Definition, Method of solution and solved problems of Homogeneous Linear Differential Equations

1. HOMOGENEOUS
LINEAR
DIFFERENTIAL
EQUATIONS

2. Definition
An equation of the form
𝑎 𝑜 𝑥 𝑛
𝑑 𝑛 𝑦
𝑑𝑥 𝑛
+ 𝑎1 𝑥 𝑛−1
𝑑 𝑛−1 𝑦
𝑑𝑥 𝑛−1
+ … … … + 𝑎 𝑛−1 𝑥
𝑑𝑦
𝑑𝑥
+ 𝑎 𝑛 𝑦 = 𝑋
Where 𝑎 𝑜 , 𝑎1,…, 𝑎 𝑛 are constants and X is a function of 𝑥 is called a
homogeneous linear differential equation of the nth order.

3. Method of Solution
1. Reduce the given homogeneous linear differential equation into linear equation
with constant coefficients by putting
𝑥 = 𝑒 𝑧, 𝐷 ≡
𝑑
𝑑𝑧
, 𝑥
𝑑𝑦
𝑑𝑥
= 𝐷𝑦 , 𝑥2
𝑑2 𝑦
𝑑𝑥2
= 𝐷 𝐷 − 1 𝑦, 𝑥3
𝑑3 𝑦
𝑑𝑥3
= 𝐷 𝐷 − 1 𝐷 − 2 𝑦
and so on
2. Given equation becomes 𝑓 𝐷 𝑦 = 𝑍, 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑧
3. Take 𝑦 = 𝑒 𝑚𝑧
as a trial solution
4. Solve the equation
5. Replace 𝑧 𝑏𝑦 log 𝑥

4. Prove that the following:
a 𝑥
𝑑𝑦
𝑑𝑥
= 𝐷𝑦 𝑏 𝑥2
𝑑2
𝑦
𝑑𝑥2
= 𝐷 𝐷 − 1 𝑦 𝑐 𝑥3
𝑑3
𝑦
𝑑𝑥3
= 𝐷 𝐷 − 1 (𝐷 − 2)𝑦
Proof:
a) Let 𝑥 = 𝑒 𝑧
.
Then log 𝑥 = log 𝑒 𝑧
= 𝑧 log 𝑒 = 𝑧 [log 𝑒 = 1]
or, 𝑧 = log 𝑥
or,
𝑑𝑧
𝑑𝑥
=
𝑑
𝑑𝑥
log 𝑥 =
1
𝑥
… … (1)
Then by chain rule , we get
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑧
.
𝑑𝑧
𝑑𝑥
=
𝑑𝑦
𝑑𝑧
.
1
𝑥
[𝑏𝑦 (1)]
𝑜𝑟,
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑧
.
1
𝑥
… … (2)
𝑜𝑟, 𝑥
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑧
𝑜𝑟, 𝑥
𝑑𝑦
𝑑𝑥
= 𝐷𝑦, 𝑤ℎ𝑒𝑟𝑒 𝐷 ≡
𝑑
𝑑𝑧
proved

5. b) We have
𝑑2 𝑦
𝑑𝑥2
=
𝑑
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
𝑑
𝑑𝑥
𝑑𝑦
𝑑𝑧
.
1
𝑥
[𝑏𝑦 (2)]
=
1
𝑥
.
𝑑
𝑑𝑥
𝑑𝑦
𝑑𝑧
+
𝑑𝑦
𝑑𝑧
.
𝑑
𝑑𝑥
1
𝑥
=
1
𝑥
.
𝑑𝑧
𝑑𝑥
.
𝑑
𝑑𝑧
𝑑𝑦
𝑑𝑧
+
𝑑𝑦
𝑑𝑧
.
𝑑
𝑑𝑥
(𝑥−1
)
=
1
𝑥
.
1
𝑥
.
𝑑2 𝑦
𝑑𝑧2
+
𝑑𝑦
𝑑𝑧
. (−𝑥−2)
=
1
𝑥2
.
𝑑2
𝑦
𝑑𝑧2
−
1
𝑥2
.
𝑑𝑦
𝑑𝑧
𝑜𝑟,
𝑑2 𝑦
𝑑𝑥2
=
1
𝑥2
𝑑2 𝑦
𝑑𝑧2
−
𝑑𝑦
𝑑𝑧
… (3)
𝑜𝑟, 𝑥2
𝑑2 𝑦
𝑑𝑥2
=
𝑑2 𝑦
𝑑𝑧2
−
𝑑𝑦
𝑑𝑧
𝑜𝑟, 𝑥2
𝑑2 𝑦
𝑑𝑥2
= 𝐷2
𝑦 − 𝐷𝑦
Therefore
𝑥2
𝑑2 𝑦
𝑑𝑥2
= 𝐷 𝐷 − 1 𝑦

