

$$y^{″}-\frac{2}{x}{y}^{\prime }+\frac{2}{x^2}y=-\frac{c}{x^2}+k\frac{e^{x^3/3}}{x^2}.$$

To find a solution for this last equation, we use Euler for the homogeneous equation $y^{″}-\frac{2}{x}{y}^{\prime }+\frac{2}{x^2}y=0$ or

$$x^2{y}^{″}-2x{y}^{\prime }+2y=0.$$

(75)

Setting $y=x^r$, we get the fundamental solutions to be $y_1=x^2$ and $y_2=x$. Hence, the general solution for the homogeneous part is $y_h=c_1{x}^2+c_{2}x$. To find a particular solution, we implement again the technique used in the previous example.

Consider

$$y^{″}-\frac{2}{x}{y}^{\prime }+\frac{2}{x^2}y=g(x)$$

(76)

with $g(x)=-\frac{c}{x^2}+k\frac{e^{x^3/3}}{x^2}$. Using one of the fundamental solutions for example $y=x$, since $b=-\frac{y^{\prime }}{y}$, then, from Equation (43) with $g(x)$ instead of $R(x)$, we get:

$${(u(y^{\prime }+by))}^{\prime }=ug(x).$$

(77)

We recall that $\frac{u^{\prime }}{u}=P-b$, where in this case from (76), $P=-\frac{2}{x}$ and $b=-\frac{y^{\prime }}{y}=-\frac{1}{x}$. Thus, $\frac{u^{\prime }}{u}=-\frac{1}{x}$, hence $u=x^{-1}$. Thus, replacing u and b into (77), we have:

$${\left(x^{-1}\left(y^{\prime }+(-\frac{1}{x})y\right)\right)}^{\prime }=\frac{1}{x}g(x),\mathit{then}\frac{1}{x}\left(y^{\prime }-\frac{1}{x}y\right)=\int \frac{1}{x}g(x)dx+\tilde{k}$$

and, for this last equation, we obtain the solution by using the first degree order integrating factor technique.

Example 6.

Consider the second order ODE

$$y^{″}+\left(2x^2-\frac{2}{x}\right)y^{\prime }+\left(x^4+\frac{2}{x^2}\right)y=0.$$

If we assume $b=x^2-\frac{1}{x}$, then $b^{\prime }=2x+\frac{1}{x^2}$. Here, we know that $P=2x^2-\frac{2}{x}$ and $Q=x^4+\frac{2}{x^2}$. Next, since $\frac{u^{\prime }}{u}=P-b$, replacing P and b, we get $\frac{u^{\prime }}{u}=x^2-\frac{1}{x}\mathit{then}u=e^{x^3/3}{x}^{-1}k$. Thus, we pick $u=\frac{e^{x^3/3}}{x}$. By using (77) i.e.,: ${(u(y^{\prime }+by))}^{\prime }=ug(x)$, if we have that $g(x)=0$, then

$${\left[\frac{e^{x^3/3}}{x}\left(y^{\prime }+\left(x^2-\frac{1}{x}\right)y\right)\right]}^{\prime }=\frac{e^{x^3/3}}{x}(0).$$

Then, integrating both sides, we obtain:

$$y^{\prime }+\left(x^2-\frac{1}{x}\right)y=cx{e}^{-x^3/3},$$

and from this last equation we can easily obtain the solution.

Recall from Section 4.1 and Section 4.2 that it is enough to know one solution either from the original homogeneous differential equation or the corresponding associated integrating factor differential equation in order to find a particular solution for the ODE.

4.3. Integrating Factor Technique for Fourth Order ODEs

We begin this subsection with a theorem.

Theorem 4.

Given the equation:

$$y^{(4)}+P{y}^{{}^{‴}}+Q{y}^{{}^{″}}+R{y}^{{}^{\prime }}+Wy=f(x).$$

(78)

If we know one solution of either (79) or (80),

$$\begin{array}{cc}\hfill u^{(4)}& +-P{u}^{{}^{‴}}+u^{{}^{″}}(Q-3P^{\prime })\hfill \\ \hfill & +u^{{}^{\prime }}(2Q^{\prime }-3P^{″}-R)\hfill \\ \hfill & +u(W+Q^{″}-P^{‴}-R^{\prime })=0,\hfill \end{array}$$

(79)

$$\begin{array}{cc}\hfill y^{(4)}& +3P{y}^{{}^{‴}}+(3P^{\prime }+3P^2+Q)y^{{}^{″}}\hfill \\ \hfill & +(P^{″}+3P^{\prime }P+2Q^{\prime }-R+2PQ+P^3)y^{{}^{\prime }}\hfill \\ \hfill & +(-R^{\prime }-PR+P^{2}Q+P^{\prime }Q+Q^{″}+2Q^{\prime }P+W)y=0,\hfill \end{array}$$

(80)

then we can find a particular solution for (78).

