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Chapter- 4
Higher order differential equation
Linear ordinary differential equation: A linear ordinary differential equation of order 𝑛 is an
equation of the form
0
𝑛
𝑛
1
𝑛− 1
𝑛− 1
𝑛− 1
𝑛
Where 𝑎 0
Homogeneous Differential Equation: A differential equation is said to be homogeneous if each
term of the equation contains at least a derivative or a dependent variable.
Homogeneous linear differential equation: A linear ordinary differential equation of order 𝑛 is an
equation of the form
0
𝑛
𝑛
1
𝑛− 1
𝑛− 1
𝑛− 1
𝑛
Where 𝑎 0
Linearly independent : The 𝑛 functions 𝑓 1
2
𝑛
are called linearly independent if there exists
constants 𝑐 1
2
𝑛
such that the linear combination 𝑐
1
1
2
2
𝑛
𝑛
= 0 implies that 𝑐
1
2
𝑛
= 0. If not all 𝑐
𝑖
are zero then the functions 𝑓
1
2
𝑛
are called linearly dependent.
Wronskian : Let 𝑓 1
2
𝑛
be 𝑛 real functions each of which has (𝑛 − 1 )𝑡ℎ derivative on a real
interval. The determinant
1
2
𝑛
1
2
1
2
𝑛
𝑛
1
𝑛− 1
2
𝑛− 1
𝑛
𝑛− 1
in which primes denote derivatives, is called wronskian of these 𝑛 functions. If 𝑊(𝑓
1
2
𝑛
then the functions are linearly independent otherwise the functions are linearly dependent.
Theorem 1 (without proof) : Every homogeneous linear differential equation of order 𝑛 has exactly
𝑛 number of independent solutions.
Theorem 2 : If 𝑓
1
2
𝑛
are𝑛 number of independent solutions of the homogeneous linear
differential equation𝑎
0
𝑑
𝑛
𝑦
𝑑𝑥
𝑛
1
𝑑
𝑛− 1
𝑦
𝑑𝑥
𝑛− 1
𝑛− 1
𝑑𝑦
𝑑𝑥
𝑛
(𝑥)𝑦 = 0 then their linear
combination 𝑐 1
1
2
2
𝑛
𝑛
is also a solution of that differential equation.
Proof : If𝑓 1
2
𝑛
are independent solutions of homogeneous linear differential equation
0
𝑑
𝑛
𝑦
𝑑𝑥
𝑛
1
𝑑
𝑛− 1
𝑦
𝑑𝑥
𝑛− 1
𝑛− 1
𝑑𝑦
𝑑𝑥
𝑛
then we obtain
0
𝑑
𝑛
𝑓
1
𝑑𝑥
𝑛
1
𝑑
𝑛− 1
𝑓
1
𝑑𝑥
𝑛− 1
𝑛− 1
𝑑𝑓
1
𝑑𝑥
𝑛
1
0
𝑑
𝑛
𝑓 2
𝑑𝑥
𝑛
1
𝑑
𝑛− 1
𝑓 2
𝑑𝑥
𝑛− 1
𝑛− 1
𝑑𝑓 2
𝑑𝑥
𝑛
2
0
𝑑
𝑛
𝑓
𝑛
𝑑𝑥
𝑛
1
𝑑
𝑛− 1
𝑓
𝑛
𝑑𝑥
𝑛− 1
𝑛− 1
𝑑𝑓
𝑛
𝑑𝑥
𝑛
𝑛
= 0 (n+1)
Now if 𝑐 1
1
2
2
𝑛
𝑛
is a solution of equation (1) then it must satisfies the equation.
0
𝑛
1
1
2
2
𝑛
𝑛
𝑛
1
𝑛− 1
1
1
2
2
𝑛
𝑛
𝑛− 1
𝑛− 1
𝑑(𝑐
1
𝑓
1
+𝑐
2
𝑓
2
+⋯+𝑐
𝑛
𝑓
𝑛
)
𝑑𝑥
𝑛
1
1
2
2
𝑛
𝑛
1
0
𝑛
1
𝑛
1
𝑛− 1
1
𝑛− 1
𝑛− 1
1
𝑛
1
2
0
𝑛
2
𝑛
1
𝑛− 1
2
𝑛− 1
𝑛− 1
2
𝑛
2
𝑛
0
𝑛
𝑛
𝑛
1
𝑛− 1
𝑛
𝑛− 1
𝑛− 1
𝑛
𝑛
𝑛
Thus the linear combination 𝑐 1
1
2
2
𝑛
𝑛
is also a solution of equation (1) called the
general solution. (Proved)
Problem 1 : Consider the differential equation
′′
′
(a) Show that 𝑒
2 𝑥
and 𝑒
3 𝑥
are linearly independent solutions of this equation on the interval−∞ < 𝑥 <
(b) Write the general solution of the given equation.