6. b) We have
𝑑3 𝑦
𝑑𝑥3
=
𝑑
𝑑𝑥
𝑑2 𝑦
𝑑𝑥2
=
𝑑
𝑑𝑥
1
𝑥2
𝑑2 𝑦
𝑑𝑧2
−
𝑑𝑦
𝑑𝑧
𝑏𝑦 3
=
1
𝑥2
.
𝑑
𝑑𝑥
𝑑2 𝑦
𝑑𝑧2
−
𝑑𝑦
𝑑𝑧
+
𝑑2 𝑦
𝑑𝑧2
−
𝑑𝑦
𝑑𝑧
.
𝑑
𝑑𝑥
1
𝑥2
=
1
𝑥2
.
𝑑𝑧
𝑑𝑥
.
𝑑
𝑑𝑧
𝑑2 𝑦
𝑑𝑧2
−
𝑑𝑦
𝑑𝑧
+
𝑑2 𝑦
𝑑𝑧2
−
𝑑𝑦
𝑑𝑧
.
𝑑
𝑑𝑥
𝑥−2
=
1
𝑥2
.
1
𝑥
.
𝑑3 𝑦
𝑑𝑧3
−
𝑑2 𝑦
𝑑𝑧2
+
𝑑2 𝑦
𝑑𝑧2
−
𝑑𝑦
𝑑𝑧
. −2𝑥−3
=
1
𝑥3
.
𝑑3 𝑦
𝑑𝑧3
−
𝑑2 𝑦
𝑑𝑧2
+
𝑑2 𝑦
𝑑𝑧2
−
𝑑𝑦
𝑑𝑧
. −2
1
𝑥3

7. =
1
𝑥3
.
𝑑3
𝑦
𝑑𝑧3
−
𝑑2
𝑦
𝑑𝑧2
− 2
1
𝑥3
.
𝑑2
𝑦
𝑑𝑧2
−
𝑑𝑦
𝑑𝑧
=
1
𝑥3
.
𝑑3 𝑦
𝑑𝑧3
−
𝑑2 𝑦
𝑑𝑧2
− 2
𝑑2 𝑦
𝑑𝑧2
+ 2
𝑑𝑦
𝑑𝑧
𝑜𝑟,
𝑑3 𝑦
𝑑𝑥3
=
1
𝑥3
(𝐷3 𝑦 − 𝐷2 𝑦 − 2𝐷2 𝑦 + 2𝐷𝑦)
𝑜𝑟, 𝑥3
𝑑3 𝑦
𝑑𝑥3
= 𝐷3 𝑦 − 𝐷2 𝑦 − 2𝐷2 𝑦 + 2𝐷𝑦 = 𝐷2 𝑦(𝐷 − 1) − 2𝐷𝑦(𝐷 − 1)
𝑜𝑟, 𝑥3
𝑑3
𝑦
𝑑𝑥3
= 𝐷 − 1 𝐷2
𝑦 − 2𝐷𝑦 = 𝐷 − 1 𝐷 𝑦 (𝐷 − 2)
Therefore
𝑥3
𝑑3 𝑦
𝑑𝑥3
= 𝐷 𝐷 − 1 (𝐷 − 2)𝑦

