Unit I - Linear Differential Equations and Applications

A differential Equation which contains the *differential

coefficients * and dependent variable in the first degree,
does not involve the product of a derivative with another
derivative or with dependent variable, and in which the
Coefficients are constants is called a Linear differential
equation with constant coefficients.
(*differential coefficients * means derivatives)
The general form of such a differential equation of order 𝑛
is
𝐝𝐧 𝐲 𝐝𝐧−𝟏 𝐲 𝐝𝐧−𝟐 𝐲 𝐝𝐲
𝐚𝟎 𝐧 + 𝐚𝟏 𝐧−𝟏 + 𝐚𝟐 𝐧−𝟐 + ⋯+ 𝐚𝐧−𝟏 + 𝐚𝐧 𝐲 = 𝐟 𝐱
𝐝𝐱 𝐝𝐱 𝐝𝐱 𝐝𝐱
 𝐝𝐧 𝐲 𝐝𝐧−𝟏 𝐲 𝐝𝐧−𝟐 𝐲 𝐝𝐲
𝐚𝟎 𝐧 + 𝐚𝟏 𝐧−𝟏 + 𝐚𝟐 𝐧−𝟐 + ⋯ + 𝐚𝐧−𝟏 + 𝐚𝐧 𝐲 = 𝐟 𝐱
𝐝𝐱 𝐝𝐱 𝐝𝐱 𝐝𝐱
𝑑
D stands for 𝐷𝑦 =
𝑑𝑦
𝑑𝑥 𝑑𝑥
𝑑2𝑦 𝑑3𝑦
=𝐷 2 𝑦 =𝐷 3 𝑦 & so on…
𝑑𝑥 2 𝑑𝑥 3

𝑎0 𝐷𝑛 𝑦 + 𝑎1 𝐷𝑛−1 𝑦 + 𝑎2 𝐷𝑛−2 𝑦 + ⋯ + 𝑎𝑛−1 𝐷𝑦 + 𝑎𝑛 𝑦 = 𝑓 𝑥

𝑎0 𝐷𝑛 + 𝑎1 𝐷𝑛−1 + 𝑎2 𝐷𝑛−2 + ⋯ + 𝑎𝑛−1 𝐷 + 𝑎𝑛 𝑦 = 𝑓 𝑥

∅ 𝐷 𝑦 = 𝑓 𝑥 Where
∅ 𝐷 = 𝑎0 𝐷𝑛 + 𝑎1 𝐷𝑛−1 + 𝑎2 𝐷𝑛−2 + ⋯ + 𝑎𝑛−1 𝐷 + 𝑎𝑛
Standard form of LDE: ∅ 𝑫 𝒚=𝒇 𝒙
∅ 𝑫 is a polynomial in 𝑫
Solution of LDE: 𝒚 = 𝒚𝒄 + 𝒚𝒑
𝑦𝑐 : Complimentary function
𝑦𝑝 : Particular Integral

Complimentary function (C.F.)𝑦𝑐 Particular Integral (P.I.) 𝑦𝑝

1)Involves 𝑛 arbitrary constants. 1)Does not involve any
arbitrary constant.
.
2)Only one method to find 2) Three methods to find
A) General method
B) Short cut method
C) Variation of
parameters method.
Method to find complimentary function 𝑦𝑐
Consider standard LDE ∅ 𝑫 𝒚 = 𝒇 𝒙
put 𝒇 𝒙 = 𝟎 ⇒ ∅ 𝑫 𝒚 = 𝟎 ⇒ ∅ 𝑫 = 𝟎

The equation ∅ 𝑫 = 𝟎 is called as an auxiliary equation

Recall ∅ 𝑫 is a polynomial 𝐷
∅ 𝐷 =0
⇒ 𝑎0 𝐷𝑛 + 𝑎1 𝐷𝑛−1 + 𝑎2 𝐷𝑛−2 + ⋯ + 𝑎𝑛−1 𝐷 + 𝑎𝑛 = 0
Factories this polynomial .It is 𝑛𝑡ℎ degree polynomial so it
has 𝑛 roots.
𝐷 − 𝑚1 𝐷 − 𝑚2 … 𝐷 − 𝑚𝑛 = 0
𝑚1 , 𝑚2 ,… 𝑚𝑛 are roots of auxiliary equation ∅ 𝐷 = 0
Case 1-Real and distinct roots

𝑑2 𝑦 𝑑𝑦
2
−5 − 6𝑦 = 0
𝑑𝑥 𝑑𝑥

𝐷2 𝑦 − 5𝐷𝑦 − 6𝑦 = 0
𝐷2 − 5𝐷 − 6 𝑦 = 0
AE is 𝐷2 − 5𝐷 − 6 = 0
𝐷2 − 6𝐷 + 𝐷 − 6 = 0
𝐷 𝐷−6 +1 𝐷−6 =0
𝐷−6 𝐷+1 =0
𝐷 = 6, −1

𝑦𝑐 = 𝑐1 𝑒 6𝑥 + 𝑐2 𝑒 −𝑥
Case 2-Repeated real roots
For 𝑚1 = 𝑚2 ,
𝑦 = 𝑐1 𝑥 + 𝑐2 𝑒 𝑚1 𝑥 + 𝑐3 𝑒 𝑚3 𝑥 + ⋯ + 𝑐𝑛 𝑒 𝑚𝑛 𝑥
For 𝑚1 = 𝑚2 = 𝑚3 ,
𝑦 = 𝑐1 𝑥 2 + 𝑐2 𝑥 + 𝑐3 𝑒 𝑚1𝑥 + 𝑐4 𝑒 𝑚4𝑥 + ⋯ + 𝑐𝑛 𝑒 𝑚𝑛 𝑥
𝐷4 − 2𝐷3 + 𝐷2 𝑦 = 0
AE is 𝐷4 − 2𝐷3 + 𝐷2 = 0
𝐷4 − 𝐷3 − 𝐷3 + 𝐷2 = 0
𝐷3 𝐷 − 1 − 𝐷2 𝐷 − 1 = 0
𝐷3 − 𝐷2 𝐷 − 1 = 0
𝐷2 𝐷 − 1 𝐷 − 1 = 0
𝐷 = 0,0,1,1
𝑦𝑐 = 𝑐1 𝑥 + 𝑐2 𝑒 0𝑥 + 𝑐3 𝑥 + 𝑐4 𝑒 𝑥
𝑦𝑐 = 𝑐1 𝑥 + 𝑐2 + 𝑐3 𝑥 + 𝑐4 𝑒 𝑥
Case 3-Imaginary (complex) roots
For 𝐷 = 𝛼 ± 𝑖𝛽 , 𝑦 = 𝑒 𝛼𝑥 𝑐1 cos 𝛽𝑥 + 𝑐2 sin 𝛽𝑥