Problem 2 : Consider the differential equation
′′
′
(a) Show that 𝑒
𝑥
and 𝑥𝑒
𝑥
are linearly independent solutions of this equation on the interval −∞ < 𝑥 <
(b) Write the general solution of the given equation.
(c) Find the solution that satisfies the conditions𝑦( 0 ) = 1 , 𝑦′( 0 ) = 4. Explain why this solution is
unique. Over what interval is it defined?
Solution : (a) Given that 𝑦
′′
′
Let 𝑦 = 𝑒
𝑥
then 𝑦′ = 𝑒
𝑥
and 𝑦′′ = 𝑒
𝑥
Substituting the values of 𝑦, 𝑦
′
, 𝑦′′ in 𝐿. 𝐻. 𝑆 of equation (1) we get
𝑥
𝑥
𝑥
Again let 𝑦 = 𝑥𝑒
𝑥
then 𝑦
′
𝑥
𝑥
and 𝑦
′′
𝑥
𝑥
𝑥
𝑥
𝑥
Substituting the values of 𝑦, 𝑦
′
, 𝑦′′ in 𝐿. 𝐻. 𝑆 of equation (1) we get
𝑥
𝑥
𝑥
𝑥
𝑥
Hence 𝑒
𝑥
and 𝑥𝑒
𝑥
are two solutions of equation (1).
Now 𝑊(𝑒
𝑥
𝑥
𝑥
𝑥
𝑥
𝑥
𝑥
2 𝑥
2 𝑥
2 𝑥
2 𝑥
≠ 0 for all 𝑥 in the interval
−∞ < 𝑥 < ∞. Thus 𝑒
𝑥
and 𝑥𝑒
𝑥
are linearly independent solutions of equation (1) on the interval
(b) The general solution of equation (1) is
1
𝑥
2
𝑥
(c) Here,
1
𝑥
2
𝑥
′
1
𝑥
2
𝑥
2
𝑥
Using initial conditions𝑦
= 4 in equation (2) and (3) we obtain
1
and
1
2
Solving equation (4) and (5) we get
1
2
Since the number of unknown constants and the number of equations from initial conditions are same
so the value of 𝑐
1
and 𝑐
2
are unique.
Thus the required solution is
𝑥
𝑥
Since 𝑐
1
and 𝑐
2
are unique so this solution is unique over the interval −∞ < 𝑥 < ∞.
Problem 3 : Consider the differential equation
2
′′
′
(a) Show that 𝑥and 𝑥
2
are linearly independent solutions of this equation on the interval 0 < 𝑥 < ∞.
(b) Write the general solution of the given equation.
(c) Find the solution that satisfies the conditions𝑦
= 2. Explain why this solution is
unique. Over what interval is it defined?
Solution : (a) Given that 𝑥
2
′′
′
Let 𝑦 = 𝑥 then 𝑦′ = 1 and 𝑦′′ = 0
Substituting the values of 𝑦, 𝑦
′
, 𝑦′′ in 𝐿. 𝐻. 𝑆 of equation (1) we get
Again let 𝑦 = 𝑥
2
then 𝑦′ = 2 𝑥 and 𝑦′′ = 2
Substituting the values of 𝑦, 𝑦
′
, 𝑦′′ in 𝐿. 𝐻. 𝑆 of equation (1) we get
2
2
Hence 𝑥and 𝑥
2
are two solutions of equation (1).