Proof. 

We proceed the same way as with the case for degree 3. If we multiply (78) by u i.e.,

$$u\left(y^{(4)}+P{y}^{{}^{‴}}+Q{y}^{{}^{″}}+R{y}^{{}^{\prime }}+Wy\right)=uf(x)$$

(81)

and propose that

$${\left[u\left(y^{‴}+b{y}^{{}^{″}}+a{y}^{{}^{\prime }}+cy\right)\right]}^{\prime }=uf(x).$$

(82)

Expanding (82) and equating (81) to (82), we obtain

$$\begin{array}{cc}\hfill y^{(4)}u& +y^{{}^{‴}}[u^{\prime }+ub]+y^{″}[u^{\prime }b+u{b}^{\prime }+ua]\hfill \\ \hfill & +y^{\prime }[u^{\prime }a+u{a}^{\prime }+uc]+y[u^{\prime }c+u{c}^{\prime }]\hfill \\ \hfill & =u{y}^{(4)}+y^{‴}uP+y^{″}uQ+y^{\prime }uR+yuW=uf(x).\hfill \end{array}$$

Thus, we have

(i)

$uP=u^{\prime }+ub,$

(ii)

$uQ=u^{\prime }b+u{b}^{\prime }+ua,$

(iii)

$uR=u^{\prime }a+u{a}^{\prime }+uc,$

(iv)

$uW=u^{\prime }c+u{c}^{\prime }={(uc)}^{\prime }.$

From (i) $ub=uP-u^{\prime }$ and from (ii) ${(ub)}^{\prime }=uQ-ua$, thus $uQ-ua={(uP-u^{\prime })}^{\prime }$ and so $ua=uQ-u^{\prime }P-u{P}^{\prime }+u^{″}$; therefore,

$${(ua)}^{\prime }=u^{\prime }Q+u{Q}^{\prime }-u^{″}P-2u^{\prime }{P}^{\prime }-u{P}^{″}+u^{‴}=uR-cu$$

(83)

and, solving for $cu$ in (83), we obtain

$$cu=uR-u^{\prime }Q-u{Q}^{\prime }+u^{″}P+2u^{\prime }{P}^{\prime }+u{P}^{″}-u^{‴}.$$

Solving for u in (83) and using the fact that $uW={(uc)}^{\prime }$, we obtain the integrating factor equation:

$$u^{(4)}-P{u}^{‴}+u^{″}(Q-3P^{\prime })+u^{\prime }(2Q^{\prime }-3P^{″}-R)+u(W+Q^{″}-P^{‴}-R^{\prime })=0.$$

Now from (i),

$$u^{\prime }=u(P-b)\mathrm{implies}\frac{u^{\prime }}{u}=P-b.$$

(84)

From (ii),

$$u^{\prime }b=u(Q-b^{\prime }-a)\mathrm{implies}\frac{u^{\prime }}{u}=\frac{Q-b^{\prime }-a}{b}=P-b$$

and then $Q-b^{\prime }-a=b(P-b)$. Thus,

$$a=Q-b^{\prime }-b(P-b).$$

(85)

From (iii),

$$\frac{u^{\prime }}{u}=\frac{R-a^{\prime }-c}{a}=P-b\mathrm{implies}R-a^{\prime }-c=a(P-b)$$

and so

$$c=R-a^{\prime }-a(P-b).$$

(86)

Finally, from (iv),

$$u^{\prime }c=u(W-c^{\prime })\mathrm{implies}\frac{u^{\prime }}{u}=\frac{W-c^{\prime }}{c}=P-b$$

and then $W-c^{\prime }=c(P-b)$.