4𝑦 ′′ − 8𝑦 ′ + 7𝑦 = 0
AE IS 4𝐷2 − 8𝐷 + 7 = 0

8 ± 64 − 112 8 ± −48 8 ± 4 3𝑖 3
𝐷= = = =1± 𝑖
8 8 8 2

3
𝛼 = 1, 𝛽 =
2
𝑥
3 3
𝑦=𝑒 𝑐1 cos 𝑥 + 𝑐2 sin 𝑥
2 2
Case 4- Repeated imaginary (complex) roots
Let 𝐷 = 𝛼 ± 𝑖𝛽 be repeated twice
𝑦 = 𝑒 𝛼𝑥 𝑐1 𝑥 + 𝑐2 cos 𝛽𝑥 + 𝑐3 𝑥 + 𝑐4 sin 𝛽𝑥

𝐷2 + 16 2 = 0
𝐷2 + 16 𝐷2 + 16 = 0
𝐷2 + 16 = 0 and 𝐷2 + 16 = 0
𝐷 = ±4i and 𝐷 = ±4i
𝐷 = ±4𝑖, ±4𝑖
𝛼 =0 ,𝛽 =4

𝑦 = 𝑒 0𝑥 𝑐1 𝑥 + 𝑐2 cos 4𝑥 + 𝑐3 𝑥 + 𝑐4 sin 4𝑥

𝑦= 𝑐1 𝑥 + 𝑐2 cos 4𝑥 + 𝑐3 𝑥 + 𝑐4 sin 4𝑥
1 (𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏 2 𝐷 + 1 2 = 𝐷2 + 2𝐷 + 1
2 (𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏 2 𝐷 − 1 2 = 𝐷2 − 2𝐷 + 1
3 𝑎2 − 𝑏 2 = 𝑎 − 𝑏 (𝑎 + 𝑏) 𝐷2 − 1 = (𝐷 − 1)(𝐷 + 1)
4 (𝑎 + 𝑏)3 𝐷+1 3
= 𝑎3 + 3𝑎2 𝑏 + 3𝑎𝑏 2 + 𝑏 3 = 𝐷3 + 3𝐷2 + 3𝐷 + 1
5 (𝑎 − 𝑏)3 𝐷−1 3
= 𝑎3 − 3𝑎2 𝑏 + 3𝑎𝑏 2 − 𝑏 3 = 𝐷3 − 3𝐷2 + 3𝐷 − 1
6 𝑎3 − 𝑏3 𝐷3 − 1
= (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏 2 ) = (𝐷 − 1)(𝐷2 + 𝐷 + 1)
7 𝑎3 + 𝑏3 = 𝐷3 + 1
(𝑎 + 𝑏)(𝑎2 − 𝑎𝑏 + 𝑏 2 ) = (𝐷 + 1)(𝐷2 − 𝐷 + 1)
 1
Particular integral 𝑦𝑝 = 𝑓 𝑥
∅ 𝐷
Methods to find particular integral 𝑦𝑝
A) General method
B) Short-cut method
C) Variation of parameters method

