This document outlines analytical methods for solving higher order ordinary differential equations (ODEs). It discusses how second and third order ODEs can be solved using self-adjoint forms and integrating factor techniques. Examples are provided to demonstrate how to identify the appropriate conditions and solve ODEs of different orders up to fourth order. Integrating factor methods allow determining a particular solution given a known solution to an associated integrating factor equation.

1. Analytical Methods for Solving Higher
Order Ordinary Differential Equation.
Rabbiya Ahmed
Reg: CIIT/SP19-BSM-013/ATK
Supervised By :Dr Maimona Rafiq
Department of Mathematics
COMSATS University Islamabad,
Attock Campus
1 / 23

2. Outlines
• Introduction
• Self-adjoint Method for Solving Higher Order ODEs
• Integrating Factor Technique for Solving Higher Order ODEs
• References
2 / 23

3. Introduction
Definition 1.
A differential equation containing the derivatives (some finite
number of derivatives ) of one or more dependent variables, with
respect to one or more independent variables, is said to be a
differential equation.
Differential equations are classified into two main parts :
1. Ordinary Differential Equations.
2. Partial Differential Equations.
Definition 2.
If an equation contains only ordinary derivatives of one or more
dependent variables with respect to a single independent variable,
it is said to be an ordinary differential equation (ODE).
3 / 23

4. Introduction
Example 3.
dy
dx + 5y = ex
Definition 4.
An equation involving partial derivatives of one or more dependent
variables with respect to two or more independent variables is
called a partial differential equation.
Example 5.
∂2u
∂x2 + ∂u
∂y = 0
4 / 23

5. Introduction
Definition 6.
A solution of a differential equation is a function that satisfies the
equation. When you substitute the function into the differential
equation, you get a true mathematical statement.
Definition 7.
An initial value problem is a problem which seeks to determine a
solution to a differential equation subject to conditions on the
unknown function and its derivative specified at one value of the
independent variable.
5 / 23

6. Introduction
Definition 8.
A boundary value problem is a problem which seeks to determine a
solution to a differential equation subject to conditions on the
unknown function specified at two or more values of the
independent variable..
Definition 9.
A differential equation Ly = 0 is said to be a self adjoint operator
if the operator L is self adjoint. Thus we have,
Ly = a0yn + a′
0y′ + a2y = (a0y′)′ + a2y
6 / 23

7. Higher Order ODEs in the Self-Adjoint Form
Second Order ODEs in the Self-Adjoint Form
Consider the second order ODE:
p0(x)u′′
(x) + p1(x)u′
(x) + p2(x)u(x) = 0. (1)
Suppose we have an operator L defined as
L = p0(x)
d2
dx2
+ p1(x)
d
dx
+ p2(x). (2)
Using the substitution p′
0 = p1, we can rewrite Label (1) as
L̃u =
d2
dx2
(p0u) −
d
dx
(p1u) + p2u = 0. (3)
For a given operator L, there exists a corresponding operator L,
known as the adjoint operator associated with L. If the condition 7 / 23

8. Second Order ODE’s in the Self-Adjoint Form
Definition 10.
A second order ODE (1) is said to be in the self-adjoint form if and
only if:
Lu = Lu =
d
dx
p0u′
+ p2u = 0,
where p0(x) 0 on (a, b), p′
0(x), ant p2(x) are continuous
functions on [a, b] and condition and
p′
0 = p1
is satisfied.
8 / 23

9. Theorem 11.
If a second order self-adjoint ODE
d
dx
(p0(x)u′
) + p2(x)u = 0
verifies the condition
p2(x) =
p0
′′
2
−
p0
′2
4p0
then the solution to equ is:
u(x) =
1
p
(p0(x)
(C1(x) + C0)
where C1 and C2 are arbitrary constants.
9 / 23