Replacing the above equation in (86) and using (85), we get

$$c=R-Q^{\prime }+b^{″}+b^{\prime }(P-b)+b{(P-b)}^{\prime }-[Q-b^{\prime }-b(P-b)](P-b)$$

(87)

and from (86), taking the derivative of (87) and using the fact that $c^{\prime }=W-c(P-b)$:

$$\begin{array}{ll}\hfill W-c(P-b)& =R^{\prime }-Q^{″}+b^{‴}+b^{″}(P-b)+b^{\prime }{(P-b)}^{\prime }+b^{\prime }{(P-b)}^{\prime }+b{(P-b)}^{″}\hfill \\ \hfill & -(Q^{\prime }-b^{″}-b^{\prime }(P-b)-b{(P-b)}^{\prime })(P-b)-(Q-b^{\prime }-b(P-b)){(P-b)}^{\prime }\hfill \\ \hfill & =W-(P-b)[R-Q^{\prime }+b^{″}+b^{\prime }(P-b)+b{(P-b)}^{\prime }\hfill \\ \hfill & -(Q-b^{\prime }-b(P-b))(P-b)].\hfill \end{array}$$

Solving for $b^{‴}$, we get

$$\begin{array}{cc}\hfill b^{‴}& =4b^{″}b+3b^{\prime 2}-6b^2{b}^{\prime }+b^4\hfill \\ \hfill & +(-R^{\prime }+Q^{″}+Q{P}^{\prime }+W-RP+Q{P}^2+2Q^{\prime }P)\hfill \\ \hfill & +b^{\prime }(-3P^{\prime }-3P^2-Q)+b(-P^{″}-3P{P}^{\prime }+R-2PQ-P^3-2Q^{\prime })\hfill \\ \hfill & -3P{b}^{″}-3b^{3}P+9b{b}^{\prime }P+b^2(3P^{\prime }+Q+3P^2).\hfill \end{array}$$

(88)

If we consider the homogeneous part of the original equation (78) i.e.,

$$y^{(4)}+P{y}^{{}^{‴}}+Q{y}^{{}^{″}}+R{y}^{{}^{\prime }}+Wy=0,$$

(89)

using the change of variable $\tilde{b}=-\frac{y^{\prime }}{y}$, then

$${\tilde{b}}^{\prime }=-\frac{y^{″}}{y}+{\tilde{b}}^2$$

(90)

and

$${\tilde{b}}^{{}^{″}}=-\frac{y^{‴}}{y}+3\tilde{b}{\tilde{b}}^{{}^{\prime }}-{\tilde{b}}^3$$

(91)

and finally computing the third derivative,

$${\tilde{b}}^{{}^{‴}}=-\frac{y^{(4)}}{y}-6{\tilde{b}}^2{\tilde{b}}^{{}^{\prime }}+4\tilde{b}{\tilde{b}}^{{}^{″}}+{\tilde{b}}^4+3{({\tilde{b}}^{{}^{\prime }})}^2.$$

(92)

Thus, comparing (88) with (92), we have that $4b^{″}b+3{b^{\prime }}^2-6b^2{b}^{{}^{\prime }}+b^4$ and $4{\tilde{b}}^{{}^{″}}\tilde{b}+3{\widetilde{b^{\prime }}}^2-6{\tilde{b}}^2{\tilde{b}}^{{}^{\prime }}+{\tilde{b}}^4$ that are common for (88) and (92). Dividing (89) by y and substituting $\tilde{b}=-\frac{y^{\prime }}{y}$ and rearranging terms, we obtain

$$\frac{y^{(4)}}{y}=-\frac{P{y}^{‴}}{y}-\frac{Q{y}^{″}}{y}+R\tilde{b}-W.$$

(93)

Replacing (93) in (92) and using (90) and (91), and $\tilde{b}=-\frac{y^{\prime }}{y}$, we obtain:

$${\tilde{b}}^{{}^{‴}}=-P{\tilde{b}}^{{}^{″}}-P{\tilde{b}}^3+3P\tilde{b}{\tilde{b}}^{{}^{\prime }}+Q{\tilde{b}}^2-Q{\tilde{b}}^{{}^{\prime }}-R\tilde{b}+W+4\tilde{b}{\tilde{b}}^{{}^{″}}-6{\tilde{b}}^2{\tilde{b}}^{{}^{\prime }}+3{({\tilde{b}}^{{}^{\prime }})}^2+{\tilde{b}}^4,$$