General method Short-cut method

1)Factorize ∅ 𝐷 into linear 1)Keep ∅ 𝐷 as a polynomial
factors only
2)Integration is involved. No 2)Applicable to few standard
constraint on 𝑓 𝑥 but functions only.
becomes useful provided
integration is simple
 1
𝑦𝑝 = 𝑓 𝑥
∅ 𝐷
A) General method: Factorize ∅ 𝐷 into linear factors
at a time we apply only one linear factor
1
𝑃𝐼 = 𝑦𝑝 = 𝑓 𝑥 = 𝑒 𝑚𝑥 𝑒 −𝑚𝑥 𝑓 𝑥 𝑑𝑥
𝐷−𝑚
1
𝑃𝐼 = 𝑦𝑝 = 𝑓 𝑥 = 𝑒 −𝑚𝑥 𝑒 𝑚𝑥 𝑓 𝑥 𝑑𝑥
𝐷+𝑚
Special case put 𝑚 = 0 in any one of above two
1
formulae: 𝑓 𝑥 = 𝑓 𝑥 𝑑𝑥
𝐷
1
Similarly 𝑓 𝑥 = 𝑓 𝑥 𝑑𝑥 𝑑𝑥 & so on
𝐷2
 1
𝑃𝐼 = 𝑦𝑝 = 𝑓 𝑥 = 𝑒 −𝑚𝑥 𝑒 𝑚𝑥 𝑓 𝑥 𝑑𝑥
𝐷+𝑚
d2 y dy e x
+ 3 + 2y = e
dx2 dx
2 ex
𝐷 + 3𝐷 + 2 𝑦 = e … Std form of LDE
AE is 𝐷2 + 3𝐷 + 2 = 0 ⇒ 𝐷 + 2 𝐷 + 1 = 0 ⇒ 𝐷 = −1, −2 ⇒
𝑦𝑐 = c1 e−x + c2 e−2x
1 e x 1 1 e x
𝑦𝑝 = e = e …(1)
𝐷+2 𝐷+1 𝐷+2 𝐷+1
1 x x
ee = 𝑒 −𝑥 𝑒 𝑥 ee 𝑑𝑥 put 𝑒 𝑥 = 𝑡 ⇒ 𝑒 𝑥 𝑑𝑥 = 𝑑𝑡
𝐷+1
x
= 𝑒 −𝑥𝑒 𝑡 𝑑𝑡 = 𝑒 −𝑥 𝑒 𝑡 = 𝑒 −𝑥 ee put in (1)
1 x x 𝑥
𝑦𝑝 = 𝑒 −𝑥 ee = 𝑒 −2𝑥 𝑒 2𝑥 𝑒 −𝑥 ee = 𝑒 −2𝑥 𝑒 𝑥 𝑒 𝑒 𝑑𝑥
𝐷+2
−2𝑥 𝑒 𝑥 𝑥
𝑦𝑝 = 𝑒 𝑒 Now 𝑦 = 𝑦𝑐 + 𝑦𝑝 ⇒ 𝑦 = c1 e−x + c2 e−2x + −2𝑥
𝑒 𝑒 𝑒
 2
1 −𝑥
𝐷 − 3𝐷 + 2 𝑦 = 𝑒 −𝑥
+ cos 𝑒
𝑒
AE is 𝐷2 − 3𝐷 + 2 = 0 ⇒ 𝐷 − 2 𝐷 − 1 = 0
𝐷 = 2,1 ⇒ 𝑦𝑐 = 𝑐1 𝑒 2𝑥 + 𝑐2 𝑒 𝑥
1 1
𝑦𝑝 = 𝑒−𝑥 + cos 𝑒 −𝑥 consider
𝐷−2 𝐷−1 𝑒
1 1 −𝑥 𝑥 −𝑥 1 −𝑥
𝐼= −𝑥 + cos 𝑒 =𝑒 𝑒 𝑒−𝑥 + cos 𝑒 𝑑𝑥
𝐷−1 𝑒 𝑒 𝑒
−𝑥
Put 𝑒 = t ⇒ −𝑒 −𝑥 𝑑𝑥 = 𝑑𝑡
1
𝐼 = 𝑒𝑥 + cos 𝑡 −𝑑𝑡 = −𝑒 𝑥 𝑒 −𝑡 + cos 𝑡 𝑑𝑡
𝑒𝑡
𝐼 = −𝑒 𝑥 −𝑒 −𝑡 + sin 𝑡 = 𝑒 𝑥 𝑒 −𝑡 − sin 𝑡
𝑥
1 −𝑥
=𝑒 −𝑥 − sin 𝑒
𝑒𝑒
 1 𝑥
1 −𝑥
𝑦𝑝 = 𝑒 −𝑥 − sin 𝑒
𝐷−2 𝑒𝑒
2𝑥 −2𝑥 𝑥
1 −𝑥
=𝑒 𝑒 𝑒 𝑒 −𝑥 − sin 𝑒 𝑑𝑥
𝑒
1
= 𝑒 2𝑥 𝑒 −𝑥 𝑒−𝑥 − sin 𝑒 −𝑥 𝑑𝑥
𝑒
Put 𝑒 −𝑥 = t ⇒ −𝑒 −𝑥 𝑑𝑥 = 𝑑𝑡
2𝑥
1 2𝑥
=𝑒 − sin 𝑡 −𝑑𝑡 = −𝑒 𝑒 −𝑡 − sin 𝑡 𝑑𝑡
𝑒𝑡
= −𝑒 2𝑥 −𝑒 −𝑡 + cos 𝑡 = 𝑒 2𝑥 𝑒 −𝑡 − cos 𝑡
2𝑥
1 2𝑥
1 −𝑥
=𝑒 𝑡
− cos 𝑡 = 𝑒 𝑒 −𝑥 − cos 𝑒
𝑒 𝑒
2𝑥 𝑥 2𝑥
1 −𝑥
𝑦 = 𝑦𝑐 + 𝑦𝑝 = 𝑐1 𝑒 + 𝑐2 𝑒 + 𝑒 −𝑥 − cos 𝑒
𝑒𝑒
B) Short-cut method: keep ∅ 𝐷 as a polynomial only
Applicable only to few standard functions.
∅ 𝑫 𝒚 = 𝒇 𝒙 standard form of LDE
1
𝑦𝑝 = 𝑓 𝑥
∅ 𝐷
1) 𝒇 𝒙 = 𝒆𝒂𝒙 where 𝑎 is a constant value.
1 1 1
𝑦𝑝 = 𝒇 𝒙 = 𝒆𝒂𝒙 = 𝒆𝒂𝒙 [Provided ∅ 𝒂 ≠ 𝟎]
∅ 𝑫 ∅ 𝑫 ∅ 𝒂
1
If ∅ 𝒂 = 𝟎 then 𝑦𝑝 = 𝑥 ′ 𝒆𝒂𝒙 [Provided ∅′ 𝒂 ≠ 𝟎]
∅ 𝒂
′ 2 1
If ∅ 𝒂 = 𝟎 then 𝑦𝑝 = 𝑥 𝒆𝒂𝒙 Provided ∅′′ 𝒂 ≠ 𝟎 ]
∅′′ 𝒂
and so on
 1 1
𝒆𝒂𝒙 = 𝒆𝒂𝒙 [Provided ∅ 𝒂 ≠ 𝟎]
∅ 𝑫 ∅ 𝒂
Example: find 𝑦𝑝 of 𝐷3 − 𝐷2 + 8𝐷 − 4 𝑦 = 𝑒 𝑥 + 2
1 𝑥
𝑦𝑝 = 3 2
𝑒 +2
𝐷 − 𝐷 + 8𝐷 − 4
1 𝑥
1 0𝑥
= 3 𝑒 + 2 × 𝑒
𝐷 − 𝐷2 + 8𝐷 − 4 𝐷3 − 𝐷2 + 8𝐷 − 4
1 𝑥
1 0𝑥
= 3 𝑒 + 2 × 𝑒
𝐷 − 𝐷2 + 8𝐷 − 4 𝐷3 − 𝐷2 + 8𝐷 − 4
∅ 𝑫 = 𝐷3 − 𝐷2 + 8𝐷 − 4
for red expression 𝑎 = 1
⇒∅ 1 =1−1+8−4=4≠0
for green expression 𝑎 = 0
⇒ ∅ 0 = 0 − 0 + 0 − 4 = −4 ≠ 0
1 𝑥 1 0𝑥 𝑒 𝑥 1
𝑦𝑝 = 𝑒 + 2 × 𝑒 = −
4 −4 4 2
 1 1
𝒆𝒂𝒙 = 𝒆𝒂𝒙 [Provided ∅ 𝒂 ≠ 𝟎]
∅ 𝑫 ∅ 𝒂
1
If ∅ 𝒂 = 𝟎 then 𝑦𝑝 = 𝑥 ′ 𝒆𝒂𝒙 [Provided ∅′ 𝒂 ≠ 𝟎]
∅ 𝒂
find 𝑦𝑝 of 𝐷2 − 4 𝑦 = 𝑒 2𝑥
∅ 𝑫 = 𝐷2 − 4 here 𝑎 = 2

∅ 𝟐 =4−4=0

∅ 𝑫 = 𝐷2 − 4 ⇒ ∅′ 𝐷 = 2𝐷

Now 𝑎 = 2
∅′ 𝑎 = ∅′ 2 = 4 ≠ 0
1 𝒂𝒙
1 2𝑥
𝑦𝑝 = 𝑥 ′ 𝒆 =𝑥 𝑒
∅ 𝒂 4
2) 𝒇 𝒙 = 𝒂𝒙 𝒐𝒓 𝒂−𝒙 where 𝑎 is a constant value.
1 1 1
i) 𝑦𝑝 = 𝒇 𝒙 = 𝒂𝒙 = 𝒂𝒙
∅ 𝑫 ∅ 𝑫 ∅ log 𝒂
1 1 −𝒙 1
ii) 𝑦𝑝 = 𝒇 𝒙 = 𝒂 = 𝒂−𝒙
∅ 𝑫 ∅ 𝑫 ∅ − log 𝒂
3 𝑥
Example 1: Find 𝑦𝑝 of 𝐷 + 1 𝑦 = 3
1 𝑥
𝑦𝑝 = 3here 𝑎 = 3
𝐷3 +1
∅ 𝑫 = 𝐷3 + 1 ⇒ ∅ log 𝒂 = ∅ log 𝟑 = log 𝟑 𝟑 +𝟏
1 𝒙
1 𝑥
𝑦𝑝 = 𝒂 = 3
∅ log 𝒂 log 𝟑 𝟑 + 𝟏
Example 2: Find 𝑦𝑝 of 𝐷2 + 2𝐷 + 1 𝑦 = 5−𝑥
1
𝑦𝑝 = 2 5−𝑥
𝐷 + 2𝐷 + 1
1 −𝑥
= 5
− log 5 2 + 2 − log 5 + 1
3) 𝒇 𝒙 = sin 𝑎𝑥 + 𝑏
1 1
𝑦𝑝 = 2
sin 𝑎𝑥 + 𝑏 = 2
sin 𝑎𝑥 + 𝑏
∅ 𝐷 ∅ −𝑎
Provided ∅ −𝑎2 ≠ 0