10. Example
Consider the order ODE
d
dx
x
2
y′
+
1
2x
y
′
−
1
8
x
y′
+
1
2x
y
= 0
substitute v = y′ + 1
2x y
Thus
d
dx
hx
2
v′
i
−
1
8x
v = 0
Now Comparing with this equ
d
dx
P0(x)u′
+ P2(x)u
P0(x) =
x
2
and P2(x) = −
1
8x
10 / 23

11. P2(x) =
P′′
0
2
−
P′
0
4P0
= 0 −
1
4
= −
1
4x
(5)
so the solution is
v(x) = c3
√
x +
c4
√
x
(6)
y′
+
1
2x
y = c3
√
x +
c4
√
x
(7)
11 / 23

12. Third Order ODE in Self-adjoint Form
Definition 12.
A third order ODE is said to be in the self-adjoint form if and only
if:
Ly = Ly = roy′
′′
+ (q(x)y)′
+ p(x)y = 0.
where r0(x) 0 and conditions
r1(x) = 2r′
0(x), q(x) = r2(x) − r′′
0 (x), p(x) = r3(x) − q′
(x).
are satisfied.
12 / 23

13. Theorem 13.
If a third order self-adjoint ODE
r(x)y′
′′
+ (q(x)y)′
+ p(x)y = 0
verifies the conditions q = r′′ − 2
3r (r′)2
and
p = −r′′
3 + 2
3r′r′′ − 10
27r2 (r′)3
, then the solution of third order
self-adjoint ODE is
y(x) =
1
3
p
r2(x)
C1x2
+ C2x + C3
where r(x) 0, p(x), q(x) are continuous differentiable functions,
and C1, C2, C3 are arbitrary constants.
13 / 23

14. Example 14.
Consider a third order ODE
x2
y′
′′
+
−
2
3
y
′
+
−
8
27x
y = 0.
where r(x) = x2. Now, we check the conditions of Theorem 13 to
solve the given ODE :
q = r′′
−
2
3r
r′
2
= −
2
3
,
and p = −
r′′′
3
+
2
3r
r′
r′′
−
10
27r2
r′
3
= −
8
27x
.
We observe that the conditions are satisfied, hence we can obtain
the analytic solution from Theorem 13 as:
y(x) =
1
3
√
x4
C1x2
+ C2x + C3
14 / 23

15. Integrating Factor techniques for Solving Higher Order
ODEs
Integrating Factor Technique for Third Order ODE
Theorem 15.
Given
y′′′
+ P(x)y′′
+ Q(x)y′
+ R(x)y = f (x),
if we know a solution to the associated integrating factor equation
u′′′
− Pu′′
+ Q − 2P′
u′
+ Q′
− P′′
− R
u = 0
or, alternatively, a solution to
y′′′
+ 2Py′′
+ P′
+ P2
+ Q
y′
+ Q′
− R + QP
y = 0,
then we can find a particular solution to third order ODE.
15 / 23

16. Example
Consider the third order ODE
u′′′
+ x2
u′′
+ 6xu′
+ 6u = 0. (8)
One solution for above equation is u1 = x2e−x3/3
and another solution is
u2 = x2e−x3/3
R x3/3
52 dx.
Comparing Theorem 15 integrating factor equation by given ode, we
observe that −P = x2, Q − 2P′ = 6x and Q − P′′ − R = 6. Solving
for P, P′, P′, Q, Q′ and R, the corresponding equation is
y′′′
− x2
y′′
+ 2xy′
− 2y = f (x)
Since
u′
u
= p − b from Theorem 15 proof ,
we can solve for b using u = e−x3/3x2. Thus, u′
u = 2
x − x2 and as
u′u = P − b = −x2 − b solving for b we get b = −2
x .
16 / 23