8. 1. Solve: 𝑥2
𝑑2 𝑦
𝑑𝑥2
− 4𝑥
𝑑𝑦
𝑑𝑥
+ 6𝑦 = 𝑥
Solution: Given equation is
𝑥2
𝑑2 𝑦
𝑑𝑥2
− 4𝑥
𝑑𝑦
𝑑𝑥
+ 6𝑦 = 𝑥 … … (1)
putting 𝑥 = 𝑒 𝑧, 𝐷 ≡
𝑑
𝑑𝑧
, 𝑥
𝑑𝑦
𝑑𝑥
= 𝐷𝑦 and𝑥2
𝑑2 𝑦
𝑑𝑥2
= 𝐷 𝐷 − 1 𝑦 𝑖𝑛 1 , 𝑤𝑒 𝑔𝑒𝑡
𝐷 𝐷 − 1 𝑦 − 4𝐷𝑦 + 6𝑦 = 𝑒 𝑧
𝑜𝑟, 𝐷2 − 5𝐷 + 6 𝑦 = 𝑒 𝑧
which is a linear equation in y

9. Let 𝑦 = 𝑒 𝑚𝑧
be a trial solution of 𝐷2
− 5𝐷 + 6 𝑦 = 0
Then we get 𝐷2
𝑦 − 5𝐷𝑦 + 6 𝑦 = 0
𝑜𝑟, 𝐷2 𝑒 𝑚𝑧 − 5𝐷𝑒 𝑚𝑧 + 6𝑒 𝑚𝑧 = 0
𝑜𝑟, 𝑚2 𝑒 𝑚𝑧 − 5𝑚𝑒 𝑚𝑧 + 6𝑒 𝑚𝑧 = 0
𝑜𝑟, (𝑚2−5𝑚 + 6) 𝑒 𝑚𝑧 = 0
Then the A.E. (Auxiliary Equation) is
𝑚2 − 5𝑚 + 6 = 0 , 𝑠𝑖𝑛𝑐𝑒 𝑒 𝑚𝑧 ≠ 0
𝑜𝑟, 𝑚2 − 2𝑚 − 3𝑚 + 6 = 0
𝑜𝑟, 𝑚 𝑚 − 2 − 3 𝑚 − 2 = 0
𝑜𝑟, 𝑚 − 2 𝑚 − 3 = 0
𝑜𝑟, 𝑚 = 2,3

10. Then C.F.= 𝑐1 𝑒 𝑚1 𝑧 + 𝑐2 𝑒 𝑚2 𝑧 = 𝑐1 𝑒2𝑧 + 𝑐2 𝑒3𝑧
𝑁𝑜𝑤 𝑃. 𝐼. =
1
𝐷2 − 5𝐷 + 6
𝑒 𝑧
=
1
12 − 5.1 + 6
𝑒 𝑧
=
1
2
𝑒 𝑧
Hence the complete solution is 𝑦 = 𝐶. 𝐹. +𝑃. 𝐼. = 𝑐1 𝑒2𝑧
+ 𝑐2 𝑒3𝑧
+
1
2
𝑒 𝑧
where 𝑧 = log 𝑥

11. 2. Solve: 𝑥2
𝑑2
𝑦
𝑑𝑥2
− 2𝑥
𝑑𝑦
𝑑𝑥
− 4𝑦 = 𝑥4
Solution: Given equation is
𝑥2
𝑑2 𝑦
𝑑𝑥2
− 2𝑥
𝑑𝑦
𝑑𝑥
− 4𝑦 = 𝑥4 … … (1)
putting 𝑥 = 𝑒 𝑧, 𝐷 ≡
𝑑
𝑑𝑧
, 𝑥
𝑑𝑦
𝑑𝑥
= 𝐷𝑦 and 𝑥2
𝑑2 𝑦
𝑑𝑥2
= 𝐷 𝐷 − 1 𝑦 𝑖𝑛 1 , 𝑤𝑒 𝑔𝑒𝑡
𝐷 𝐷 − 1 𝑦 − 2𝐷𝑦 − 4𝑦 = 𝑒4𝑧
𝑜𝑟, 𝐷2 − 3𝐷 − 4 𝑦 = 𝑒4𝑧
which is a linear equation in y