If ∅ −𝑎2 = 0 then
1 1
𝑦𝑝 = 2
sin 𝑎𝑥 + 𝑏 = 𝑥 ′ 2
sin 𝑎𝑥 + 𝑏
∅ 𝐷 ∅ −𝑎
Provided ∅′ −𝑎2 ≠ 0
If ∅′ −𝑎2 = 0 then
1 2
1
𝑦𝑝 = 2
sin 𝑎𝑥 + 𝑏 = 𝑥 ′′ 2
sin 𝑎𝑥 + 𝑏
∅ 𝐷 ∅ −𝑎
Provided ∅′′ −𝑎2 ≠ 0 and so on
4) 𝒇 𝒙 = cos 𝑎𝑥 + 𝑏
1 1
𝑦𝑝 = 2
cos 𝑎𝑥 + 𝑏 = 2
cos 𝑎𝑥 + 𝑏
∅ 𝐷 ∅ −𝑎
Provided ∅ −𝑎2 ≠ 0

If ∅ −𝑎2 = 0 then
1 1
𝑦𝑝 = 2
cos 𝑎𝑥 + 𝑏 = 𝑥 ′ 2
cos 𝑎𝑥 + 𝑏
∅ 𝐷 ∅ −𝑎
Provided ∅′ −𝑎2 ≠ 0
If ∅′ −𝑎2 = 0 then
1 2
1
𝑦𝑝 = 2
cos 𝑎𝑥 + 𝑏 = 𝑥 ′′ 2
cos 𝑎𝑥 + 𝑏
∅ 𝐷 ∅ −𝑎
Provided ∅′′ −𝑎2 ≠ 0 and so on
Find 𝑦𝑝 of 𝐷3 + 6𝐷2 + 12𝐷 + 8 𝑦 = cos 2𝑥
1
𝑦𝑝 = cos 2𝑥 [ Note: 𝐷3 = 𝐷2 × 𝐷, 𝑎 = 2 ⇒ 𝑎2 = 4 ]
𝐷3 +6𝐷 2 +12𝐷+8
1
𝑦𝑝 = cos 2𝑥
−4 𝐷 + 6 −4 + 12𝐷 + 8
1
= cos 2𝑥
−4𝐷 − 24 + 12𝐷 + 8
1 1 1
= cos 2𝑥 = cos 2𝑥
8𝐷 − 16 8 𝐷−2
1 𝐷+2 1 𝐷+2
= cos 2𝑥 = 2
cos 2𝑥
8 𝐷−2 𝐷+2 8 𝐷 −4
1 𝐷+2 1 𝐷+2 −1
= cos 2𝑥 = cos 2𝑥 = 𝐷 + 2 cos 2𝑥
8 −4 − 4 8 −8 64
−1 −1
= 𝐷 cos 2𝑥 + 2 cos 2𝑥 = −2 sin 2𝑥 + 2 cos 2𝑥
64 64
1
𝑦𝑝 = sin 2𝑥 − cos 2𝑥
32
Find 𝑦𝑝 of 𝐷4 − 𝑚4 𝑦 = sin 𝑚𝑥
1
𝑦𝑝 = 4 4
sin 𝑚𝑥
𝐷 −𝑚
[ Note: 𝐷4 = 𝐷2 × 𝐷2 , 𝑎 = 𝑚 ⇒ 𝑎2 = 𝑚2 ]
1
𝑦𝑝 = sin 𝑚𝑥 put 𝐷2 = −𝑚2
𝐷2 𝐷2 −𝑚4
1 1
𝑦𝑝 = 2 2 4
sin 𝑚𝑥 = 4 4
sin 𝑚𝑥
−𝑚 −𝑚 − 𝑚 𝑚 −𝑚
∅ 𝐷 = 𝐷4 − 𝑚4 ⇒ ∅′ 𝐷 = 4𝐷3
1 1
𝑦𝑝 = 𝑥 3
sin 𝑚𝑥 = 𝑥 2
sin𝑚𝑥
4𝐷 4𝐷 −𝑚
−𝑥 1 −𝑥
= 2
sin 𝑚𝑥 = sin 𝑚𝑥
4𝑚 𝐷 4𝑚2
−𝑥 − cos 𝑚𝑥 𝑥
𝑦𝑝 = 2
= 3
cos 𝑚𝑥
4𝑚 𝑚 4𝑚
5) 𝒇 𝒙 = sinh 𝑎𝑥 + 𝑏
1 1
𝑦𝑝 = 2
sinh 𝑎𝑥 + 𝑏 = 2
sinh 𝑎𝑥 + 𝑏
∅ 𝐷 ∅ 𝑎
Provided ∅ 𝑎2 ≠ 0

If ∅ 𝑎2 = 0 then
1 1
𝑦𝑝 = 2
sinh 𝑎𝑥 + 𝑏 = 𝑥 ′ 2 sinh 𝑎𝑥 + 𝑏
∅ 𝐷 ∅ 𝑎
Provided ∅′ 𝑎2 ≠ 0
If ∅′ 𝑎2 = 0 then
1 2
1
𝑦𝑝 = 2
sinh 𝑎𝑥 + 𝑏 = 𝑥 ′′ 2 sinh 𝑎𝑥 + 𝑏
∅ 𝐷 ∅ 𝑎
Provided ∅′′ 𝑎2 ≠ 0 and so on
6) 𝒇 𝒙 = cosh 𝑎𝑥 + 𝑏
1 1
𝑦𝑝 = 2
cosh 𝑎𝑥 + 𝑏 = 2
cosh 𝑎𝑥 + 𝑏
∅ 𝐷 ∅ 𝑎
Provided ∅ 𝑎2 ≠ 0