Now 𝑊
2
2
2
2
2
≠ 0 for all 𝑥on the interval
0 < 𝑥 < ∞. Thus 𝑥and 𝑥
2
are linearly independent solutions of equation (1) on the interval 0 < 𝑥 <
(b) The general solution of equation (1) is
1
2
2
(c) Here,
1
2
2
1
2
Using initial conditions𝑦( 1 ) = 3 , 𝑦′( 1 ) = 2 in equation (2) and (3) we obtain
1
2
and
Letting 𝑐 = 1 and recall 𝑤 = 𝑣′ we obtain
′
2
Again integrating we get
Thus the linearly independent solution is 𝑦 = 𝑣𝑥 = 𝑥 (𝑥 −
1
𝑥
2
Hence the general solution is
1
2
2
Problem 5 : Given that 𝑦 = 𝑥 + 1 is a solution of (𝑥 + 1 )
2
′′
′
independent solution by reducing the order.
Solution : Given (𝑥 + 1 )
2
′′
′
Let𝑦 = 𝑣(𝑥 + 1 ) then𝑦
′
= 𝑣 + (𝑥 + 1 )𝑣′ and 𝑦
′′
′
′′
′
′′
Substituting these values in equation (1) we get
2
′′
′
′
3
′′
2
′′
Letting 𝑤 = 𝑣′ we obtain the first – order homogeneous linear equation
Treating this as a separable equation, we obtain
By integrating we obtain
ln 𝑤 = ln(𝑥 + 1 ) + ln 𝑐
⇒ ln 𝑤 = ln(𝑥 + 1 ) + ln 𝑐
Letting 𝑐 = 1 and recall 𝑤 = 𝑣′ we obtain
′
Again integrating we get
2
Thus the linearly independent solution is 𝑦 = 𝑣(𝑥 + 1 ) = (𝑥 + 1 )
(𝑥+ 1 )
2
2
(𝑥+ 1 )
3
2
Hence the general solution is
1
2
3
Let 𝑦 = 𝑒
𝑚𝑥
be the trail solution of equation (1).
Then the auxiliary equation of equation (1) is
2
⇒ (m − 1 )(m − 2 ) = 0
Thus the required general solution is
1
𝑥
2
2 𝑥
Problem 7 : Solve the differential equation 𝑦
′′
′
Solution : Given the differential equation
′′
′
Let 𝑦 = 𝑒
𝑚𝑥
be the trail solution of equation (1).
Then the auxiliary equation of equation (1) is
2
⇒ (m − 2 )(m − 3 ) = 0
Thus the required general solution is
1
2 𝑥
2
3 𝑥
Problem 8 : Solve the differential equation 𝑦
′′′
′′
′
Solution : Given the differential equation
′′′
′′
′
Let 𝑦 = 𝑒
𝑚𝑥
be the trail solution of equation (1).
Then the auxiliary equation of equation (1) is
3
2
3
2
2
2
m − 1
m − 1
m − 1
m − 1
2
m − 1
Thus the required general solution is
1
𝑥
2
−𝑥
3
3 𝑥
Case II : Repeated real roots
Problem 9 : Solve the differential equation 𝑦
′′
′
Solution : Given the differential equation
′′
′
Let 𝑦 = 𝑒
𝑚𝑥
be the trail solution of equation (1).
Then the auxiliary equation of equation (1) is
2
m − 3
Then the two solutions are 𝑒
3 𝑥
and𝑒
3 𝑥
which are linearly dependent.
Thus their linear combination 𝑐
1
3 𝑥
2
3 𝑥
1
2
3 𝑥
0
3 𝑥
is not the general solution.
Since we already know the one solution is 𝑒
3 𝑥
, we let 𝑦 = 𝑣𝑒
3 𝑥
, then
′
3 𝑥
′
3 𝑥
′′
3 𝑥
′′
3 𝑥
′
3 𝑥
Substituting into equation (1) we get
3 𝑥
′′
3 𝑥
′
3 𝑥
3 𝑥
′
3 𝑥
3 𝑥
3 𝑥
′′
′′
′
Choosing 𝑐 = 1
′
Thus 𝑥𝑒
3 𝑥
is a linearly independent solution of equation (1)
Hence the general solution is
1
3 𝑥
2
3 𝑥
Problem 10 : Solve the differential equation 𝑦
′′
′
Solution : Given the differential equation
′′
′
Let 𝑦 = 𝑒
𝑚𝑥
be the trail solution of equation (1).
Then the auxiliary equation of equation (1) is
2
′′′
′′
′
Let 𝑦 = 𝑒
𝑚𝑥
be the trail solution of equation (1).