12. Let 𝑦 = 𝑒 𝑚𝑧 be a trial solution of 𝐷2 − 3𝐷 − 4 𝑦 = 0
Then we get 𝐷2 𝑦 − 3𝐷𝑦 − 4 𝑦 = 0
𝑜𝑟, 𝐷2 𝑒 𝑚𝑧 − 3𝐷𝑒 𝑚𝑧 − 4𝑒 𝑚𝑧 = 0
𝑜𝑟, 𝑚2 𝑒 𝑚𝑧 − 3𝑚𝑒 𝑚𝑧 − 4𝑒 𝑚𝑧 = 0
𝑜𝑟, (𝑚2
−3𝑚 − 4) 𝑒 𝑚𝑧
= 0
Then the A.E. (Auxiliary Equation) is
𝑚2 − 3𝑚 − 4 = 0 , 𝑠𝑖𝑛𝑐𝑒 𝑒 𝑚𝑧 ≠ 0
𝑜𝑟, 𝑚2
− 4𝑚 + 𝑚 − 4 = 0
𝑜𝑟, 𝑚 𝑚 − 4 + 1 𝑚 − 4 = 0
𝑜𝑟, 𝑚 − 4 𝑚 + 1 = 0
𝑜𝑟, 𝑚 = 4, −1

13. Then C.F.= 𝑐1 𝑒 𝑚1 𝑧 + 𝑐2 𝑒 𝑚2 𝑧 = 𝑐1 𝑒4𝑧 + 𝑐2 𝑒−𝑧
𝑁𝑜𝑤 𝑃. 𝐼. =
1
𝐷2 − 3𝐷 − 4
𝑒4𝑧 [case of failure]
= 𝑧
1
2𝐷 − 3
𝑒4𝑧 [multiplying by z and differentiating w. r. t. D]
= 𝑧
1
2. 4 − 3
𝑒4𝑧
=
1
5
𝑧 𝑒4𝑧
Hence the complete solution is 𝑦 = 𝐶. 𝐹. +𝑃. 𝐼. = 𝑐1 𝑒4𝑧 + 𝑐2 𝑒−𝑧 +
1
5
𝑧 𝑒4𝑧
where 𝑧 = log 𝑥

14. 3. Solve: 𝑥3
𝑑3 𝑦
𝑑𝑥3
− 𝑥2
𝑑2 𝑦
𝑑𝑥2
+ 2𝑥
𝑑𝑦
𝑑𝑥
− 2𝑦 = 𝑥3 + 3𝑥
Solution: Given that
𝑥3
𝑑3 𝑦
𝑑𝑥3
− 𝑥2
𝑑2 𝑦
𝑑𝑥2
+ 2𝑥
𝑑𝑦
𝑑𝑥
− 2𝑦 = 𝑥3 + 3𝑥 … … (1)
putting 𝑥 = 𝑒 𝑧, 𝐷 ≡
𝑑
𝑑𝑧
, 𝑥
𝑑𝑦
𝑑𝑥
= 𝐷𝑦, 𝑥2
𝑑2 𝑦
𝑑𝑥2
= 𝐷 𝐷 − 1 𝑦 𝑎𝑛𝑑
𝑥3
𝑑3 𝑦
𝑑𝑥3
= 𝐷 𝐷 − 1 𝐷 − 2 𝑦 𝑖𝑛 1 , 𝑤𝑒 𝑔𝑒𝑡
𝐷 𝐷 − 1 𝐷 − 2 𝑦 − 𝐷 𝐷 − 1 𝑦 + 2𝐷𝑦 − 2𝑦 = 𝑒3𝑧 + 3𝑒 𝑧
𝑜𝑟, 𝐷3 − 4𝐷2 + 5𝐷 − 2 𝑦 = 𝑒3𝑧 + 3𝑒 𝑧 [on simplification]
which is a linear equation in y