If ∅ 𝑎2 = 0 then
1 1
𝑦𝑝 = 2
cosh 𝑎𝑥 + 𝑏 = 𝑥 ′ 2 cosh 𝑎𝑥 + 𝑏
∅ 𝐷 ∅ 𝑎
Provided ∅′ 𝑎2 ≠ 0
If ∅′ 𝑎2 = 0 then
1 2
1
𝑦𝑝 = 2
cosh 𝑎𝑥 + 𝑏 = 𝑥 ′′ 2 cosh 𝑎𝑥 + 𝑏
∅ 𝐷 ∅ 𝑎
Provided ∅′′ 𝑎2 ≠ 0 and so on
Find P.I. of 𝐷3 + 2𝐷2 + 5𝐷 − 8 𝑦 = sinh 2𝑥 + 3
1
𝑦𝑝 = 3 2
sinh 2𝑥 + 3 𝑎 = 2
𝐷 + 2𝐷 + 5𝐷 − 8
1 2 =4
𝑦𝑝 = sinh 2𝑥 + 3 𝑝𝑢𝑡 𝐷
𝐷2 × 𝐷 + 2𝐷2 + 5𝐷 − 8
1
𝑦𝑝 = sinh 2𝑥 + 3
4 × 𝐷 + 2 4 + 5𝐷 − 8
1
𝑦𝑝 = sinh 2𝑥 + 3
4𝐷 + 8 + 5𝐷 − 8
1 1 1
𝑦𝑝 = sinh 2𝑥 + 3 = sinh 2𝑥 + 3
9𝐷 9 𝐷
1 1 cosh 2𝑥 + 3
𝑦𝑝 = sinh 2𝑥 + 3 =
9 9 2
1
= cosh 2𝑥 + 3
18
1 2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵 = 𝑠𝑖𝑛 𝐴 + 𝐵 + sin 𝐴 − 𝐵
2 2𝑐𝑜𝑠𝐴𝑐𝑜𝑠𝐵 = cos 𝐴 + 𝐵 + cos 𝐴 − 𝐵
3 2 sin 𝐴 sin 𝐵 = cos 𝐴 − 𝐵 − cos 𝐴 + 𝐵
4 2
1 − cos 2𝜃
𝑠𝑖𝑛 𝜃 =
2
5) 2
1 + cos 2𝜃
𝑐𝑜𝑠 𝜃 =
2
6) Hyperbolic functions:
𝑒 𝑥 −𝑒 −𝑥 𝑒 𝑥 +𝑒 −𝑥
sinh 𝑥 = cosh 𝑥 =
2 2
𝑑 𝑑
sinh 𝑥
= cosh 𝑥 cosh 𝑥= sinh 𝑥
𝑑𝑥 𝑑𝑥
sinh 𝑥 = cosh 𝑥 cosh 𝑥 = sinh 𝑥 ,
𝑑2 𝑥 𝜋
+ 9𝑥 = 4 cos + 𝑡 given :𝑥 = 0 𝑎𝑡 𝑡 = 0 and
𝑑𝑡 2 3
𝜋 2
𝜋
𝑥 = 2 𝑎𝑡 𝑡 = 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛: 𝐷 + 9 𝑥 = 4 cos + 𝑡
6 3
𝐷2 + 9 = 0 ⇒ 𝐷 = ±3𝑖 ⇒ 𝑥𝑐 = 𝑐1 cos 3𝑡 + 𝑐2 sin 3𝑡
1 𝜋 1 𝜋
𝑥𝑝 = 2 4 cos + 𝑡 = 4 2 cos + 𝑡
𝐷 +9 3 𝐷 +9 3
1 1
Formula:
∅ 𝐷 2 cos 𝑎𝑥 + 𝑏 = 2 cos 𝑎𝑥 + 𝑏
∅ −𝑎
𝜋 1 𝜋 1 𝜋
𝑎 =1𝑏 = , 𝑥𝑝 = 4 cos + 𝑡 = cos +𝑡
3 −1+9 3 2 3
1 𝜋
𝑥 = 𝑐1 cos 3𝑡 + 𝑐2 sin 3𝑡 + cos + 𝑡 …(1)
2 3
1 𝜋 1
𝑥 = 0 𝑎𝑡 𝑡 = 0 ⇒ 0 = 𝑐1 + cos ⇒ 𝑐1 = −
2 3 4
𝜋 1 𝜋 𝜋
𝑥 = 2 𝑎𝑡 𝑡 = ⇒ 2 = − cos + 𝑐2 sin ⇒ 𝑐2 = 2
6 4 2 2
Find 𝑦𝑝 of 𝐷4 + 10𝐷2 + 9 𝑦 = 96 sin 2𝑥 cos 𝑥
𝐷4 + 10𝐷2 + 9 𝑦 = 48 × 2 × sin 2𝑥 cos 𝑥
𝐷4 + 10𝐷2 + 9 𝑦 = 48 × 2 × sin 2𝑥 cos 𝑥
Formula: 2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵 = 𝑠𝑖𝑛 𝐴 + 𝐵 + sin 𝐴 − 𝐵
𝐷4 + 10𝐷2 + 9 𝑦 = 48 × sin 3𝑥 + sin 𝑥
1
𝑦𝑝 = 48 4 2
sin 3𝑥 + sin 𝑥
𝐷 + 10𝐷 + 9
1 1
𝑦𝑝 = 48 4 2
sin 3𝑥 + 4 2
sin 𝑥
𝐷 + 10𝐷 + 9 𝐷 + 10𝐷 + 9
1 1
= 48 sin 3𝑥 + sin 𝑥
−9 −9 + 10 −9 + 9 −1 −1 + 10 −1 + 9
Case of failure for both expressions
1 1
= 48 𝑥 3 sin 3𝑥 + 𝑥 3 sin 𝑥
4𝐷 + 20𝐷 4𝐷 + 20𝐷
1 1
= 48 𝑥 sin 3𝑥 + 𝑥 sin 𝑥
−36𝐷 + 20𝐷 −4𝐷 + 20𝐷
1 1 𝑥 𝑥
= 48 𝑥 sin 3𝑥 + 𝑥 sin 𝑥 = 48 cos 3𝑥 − cos 𝑥
−16𝐷 16𝐷 48 16
 1
7)𝑓 𝑥 = 𝑥𝑚 𝑦𝑝 = 𝑥𝑚 = ∅ 𝐷 −1 𝑥 𝑚
∅ 𝐷
Step 1:- Express ∅ 𝐷 polynomial in the form 1 − 𝑔 𝐷
or 1 + 𝑔 𝐷 where 𝑔 𝐷 is some function of 𝐷.
To do this follow following procedure
a)Take constant term common from the denominator
b) If constant term is absent then take minimum power of
𝐷 common from the denominator.
Step 2:- Use appropriate formula from following list.
1
a) = 1 + 𝑥 + 𝑥2 + 𝑥3 + ⋯
1−𝑥
1
b) = 1 − 𝑥 + 𝑥2 − 𝑥3 + ⋯
1+𝑥
Where 𝑥 = 𝑔 𝐷
Step 3:- By using the previous expansion we get various
powers of 𝐷 in the numerator.
Then we arrange ∅ 𝐷 −1 in ascending powers of 𝐷 and
operate every power of 𝐷 on 𝑥 𝑚 term by term.
Important note:- 𝑚 + 1 𝑡ℎ and higher derivatives of 𝑥 𝑚
will be zero. So we need not have to consider terms
beyond 𝐷𝑚 .
Find P. I. of 𝐷2 − 𝐷 + 1 𝑦 = 𝑥 3 − 3𝑥 2 + 1
1
𝑦𝑝 = 2 𝑥 3 − 3𝑥 2 + 1
𝐷 −𝐷+1
1 3 − 3𝑥 2 + 1
= 𝑥
1 − 𝐷 − 𝐷2
1
= 1 + 𝑥 + 𝑥 2 + 𝑥 3 + ⋯ where 𝑥 = 𝐷 − 𝐷2
1−𝑥
 1 3 − 3𝑥 2 + 1
𝑦𝑝 = 𝑥
1 − 𝐷 − 𝐷2
= 1 + 𝐷 − 𝐷2 + 𝐷 − 𝐷2 2 + 𝐷 − 𝐷2 3 + ⋯ (𝑥 3
Find P. I. of 𝐷2 − 𝐷 𝑦 = 𝑥 4 − 1
1
𝑦𝑝 = 2 (𝑥 4 −1)
𝐷 −𝐷
1 4
1
= (𝑥 −1) = (𝑥 4 −1)
𝐷 𝐷−1 −𝐷 1 − 𝐷
−1 1
= (𝑥 4 −1)
𝐷 1−𝐷
−1
= (1 + 𝐷 + 𝐷2 + 𝐷3 + 𝐷4 + ⋯ )(𝑥 4 −1)
𝐷
−1
= 𝑥 4 − 1 + 𝐷 𝑥 4 − 1 + 𝐷2 𝑥 4 − 1 + 𝐷3 𝑥 4 − 1
𝐷
8) 𝑓 𝑥 = 𝑒 𝑎𝑥 𝑉 𝑉:any function of 𝑥
1 𝑎𝑥 𝑎𝑥
1
𝑦𝑝 = 𝑒 𝑉=𝑒 𝑉
∅ 𝐷 ∅ 𝐷+𝑎
1
𝑦𝑝 = 2 𝑒 −2𝑥 sin 2𝑥
𝐷 + 5𝐷 + 6
−2𝑥
1
=𝑒 2
sin 2𝑥
𝐷−2 +5 𝐷−2 +6
−2𝑥
1
=𝑒 2
sin 2𝑥
𝐷 − 4𝐷 + 4 + 5𝐷 − 10 + 6
−2𝑥
1 −2𝑥
1 −2𝑥
1
=𝑒 2
sin 2𝑥 = 𝑒 sin 2𝑥 = 𝑒 sin 2𝑥
𝐷 +𝐷 −4 + 𝐷 𝐷−4
−2𝑥
𝐷+4 −2𝑥
𝐷+4
=𝑒 2
sin 2𝑥 = 𝑒 sin 2𝑥
𝐷 − 16 −4 − 16
−𝑒 −2𝑥 −𝑒 −2𝑥
= 𝐷 + 4 sin 2𝑥 = 𝐷 sin 2𝑥 + 4 sin 2𝑥
20 20
−𝑒 −2𝑥 −𝑒 −2𝑥
= 2 cos 2𝑥 + 4 sin 2𝑥 = cos 2𝑥 + 2 sin 2𝑥
20 10
8) 𝑓 𝑥 = 𝑥𝑉 𝑉:any function of 𝑥
1 ∅′ 𝐷 1
𝑦𝑝 = 𝑥𝑉 = 𝑥 − 𝑉
∅ 𝐷 ∅ 𝐷 ∅ 𝐷
1
Provided 𝑉 is not a case of failure.
∅ 𝐷
Find P.I. of 𝐷2 − 4 𝑦 = 𝑥 sinh 𝑥
1 ∅′ 𝐷 1
𝑦𝑝 = 2 𝑥 sinh 𝑥 = 𝑥 − sinh 𝑥
𝐷 −4 ∅ 𝐷 ∅ 𝐷
2𝐷 1 2𝐷 1
= 𝑥− 2 2
sinh 𝑥 = 𝑥 − 2 sinh 𝑥
𝐷 −4 𝐷 −4 𝐷 −4 1−4
1 1
Formula: ∅ 𝐷2 sinh 𝑎𝑥 + 𝑏 = ∅ 𝑎2 sinh 𝑎𝑥 + 𝑏
−1 2𝐷 −1 2𝐷
= 𝑥− 2 sinh 𝑥 = 𝑥 sinh 𝑥 − 2 sinh 𝑥
3 𝐷 −4 3 𝐷 −4
−1 2𝐷 −1 2𝐷
= 𝑥 sinh 𝑥 − sinh 𝑥 = 𝑥 sinh 𝑥 + sinh 𝑥
3 1−4 3 3
−1 2 −1 2
= 𝑥 sinh 𝑥 + Dsinh 𝑥 = 𝑥 sinh 𝑥 + cosh 𝑥
3 3 3 3
Variation of parameters method:
Applicable only to find 𝒚𝒑 of second order differential eqn
Let the second order DE be
𝑑2 𝑦 𝑑𝑦
𝑎0 2 + 𝑎1 + 𝑎2 = 𝑓 𝑥
𝑑𝑥 𝑑𝑥
Step1- Find It’s complimentary 𝒚𝒄 = 𝒄𝟏 𝒚𝟏 + 𝒄𝟐 𝒚𝟐