17. Since a = Q − b′ − b(P − b). replacing b, P, Q and b′, we get
a = 2
x2 . This, replacing u, b and a in [u(y′ + by′ + ay)]′ = uf (x)
and integrating both sides we get:
e−x3/3
x2
y′′
−
2
x′
y′
+
2
x2
y
=
Z
e−x3/3
x2
f (x)dx.
Suppose f (x) = c, then
e−x3/3
x2
y′′
−
2
x
y′
+
2
x2
y
= −ce−x3/3
+ k
and we have that:
y′′
−
2
x
y′
+
2
x2
y = −
c
x2
+ k
ex3/3
x2
To find a solution for this last equation, we use Euler for the howogen-
tous equation y′′ − 2
x y + 2
x2 y = 0 ar
x2
y′′
− 2xy′
+ 2y = 0. (9)
17 / 23

18. Setting y = xr , we get the fundamental solutions to be y1 = x2 and
y2 = x. Hence, the general solution for the homogeneous part is
yh = c1x2 + c2x. To find a particular solution, we divide the above
equation by x2
y′′
−
2
x
y′
+
2
x2
y = g(x) (10)
with g(x) = − c
x2 + k c3/3
x2 . Using one of the fundamental solutions
for example y = x, since b = −y
y , then, from Equation,
u y′
+ by
′
= ug(x). (11)
We know that u′
u = P − b, where in this case from above equation,
P = −2
x and b = −y′
y = −1
x . Thus, u′
u = −1
x , hence u = x−1.
Thus, replacing u and b into above equation, we have.
x−1
y′
+
−
1
x
y
′
=
1
x
g(x), then
1
x
y′
−
1
x
y
=
Z
1
x
g(x)dx+k
and, for this last equation, we obtain the solution by using the first
degree order integrating factor technique. 18 / 23

19. Integrating Factor techniques for solving fourth order
ODE
Theorem 16.
Given the equation:
y(4)
+ Py′′′
+ Qy′′
+ Ry′
+ Wy = f (x). (12)
If we know one solution of either
u(4)
− Pu′′′
+ u′′
Q − 3P′
+ u′
2Q′
− 3P′′
− R
+ u W + Q′′
− P′′
− R′
= 0
y(4)
+ 3Py′′
+ 3P′
+ 3P2
+ Q
y′′
+ P′′
+ 3P′
P + 2Q′
− R + 2PQ + P3
y′
+ −R′
− PR + P2
Q + P′
Q + Q′′
+ 2Q′
P + W
y = 0
19 / 23

20. Example
Consider the fourth order ODE
y(4)
− x2
y′′
+ 3xy′′
− 6y′
+
6
x
y = f (x)
In this example, we hate P = −x2, Q = 3x, R = −6 and W = 6
x .
Therefore, −P = x2, Q − 3P′ = 9x, 2Qr − 3Pv − R = 18 and
W + Qn − pw − Rr = 6
x . Hence,
u(4)
−Pu′′′
+u′′
Q − 3P′
+u′
2Q′
− 3P′′
− R
+u W + Q′′
− P′′′
− R′
would be
u(4)
− x2
u′′′
+ u′′
(9x) + u′
(18) + u
6
x
= 0.
One solution of this equation is u = x3e−x3/3. Replacing u′
u = 3
x −x2
and since u′
u = P − b, then u′
u = −x2 − b. Therefore, b = −3
x and
as a = Q − b(P − b) − b, this means a = 6
x2 . In addition, since
c = R − a′ − a(P − b), it implies that c = − 6
x3 . Then, following
20 / 23

21. Replacing u, a, b and c into the above equation and integrating we
have:
x3
e−x3/3
y′′
−
3
x
y′
+
6
x2
y′
−
6
x3
y
=
Z
x3
e−x3/3
f (x)dx (14)
To find this solution, we first note that y′′′ − 3
x y′′ + 6
x2 y′ − 6
x3 y = 0
can be converted to an Euler equation
x3
y′′′
− 3x2
y′
+ 6xy′
− 6y = 0
Setting y = xr , we obtain y1 = x, y2 = x2 and y3 = x3 fo be the
solutions of the third order ODE
y′′
−
3
x
y′′
+
6
x2
y′
−
6
x3
y.
Thus, we can solve for (14) by using the same methodology as we
did with degree 3.
21 / 23

22. Bibliography:
22 / 23

23. 23 / 23