Then the auxiliary equation of equation (1) is
3
2
3
Thus the required general solution is
1
2 𝑥
2
2 𝑥
3
2
2 𝑥
1
2
3
2
2 𝑥
Problem 13 : Solve the differential equation 𝑦
𝑖𝑣
′′
Solution : Given the differential equation
𝑖𝑣
′′
Let 𝑦 = 𝑒
𝑚𝑥
be the trail solution of equation (1).
Then the auxiliary equation of equation (1) is
4
2
2
2
2
Thus the required general solution is
1
cos 2 𝑥 + 𝑐
2
sin 2 𝑥 + 𝑐
3
𝑥 cos 2 𝑥 + 𝑐
4
𝑥 sin 2 𝑥 = (𝑐
1
3
𝑥)cos 2 𝑥 + (𝑐
2
4
𝑥) sin 2 𝑥
Problem 14 : Solve the initial value problem
′′′
′′
′
′
′′
Solution : Given the differential equation
′′′
′′
′
Let 𝑦 = 𝑒
𝑚𝑥
be the trail solution of equation (1).
Then the auxiliary equation of equation (1) is
3
2
2
2
The general solution of equation (1) is
1
2 𝑥
2
cos 2 𝑥 + 𝑐
3
sin 2 𝑥) (2)
′
1
2 𝑥
2
sin 2 𝑥 + 2 𝑐
3
cos 2 𝑥) (3)
′′
1
2 𝑥
2
cos 2 𝑥 − 4 𝑐
3
sin 2 𝑥) (4)
Applying the initial conditions 𝑦( 0 ) = 2 , 𝑦
′
( 0 ) = 0 and𝑦
′′
( 0 ) = 0 in equations (2), (3) and (4) we
get
1
2
1
3
and 0 = 4 𝑐
1
2
Solving equations (5), (6) and (7) we get
1
2
= 1 and 𝑐
3
Thus the required solution is
2 𝑥
Problem 15 : The roots of the auxiliary equation corresponding to a certain 11 th order homogeneous
linear differential equation with constant coefficients are
4 , 4 , 4 , 4 , 5 , 2 + 3 𝑖, 2 − 3 𝑖, 2 + 3 𝑖, 2 − 3 𝑖, 4 + 5 𝑖, 4 − 5 𝑖 write the general solution.
Solution : The general solution is
1
2
3
2
4
3
4 𝑥
5
5 𝑥
2 𝑥
6
7
cos 3 𝑥 +
8
9
sin 3 𝑥
4 𝑥
10
cos 5 𝑥 + 𝑐
11
sin 5 𝑥
Non homogeneous linear equation with constant coefficients :
𝑛
1
𝑛− 1
𝑛
)𝑦 = 𝐹(𝑥) where 𝐷 =
𝑑
𝑑𝑥
Problem 16 : Find the particular integral for 𝑭
𝒂𝒙
Solution : by successive differentiation we obtain
𝑎𝑥
𝑎𝑥
𝑎𝑥
𝑎𝑥
2
𝑎𝑥
2
𝑎𝑥
𝑛
𝑎𝑥
𝑛
𝑎𝑥
(n)
Multiplying equation (1), (2), …,(n) by 𝑎
𝑛
𝑛− 1
, ⋯ , 1 respectively and adding we get
𝑎𝑥
𝑎𝑥
The auxiliary equation of corresponding homogeneous equation of (1) is
2
2
∴ Complementary function, 𝑦 𝑐
1
2 𝑥
2
2 𝑥
Particular integral,
𝑝
2
2 𝑥
2 𝑥
2
2 𝑥
2
2 𝑥
Thus the required general solution is
𝑐
𝑝
1
2 𝑥
2
2 𝑥
2
2 𝑥
Problem 20 : Solve the equation (𝐷