Step2- let it’s particular integral be 𝑦𝑝 = 𝑢𝑦1 + 𝑣𝑦2

−𝑦2 𝑦1
Where 𝑢 = 𝑓 𝑥 𝑑𝑥 𝑣= 𝑓 𝑥 𝑑𝑥 and
𝑊 𝑊

𝑦1 𝑦2 ′−𝑦 𝑦 ′
𝑊= 𝑦 ′ 𝑦2 ′ = 𝑦 𝑦
1 2 2 1
1
 𝐷2 − 4𝐷 + 4 𝑦 = 𝑒 2𝑥 𝑠𝑒𝑐 2 𝑥 𝐷 = 2,2
𝑦𝑐 = 𝑐1 𝑥 + 𝑐2 𝑒 2𝑥 compare with 𝒚𝒄 = 𝒄𝟏 𝒚𝟏 + 𝒄𝟐 𝒚𝟐
2𝑥 2𝑥
𝑦1 = 𝑥𝑒 2𝑥 , 𝑦2 = 𝑒 2𝑥 𝑊= 𝑥𝑒 𝑒
2𝑥𝑒 2𝑥 + 𝑒 2𝑥 2𝑒 2𝑥
𝑊= 2𝑥𝑒 4𝑥 − 2𝑥𝑒 4𝑥 − 𝑒 4𝑥 = −𝑒 4𝑥 𝑦𝑝 = 𝑢𝑦1 + 𝑣𝑦2
−𝑦2
𝑢= 𝑓 𝑥 𝑑𝑥
𝑊
−𝑒 2𝑥 2𝑥 2 𝑥𝑑𝑥
= 𝑒 𝑠𝑒𝑐 = 𝑠𝑒𝑐 2 𝑥𝑑𝑥 = tan 𝑥
−𝑒 4𝑥
𝑦1 𝑥𝑒 2𝑥 2𝑥 2𝑥 = 2 𝑥𝑑𝑥
𝑣= 𝑓 𝑥 𝑑𝑥 = 𝑒 𝑠𝑒𝑐 −𝑥 𝑠𝑒𝑐
𝑊 −𝑒 4𝑥

𝑑𝑥
=− 𝑥 𝑠𝑒𝑐 2 𝑥𝑑𝑥 − 𝑠𝑒𝑐 2 𝑥𝑑𝑥 = − 𝑥 tan 𝑥 − tan 𝑥
𝑑𝑥
= −𝑥 tan 𝑥 + log sec 𝑥
𝑦𝑝 = tan 𝑥 𝑥𝑒 2𝑥 + −𝑥 tan 𝑥 + log sec 𝑥 𝑒 2𝑥
Equations reducible to standard LDE
Type I-Cauchy’s or Euler’s homogeneous LDE