(94)

which has four terms in common with (88). We observe that from (88) and (94) the corresponding terms are $4\tilde{b}{\tilde{b}}^{{}^{″}}-6{\tilde{b}}^2{\tilde{b}}^{{}^{\prime }}+3{\widetilde{b^{\prime }}}^2+{\tilde{b}}^4$ and $4b{b}^{″}-6b^2{b}^{\prime }+3{\widetilde{b^{\prime }}}^2+b^4$. The other terms in both expressions suggest that $-P\to -3P$ for coefficient of ${\tilde{b}}^{{}^{″}}$ and $b^{{}^{″}}$, respectively. Then, from (94), $-P{\tilde{b}}^3\to -3P{b}^3$, $3P\tilde{b}{\tilde{b}}^{{}^{\prime }}\to 9Pb{b}^{\prime }$, $Q{\tilde{b}}^2\to (3P^{\prime }+Q+3P^2)b^2$, $-Q{\tilde{b}}^{{}^{\prime }}\to (-3P^{\prime }-3P^2-Q)b^{\prime }$, $-R\tilde{b}\to (-P^{″}-3P{P}^{\prime }+R-2PQ-P^3-2Q^{\prime })b$ and $W\to (-R^{\prime }+Q^{″}+Q{P}^{\prime }+W-RP+Q{P}^2+2Q^{\prime }P)$. This suggests that as (94) comes doing $-\frac{y^{\prime }}{y}=\tilde{b}$ from the homogeneous part of (78), then, similarly, (88) must come from doing $b=-\frac{y^{\prime }}{y}$ on (80).

Thus, if we know one solution of (80), by using the change of variable $b=-\frac{y^{\prime }}{y}$, we have one solution of (88) which means we have a solution for b, consequently for a and c from (i), (ii), (iii) and (iv)—hence the proof. ☐

Example 7.

Consider the fourth order ODE

$$y^{(4)}-x^2{y}^{‴}+3x{y}^{″}-6y^{\prime }+\frac{6}{x}y=f(x).$$

In this example, we have $P=-x^2,Q=3x,R=-6$ and $W=\frac{6}{x}$. Therefore, $-P=x^2,Q-3P^{\prime }=9x,2Q^{\prime }-3P^{″}-R=18$ and $W+Q^{″}-P^{‴}-R^{\prime }=\frac{6}{x}$. Hence,

$$u^{(4)}-P{u}^{‴}+u^{″}(Q-3P^{\prime })+u^{\prime }(2Q^{\prime }-3P^{″}-R)+u(W+Q^{″}-P^{‴}-R^{\prime })=0$$

would be

$$u^{(4)}-x^2{u}^{‴}+u^{″}(9x)+u^{\prime }(18)+u\frac{6}{x}=0.$$

One solution of this equation is $u=x^3{e}^{-x^3/3}$. Replacing $\frac{u^{\prime }}{u}=\frac{3}{x}-x^2$ and since $\frac{u^{\prime }}{u}=P-b$, then $\frac{u^{\prime }}{u}=-x^2-b$. Therefore, $b=-\frac{3}{x}$ and as $a=Q-b(P-b)-b^{\prime }$, this means $a=\frac{6}{x^2}$. In addition, since $c=R-a^{\prime }-a(P-b)$, it implies that $c=-\frac{6}{x^3}$. Then, following the proof of Theorem 4, we have:

$${\left[u\left(y^{{}^{‴}}+b{y}^{{}^{″}}+a{y}^{{}^{\prime }}+cy\right)\right]}^{\prime }=uf(x).$$

Replacing $u,a,b$ and c into the above equation and integrating we have:

$$x^3{e}^{-x^3/3}\left(y^{{}^{‴}}-\frac{3}{x}{y}^{{}^{″}}+\frac{6}{x^2}{y}^{{}^{\prime }}-\frac{6}{x^3}y\right)=\int x^3{e}^{-x^3/3}f(x)dx.$$

(95)

To find this solution, we first note that $y^{{}^{‴}}-\frac{3}{x}{y}^{{}^{″}}+\frac{6}{x^2}{y}^{{}^{\prime }}-\frac{6}{x^3}y=0$ can be converted to an Euler equation

$$x^3{y}^{‴}-3x^2{y}^{″}+6x{y}^{\prime }-6y=0.$$

(96)

Setting $y=x^r$, we obtain $y_1=x,y_2=x^2$ and $y_3=x^3$ to be the solutions of the third order ODE

$$y^{‴}-\frac{3}{x}{y}^{″}+\frac{6}{x^2}{y}^{\prime }-\frac{6}{x^3}y.$$

Thus, we can solve for (95) by using the same methodology as we did with degree 3.

5. Conclusions

In this work, we introduce two new analytic methods for solving some particular classes of higher order ODEs, without the need for solving them numerically. In particular, we developed a self-adjoint type formulation and integrating-factor approaches for solving higher order ODEs including odd order ODEs. The analytic solutions produced using these methodologies are exact unlike numerical solutions that have approximation errors. The methodologies presented in this work may serve as a reference for solving other higher order ODEs.