15. Let 𝑦 = 𝑒 𝑚𝑧 be a trial solution of 𝐷3 − 4𝐷2 + 5𝐷 − 2 𝑦 = 0
Then we get 𝐷3 𝑦 − 4𝐷2 𝑦 + 5𝐷𝑦 − 2𝑦 = 0
𝑜𝑟, 𝐷3 𝑒 𝑚𝑧 − 4𝐷2 𝑒 𝑚𝑧 + 5𝐷𝑒 𝑚𝑧 − 2𝑒 𝑚𝑧 = 0
𝑜𝑟, 𝑚3 𝑒 𝑚𝑧 − 4𝑚2 𝑒 𝑚𝑧 + 5𝑚𝑒 𝑚𝑧 − 2𝑒 𝑚𝑧 = 0
𝑜𝑟, ( 𝑚3−4𝑚2 + 5𝑚 − 2) 𝑒 𝑚𝑧 = 0
Then the A.E. (Auxiliary Equation) is
𝑚3−4𝑚2 + 5𝑚 − 2 = 0, 𝑠𝑖𝑛𝑐𝑒 𝑒 𝑚𝑧 ≠ 0
𝑜𝑟, 𝑚3−𝑚2 − 3𝑚2 + 3𝑚 + 2𝑚 − 2 = 0
𝑜𝑟, 𝑚2
𝑚 − 1 − 3𝑚 𝑚 − 1 + 2 𝑚 − 1 = 0
𝑜𝑟, 𝑚 − 1 𝑚2 − 3𝑚 + 2 = 0
𝑜𝑟, 𝑚 − 1 𝑚 − 1 (𝑚 − 2) = 0
𝑜𝑟, 𝑚 = 1, 1, 2

16. Then C.F.= (𝑐1+𝑐2 𝑥) 𝑒 𝑚1 𝑧 + 𝑐3 𝑒 𝑚3 𝑧 = (𝑐1+𝑐2 𝑥) 𝑒 𝑧 + 𝑐3 𝑒2𝑧
𝑁𝑜𝑤 𝑃. 𝐼. =
1
𝐷3 − 4𝐷2 + 5𝐷 − 2
(𝑒3𝑧
+ 3𝑒 𝑧
)
=
1
𝐷3 − 4𝐷2 + 5𝐷 − 2
𝑒3𝑧
+
1
𝐷3 − 4𝐷2 + 5𝐷 − 2
3𝑒 𝑧
=
1
33 − 4 . 32 + 5 . 3 − 2
𝑒3𝑧 + 𝑧
1
3𝐷2 − 8 𝐷 + 5
3𝑒 𝑧
=
1
4
𝑒3𝑧
+ 𝑧2
1
6 𝐷 − 8
3𝑒 𝑧
=
1
4
𝑒3𝑧
+ 𝑧2
1
6 . 1 − 8
3𝑒 𝑧
=
1
4
𝑒3𝑧
−
3
2
𝑧2
𝑒 𝑧
Hence the complete solution is 𝑦 = (𝑐1+𝑐2 𝑥) 𝑒 𝑧 + 𝑐3 𝑒2𝑧 +
1
4
𝑒3𝑧 −
3
2
𝑧2 𝑒 𝑧
where 𝑧 = log 𝑥