𝑑 𝑛𝑦 𝑑 𝑛−1 𝑦 𝑑 𝑛−2 𝑦
𝑎0 𝑥 𝑛 𝑛 + 𝑎1 𝑥 𝑛−1 𝑛−1 + 𝑎2 𝑥 𝑛−2 𝑛−2 + ⋯
𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑𝑦
+ 𝑎𝑛−1 𝑥 + 𝑎𝑛 𝑦 = 𝐹(𝑥)
𝑑𝑥
𝑑
Put 𝑥 = 𝑒𝑧 ⇒ 𝑧 = log 𝑥 𝐷=
𝑑𝑧
𝑑𝑦 𝑑 2𝑦
𝑥 = 𝐷𝑦, 𝑥2 2
= 𝐷 𝐷−1 𝑦,
𝑑𝑥 𝑑𝑥
𝑑 3𝑦
𝑥3 3
= 𝐷 𝐷 − 1 𝐷 − 2 ]𝑦
𝑑𝑥
𝑑 𝑟𝑦
𝑥𝑟 𝑟
= 𝐷 𝐷 − 1 𝐷 − 2 … (𝐷 − (𝑟 − 1)]𝑦
𝑑𝑥
Type II - Legender’s or linear LDE
𝑛 𝑛−1
𝑛
𝑑 𝑦 𝑛−1
𝑑 𝑦
𝑎0 𝑎𝑥 + 𝑏 𝑛
+ 𝑎1 𝑎𝑥 + 𝑏
𝑑𝑥 𝑑𝑥 𝑛−1
𝑑 𝑛−2 𝑦 𝑑𝑦
𝑛−2
+ 𝑎2 𝑎𝑥 + 𝑏 𝑛−2
+ ⋯ + 𝑎𝑛−1 𝑎𝑥 + 𝑏 + 𝑎𝑛 𝑦
𝑑𝑥 𝑑𝑥
= 𝐹(𝑥)
𝑑
Put 𝑎𝑥 + 𝑏 = 𝑒 𝑧 ⇒ 𝑧 = log 𝑎𝑥 + 𝑏 𝐷=
𝑑𝑧
2𝑦
𝑑𝑦 𝑑
𝑎𝑥 + 𝑏 = 𝑎𝐷𝑦, 𝑎𝑥 + 𝑏 2 2
= 𝑎 2
𝐷 𝐷−1 𝑦,
𝑑𝑥 𝑑𝑥
𝑑 3𝑦
(𝑎𝑥 + 𝑏)3 3 = 𝑎3 𝐷 𝐷 − 1 𝐷 − 2 ]𝑦
𝑑𝑥
𝑑 𝑟𝑦
(𝑎𝑥 + 𝑏)𝑟 𝑟 = 𝑎𝑟 𝐷 𝐷 − 1 𝐷 − 2 … (𝐷 − (𝑟 − 1)]𝑦
𝑑𝑥
 𝑑 2𝑦 𝑑𝑦
3 2
𝑥 2
+ 3𝑥 + 𝑥𝑦 = sin log 𝑥
𝑑𝑥 𝑑𝑥
𝑑 2𝑦 𝑑𝑦 sin log 𝑥
2
𝑥 2
+ 3𝑥 +𝑦 =
𝑑𝑥 𝑑𝑥 𝑥
𝑑
𝑥= 𝑒𝑧 ⇒ 𝑧 = log 𝑥 𝐷=
𝑑𝑧
𝑑𝑦 𝑑 2𝑦
𝑥 = 𝐷𝑦, 𝑥2 2
= 𝐷 𝐷−1 𝑦,
𝑑𝑥 𝑑𝑥
sin 𝑧
𝐷 𝐷 − 1 𝑦 + 3𝐷𝑦 + 𝑦 = 𝑧
𝑒
𝐷2 − 𝐷 + 3𝐷 + 1 𝑦 = 𝑒 −𝑧 sin 𝑧
𝐷2 + 2𝐷 + 1 𝑦 = 𝑒 −𝑧 sin 𝑧
𝐷 + 1 2 = 0 ⇒ 𝐷 = −1, −1 ⇒ 𝑦𝑐 = 𝑐1 𝑧 + 𝑐2 𝑒 −𝑧
 1
𝑦𝑝 = 2 𝑒 −𝑧 sin 𝑧
𝐷 + 2𝐷 + 1
1 𝑎𝑥 𝑎𝑥
1
𝑦𝑝 = 𝑒 𝑉=𝑒 𝑉
∅ 𝐷 ∅ 𝐷+𝑎
−𝑧
1
=𝑒 2
sin 𝑧
(𝐷 − 1) +2(𝐷 − 1) + 1
−𝑧
1
=𝑒 2
sin 𝑧
𝐷 − 2𝐷 + 1 + 2𝐷 − 2 + 1
−𝑧
1 −𝑧
1 −𝑧 sin 𝑧
=𝑒 sin 𝑧 = 𝑒 sin 𝑧 = −𝑒
𝐷2 −1
𝑦 = 𝑐1 𝑧 + 𝑐2 𝑒 −𝑧 − 𝑒 −𝑧 sin 𝑧
𝑐1 𝑧 + 𝑐2 − sin 𝑧
𝑦=
𝑒𝑧
𝑐1 log 𝑥 + 𝑐2 − sin log 𝑥
𝑦=
𝑥
 𝑑 2𝑦 𝑑𝑦
2
2𝑥 + 3 2
− 2 2𝑥 + 3 − 12𝑦 = 6𝑥
𝑑𝑥 𝑑𝑥
Put 2𝑥 + 3 = 𝑒 𝑧 ⇒ 𝑧 = log(2𝑥 + 3)
𝑑𝑦 𝑑 2𝑦
𝑎𝑥 + 𝑏 = 𝑎𝐷𝑦, 𝑎𝑥 + 𝑏 2 = 𝑎 2 𝐷 𝐷−1 𝑦
𝑑𝑥 𝑑𝑥 2
Comparing with above formula 𝑎 = 2
𝑒𝑧 − 3
4 𝐷 𝐷 − 1 𝑦 − 2 2𝐷𝑦 − 12𝑦 = 6
2
4𝐷2 − 4𝐷 − 4𝐷 − 12 𝑦 = 3𝑒 𝑧 − 9
4𝐷2 − 8𝐷 − 12 𝑦 = 3𝑒 𝑧 − 9
4𝐷2 − 8𝐷 − 12 = 0 ⇒ 𝐷2 − 2𝐷 − 3 = 0 ⇒ 𝐷 = 3, −1
𝑦𝑐 = 𝑐1 𝑒 3𝑧 + 𝑐2 𝑒 −𝑧
1 𝑧−9
𝑦𝑝 = 3𝑒
4𝐷2 − 8𝐷 − 12
 1 𝑧
𝑦𝑝 = 2
3𝑒 −9
4𝐷 − 8𝐷 − 12
1 𝑧
1 0𝑧
= 3𝑒 − 9 × 𝑒
4𝐷2 − 8𝐷 − 12 4𝐷2 − 8𝐷 − 12
1 𝒂𝒙 1
𝒆 𝒆𝒂𝒙 𝑏𝑙𝑎𝑐𝑘 𝑡𝑒𝑟𝑚 𝑎 = 1 &. 𝑟𝑒𝑑 𝑎 = 0
=
∅ 𝑫 ∅ 𝒂
1 𝑧
1
=3 𝑒 −9 𝑒 0𝑧
4 1 − 8 1 − 12 0 − 0 − 12
−3 𝑧 9 −3 𝑧 3
= 𝑒 + = 𝑒 +
16 12 16 4
3𝑧 −𝑧
3 𝑧 3
𝑦 = 𝑐1 𝑒 + 𝑐2 𝑒 − 𝑒 +
16 4
2𝑥 + 3 = 𝑒 𝑧 ⇒ 2𝑥 + 3 3 = 𝑒 3𝑧
3
𝑐2 3 3
𝑦 = 𝑐1 2𝑥 + 3 + − 2𝑥 + 3 +
2𝑥 + 3 16 4
Simultaneous linear differential equations:
𝑑𝑥 𝑑𝑦 𝑑
+ 𝑦 = 𝑒 𝑡 , + 𝑥 = 𝑒 −𝑡 introduce 𝐷 =
𝑑𝑡 𝑑𝑡 𝑑𝑡
𝑡 −𝑡
𝐷𝑥 + 𝑦 = 𝑒 … 1 , 𝐷𝑦 + 𝑥 = 𝑒 … (2)
Operate 𝐷 on (1) 𝐷2 𝑥 + 𝐷𝑦 = 𝐷𝑒 𝑡
𝐷2 𝑥 + 𝐷𝑦 = 𝑒 𝑡 …(3) (3)-(2) gives
𝐷2 𝑥 + 𝐷𝑦 = 𝑒 𝑡 − 𝐷𝑦 + 𝑥 = 𝑒 −𝑡 = 𝐷 2 𝑥 − 𝑥 = 𝑒 𝑡 − 𝑒 −𝑡
𝐷2 − 1 𝑥 = 𝑒 𝑡 − 𝑒 −𝑡 𝐷 = 1, −1 𝑥𝑐 = 𝑐1 𝑒 𝑡 + 𝑐2 𝑒 −𝑡
1 1 1
𝑥𝑝 = 𝑒 𝑡 − 𝑒 −𝑡 = 𝑒 𝑡
− 𝑒 −𝑡
case of failure
𝐷2 −1 𝐷2 −1 𝐷2 −1
∅ 𝐷 = 𝐷2 − 1 ⇒ ∅′ 𝐷 = 2𝐷
𝑡 𝑡 𝑡 −𝑡 𝑡 𝑡
𝑥𝑝 = 𝑒 − 𝑒 = 𝑒 + 𝑒 −𝑡
2 −2 2
𝑡
𝑥 = 𝑐1 𝑒 𝑡 + 𝑐2 𝑒 −𝑡 + 𝑒 𝑡 + 𝑒 −𝑡 put in (1) 𝐷𝑥 + 𝑦 = 𝑒 𝑡 ⇒
2
𝑡 𝑡
𝐷 𝑐1 𝑒 + 𝑐2 𝑒 + 𝑒 + 𝑒 −𝑡 + 𝑦 = 𝑒 𝑡
𝑡 −𝑡
2
𝑡 −𝑡
𝑡 𝑡 −𝑡
1 𝑡
𝑐1 𝑒 − 𝑐2 𝑒 + 𝑒 − 𝑒 + 𝑒 + 𝑒 −𝑡 + 𝑦 = 𝑒 𝑡
2 2
Symmetrical simultaneous Differential Equations
𝑑𝑥 𝑑𝑦 𝑑𝑧
Equations of the type = = are called as
𝑃 𝑄 𝑅
symmetrical simultaneous Differential equations where
𝑃, 𝑄, 𝑅 are functions of 𝑥, 𝑦, 𝑧.
Solutions of such system consists of two independent
relations of the type 𝐹1 𝑥, 𝑦, 𝑧 = 𝑐1 , 𝐹2 𝑥, 𝑦, 𝑧 = 𝑐2
There are two methods to solve such a system.
1) Method of combination or grouping
2) Method of multipliers
Method of combination or grouping:- 1)We combine two
ratios in such a way that they will contain exactly two
variables. 2)By getting variable separable form we solve
differential equation and get relation.
 𝑥𝑑𝑥 𝑑𝑦 𝑑𝑧
Example 1: = =
𝑦3𝑧 𝑥2𝑧 𝑦3
𝑥𝑑𝑥 𝑑𝑦
3
= 2
𝑦 𝑧 𝑥 𝑧
𝑥 4 𝑦 4
𝑥 3 𝑑𝑥 = 𝑦 3 𝑑𝑦 ⇒ 𝑥 3 𝑑𝑥 = 𝑦 3 𝑑𝑦 ⇒ = +𝑐
4 4
𝑥 4 − 𝑦 4 = 𝑐1 ….(1)