3
2
𝑥
3 𝑥
Solution : Given that (𝐷
3
2
𝑥
3 𝑥
The auxiliary equation of corresponding homogeneous equation of (1) is
3
2
3
∴ Complementary function, 𝑦 𝑐
1
𝑥
2
𝑥
3
2
𝑥
Particular integral,
𝑝
3
2
𝑥
3
2
3 𝑥
2
𝑥
3 𝑥
2
𝑥
3 𝑥
3
𝑥
3 𝑥
3
𝑥
3 𝑥
Thus the required general solution is
𝑐
𝑝
1
𝑥
2
𝑥
3
2
𝑥
3
𝑥
3 𝑥
Particular integral for 𝑭(𝒙) = 𝒔𝒊𝒏 𝒂𝒙, 𝒄𝒐𝒔 𝒂𝒙
𝑝
2
sin 𝑎𝑥 , cos 𝑎𝑥 =
2
sin 𝑎𝑥 , cos 𝑎𝑥 ; 𝑓(𝑎
2
Problem 21 : Solve the equation (𝐷
2
𝑥
Solution : Given that
2
𝑥
The auxiliary equation of corresponding homogeneous equation of (1) is
2
∴ Complementary function,
𝑐
−
𝑥
2
(𝑐
1
cos
2
sin
Particular integral,
𝑝
2
𝑥
2
sin 2 𝑥
𝑥
2
sin 2 𝑥
𝑥
sin 2 𝑥 =
𝑥
sin 2 𝑥
𝑥
2
sin 2 𝑥 =
𝑥
2
sin 2 𝑥
𝑥
sin 2 𝑥 =
𝑥
2 cos 2 𝑥 + 3 sin 2 𝑥
Thus the required general solution is
𝑐
𝑝
−
𝑥
2
(𝑐
1
cos
2
sin
𝑥
2 cos 2 𝑥 + 3 sin 2 𝑥
Problem 22 : Solve the equation (𝐷
2
2
𝑦 = cos 3 𝑥
Solution : Given that (𝐷
2
2
𝑦 = cos 3 𝑥 (1)
The auxiliary equation of corresponding homogeneous equation of (1) is
2
2
2
∴ Complementary function,
𝑐
1
2
𝑥) cos 𝑥 + (𝑐
3
4
𝑥) sin 𝑥
Particular integral,
𝑝
2
2
cos 3 𝑥
𝑐
1
2
−𝑥
3
−𝑥
Particular integral,
𝑝
3
2
2 𝑥
3
2
2
3
2
2 𝑥
2
2
2
2 𝑥
− 2
2
− 2
2 𝑥
2
2
2
2 𝑥
2
2 𝑥
3
2
2
2 𝑥
3
2
Thus the required general solution is
𝑐
𝑝
1
2
−𝑥
3
−𝑥
2 𝑥
3
2
Particular integral for 𝑭
𝒂𝒙
𝑝
𝑎𝑥
𝑎𝑥
Problem 25 : Solve the equation (𝐷
2
3 𝑥
3 𝑥
Solution : Given that
2
3 𝑥
3 𝑥
The auxiliary equation of corresponding homogeneous equation of (1) is
2
∴ Complementary function,
𝑐
1
3 𝑥
2
− 3 𝑥
Particular integral,
𝑝
2
3 𝑥
2
3 𝑥
3 𝑥
3 𝑥
2
3 𝑥
3 𝑥
2
3 𝑥
3 𝑥
𝐷
6
3 𝑥
3 𝑥
𝐷
6
− 1
3 𝑥
3 𝑥
𝐷
6
3 𝑥
3 𝑥
3 𝑥
3 𝑥
2
3 𝑥
3 𝑥
2
Thus the required general solution is
𝑐
𝑝
1
3 𝑥
2
− 3 𝑥
3 𝑥
3 𝑥
2
Problem 26 : Solve the equation (𝐷
2
3 𝑥
sin 2 𝑥
Solution : Given that (𝐷
2
3 𝑥
sin 2 𝑥 (1)
The auxiliary equation of corresponding homogeneous equation of (1) is
2
∴ Complementary function,
𝑐
1
√ 11 𝑥
2
−√ 11 𝑥
Particular integral,
𝑝
2
3 𝑥
sin 2 𝑥 = 𝑒
3 𝑥
2
sin 2 𝑥
3 𝑥
2
sin 2 𝑥 = 𝑒
3 𝑥
2
sin 2 𝑥
3 𝑥
2
sin 2 𝑥 = 𝑒
3 𝑥
sin 2 𝑥
3 𝑥
sin 2 𝑥 = 𝑒
3 𝑥
2
sin 2 𝑥