17. 4. Solve: 𝑥2
𝑑2 𝑦
𝑑𝑥2
− 𝑥
𝑑𝑦
𝑑𝑥
+ 4𝑦 = cos log 𝑥 + 𝑥2
sin(log 𝑥)
Solution: Given that
𝑥2
𝑑2
𝑦
𝑑𝑥2
− 𝑥
𝑑𝑦
𝑑𝑥
+ 4𝑦 = cos log 𝑥 + 𝑥2 sin log 𝑥 … . . (𝑖)
putting 𝑥 = 𝑒 𝑧, 𝐷 ≡
𝑑
𝑑𝑧
, 𝑥
𝑑𝑦
𝑑𝑥
= 𝐷𝑦 and 𝑥2
𝑑2 𝑦
𝑑𝑥2
= 𝐷 𝐷 − 1 𝑦 𝑖𝑛 𝑖 , 𝑤𝑒 𝑔𝑒𝑡
𝐷 𝐷 − 1 𝑦 − 𝐷𝑦 + 4𝑦 = cos log 𝑒 𝑧 + 𝑒2𝑧 sin(log 𝑒 𝑧)
𝑜𝑟, 𝐷2 − 2𝐷 + 4 𝑦 = cos 𝑧 log 𝑥 + 𝑒2𝑧 sin 𝑧 log 𝑒 = cos 𝑧 + 𝑒2𝑧 sin 𝑧
which is a linear equation in y

18. Let 𝑦 = 𝑒 𝑚𝑧 be a trial solution of 𝐷2 − 2𝐷 + 4 𝑦 = 0
Then we get 𝐷2 𝑦 − 2𝐷𝑦 + 4 𝑦 = 0
𝑜𝑟, 𝐷2 𝑒 𝑚𝑧 − 2𝐷𝑒 𝑚𝑧 + 4𝑒 𝑚𝑧 = 0
𝑜𝑟, 𝑚2 𝑒 𝑚𝑧 − 2𝑚𝑒 𝑚𝑧 + 4𝑒 𝑚𝑧 = 0
𝑜𝑟, (𝑚2
−2𝑚 + 4) 𝑒 𝑚𝑧
= 0
Then the A.E. (Auxiliary Equation) is
𝑚2 − 2𝑚 + 4 = 0 , 𝑠𝑖𝑛𝑐𝑒 𝑒 𝑚𝑧 ≠ 0
𝑜𝑟, 𝑚2 − 2𝑚 + 1 + 3 = 0
𝑜𝑟, 𝑚2 −2𝑚 + 1 = −3
𝑜𝑟, 𝑚 − 1 2 = 3 𝑖2
𝑜𝑟, 𝑚 − 1 = ± 3 𝑖
𝑜𝑟, 𝑚 = 1 ± 3 𝑖 [ℎ𝑒𝑟𝑒 α = 1, 𝛽 = 3]

19. Then C.F.= 𝑒 𝛼 𝑧
(𝑐1 cos 𝛽𝑧 + 𝑐2 sin 𝛽𝑧) = 𝑒 𝑧
(𝑐1 cos 3𝑧 + 𝑐2 sin 3𝑧)
𝑁𝑜𝑤 𝑃. 𝐼. =
1
𝐷2 − 2𝐷 + 4
cos 𝑧 + 𝑒2𝑧
sin 𝑧
=
1
𝐷2 − 2𝐷 + 4
cos 𝑧 +
1
𝐷2 − 2𝐷 + 4
𝑒2𝑧 sin 𝑧
=
1
−12 − 2𝐷 + 4
cos 𝑧 + 𝑒2𝑧
1
(𝐷 + 2)2−2(𝐷 + 2) + 4
sin 𝑧
=
1
3 − 2𝐷
cos 𝑧 + 𝑒2𝑧
1
𝐷2 + 2𝐷 + 4
sin 𝑧
=
3 + 2𝐷
3 + 2𝐷 (3 − 2𝐷)
cos 𝑧 + 𝑒2𝑧
1
−12 + 2𝐷 + 4
sin 𝑧
=
3 + 2𝐷
9 − 4𝐷2
cos 𝑧 + 𝑒2𝑧
1
3 + 2𝐷
sin 𝑧

20. =
3 + 2𝐷
9 − 4(−12)
cos 𝑧 + 𝑒2𝑧
3 − 2𝐷
3 + 2𝐷 (3 − 2𝐷)
sin 𝑧
=
1
9 + 4
(3𝑐𝑜𝑠 𝑧 + 2𝐷 𝑐𝑜𝑠 𝑧) +𝑒2𝑧
3 − 2𝐷
9 − 4𝐷2
sin 𝑧
=
1
13
3𝑐𝑜𝑠 𝑧 − 2 sin 𝑧 + 𝑒2𝑧
3 − 2𝐷
9 − 4(−12)
sin 𝑧
=
1
13
3𝑐𝑜𝑠 𝑧 − 2 sin 𝑧 + 𝑒2𝑧
1
9 + 4
(3 sin 𝑧 − 2𝐷 sin 𝑧)
=
1
13
3𝑐𝑜𝑠 𝑧 − 2 sin 𝑧 + 𝑒2𝑧
1
13
(3 sin 𝑧 − 2 cos 𝑧)
Hence the complete solution is
𝑦 = 𝑒 𝑧 𝑐1 cos 3𝑧 + 𝑐2 sin 3𝑧 +
1
13
3𝑐𝑜𝑠 𝑧 − 2 sin 𝑧 + 𝑒2𝑧 1
13
(3 sin 𝑧 − 2 cos 𝑧)
where 𝑧 = log 𝑥

21. 5. Solve: 𝑥2
𝑑2 𝑦
𝑑𝑥2
+ 2𝑥
𝑑𝑦
𝑑𝑥
= log 𝑥
Solution: Given equation is
𝑥2
𝑑2 𝑦
𝑑𝑥2
+ 2𝑥
𝑑𝑦
𝑑𝑥
= log 𝑥 … … (𝑖)
putting 𝑥 = 𝑒 𝑧
, 𝐷 ≡
𝑑
𝑑𝑧
, 𝑥
𝑑𝑦
𝑑𝑥
= 𝐷𝑦 and𝑥2
𝑑2 𝑦
𝑑𝑥2
= 𝐷 𝐷 − 1 𝑦 𝑖𝑛 𝑖 , 𝑤𝑒 𝑔𝑒𝑡
𝐷 𝐷 − 1 𝑦 + 2𝐷𝑦 = log 𝑒 𝑧
𝑜𝑟, 𝐷2
+ 𝐷 𝑦 = 𝑧 log 𝑒
𝑜𝑟, 𝐷2 + 𝐷 𝑦 = 𝑧
which is a linear equation in y

22. Let 𝑦 = 𝑒 𝑚𝑧
be a trial solution of 𝐷2
+ 𝐷 𝑦 = 0
Then we get 𝐷2 𝑦 + 𝐷𝑦 = 0
𝑜𝑟, 𝐷2 𝑒 𝑚𝑧 + 𝐷𝑒 𝑚𝑧 = 0
𝑜𝑟, 𝑚2 𝑒 𝑚𝑧 + 𝑚𝑒 𝑚𝑧 = 0
𝑜𝑟, (𝑚2
+𝑚) 𝑒 𝑚𝑧
= 0
Then the A.E. (Auxiliary Equation) is
𝑚2 + 𝑚 = 0 , 𝑠𝑖𝑛𝑐𝑒 𝑒 𝑚𝑧 ≠ 0
𝑜𝑟, 𝑚(𝑚 + 1) = 0
𝑜𝑟, 𝑚 = 0, −1
Then C.F.= 𝑐1 𝑒 𝑚1 𝑧
+ 𝑐2 𝑒 𝑚2 𝑧
= 𝑐1 + 𝑐2 𝑒−𝑧

23. 𝑁𝑜𝑤 𝑃. 𝐼. =
1
𝐷2 + 𝐷
𝑧 =
1
𝐷 (1 + 𝐷)
𝑒 𝑧
=
1
𝐷
(1 + 𝐷)−1
𝑧 =
1
𝐷
1 − 𝐷 − ⋯ 𝑧
=
1
𝐷
𝑧 − 1 =
𝑧2
2
− 𝑧
Hence the complete solution is 𝑦 = 𝑐1 + 𝑐2 𝑒−𝑧 +
𝑧2
2
− 𝑧
where 𝑧 = log 𝑥