𝑥𝑑𝑥 𝑑𝑧
3
= 3
𝑦 𝑧 𝑦
𝑥2 𝑧2
𝑥𝑑𝑥 = 𝑧𝑑𝑧 ⇒ 𝑥𝑑𝑥 = 𝑧𝑑𝑧 ⇒ = + 𝑐′
2 2
𝑥 2 − 𝑧 2 = 𝑐2 ….(2)
 𝑑𝑥 𝑑𝑦 𝑑𝑧
Example 2: = = 2 2
𝑦 −𝑥 𝑥𝑒 𝑥 +𝑦
𝑑𝑥 𝑑𝑦
=
𝑦 −𝑥
−𝑥 2 𝑦 2
−𝑥𝑑𝑥 = 𝑦𝑑𝑦 −𝑥𝑑𝑥 = 𝑦𝑑𝑦 ⇒ = + 𝑐′
2 2
𝑥 2 + 𝑦 2 = 𝑐1
𝑑𝑦 𝑑𝑧 𝑑𝑦 𝑑𝑧 𝑥 2 +𝑦 2
= 𝑥 2 +𝑦2 ⇒ = 𝑥 2 +𝑦2 ⇒ 𝑒 𝑑𝑦 = −𝑑𝑧
−𝑥 𝑥𝑒 −1 𝑒

𝑒 𝑐1 𝑑𝑦 = −𝑑𝑧 ⇒ 𝑒 𝑐1 𝑑𝑦 + 𝑑𝑧 = 𝑒 𝑐1 𝑦 + 𝑧 = 𝑐2

𝑒 𝑥 2 +𝑦 2 𝑦 + 𝑧 = 𝑐2
Method of multipliers
𝑑𝑥 𝑑𝑦 𝑑𝑧
= =
𝑃 𝑄 𝑅
In this method we try to find three multipliers 𝑙, 𝑚, 𝑛 such
that the term
𝑙𝑃 + 𝑚𝑄 + 𝑛𝑅 = 0 and the term 𝑙𝑑𝑥 + 𝑚𝑑𝑦 + 𝑛𝑑𝑧
is in exact form.
𝑑𝑥 𝑑𝑦 𝑑𝑧 𝑙𝑑𝑥+𝑚𝑑𝑦+𝑛𝑑𝑧
Now, = = =
𝑃 𝑄 𝑅 𝑙𝑃+𝑚𝑄+𝑛𝑅
So we get 𝑙𝑑𝑥 + 𝑚𝑑𝑦 + 𝑛𝑑𝑧 = 0
As 𝑙𝑑𝑥 + 𝑚𝑑𝑦 + 𝑛𝑑𝑧 is in exact form , integrate both sides
of equation 𝑙𝑑𝑥 + 𝑚𝑑𝑦 + 𝑛𝑑𝑧 = 0 to get the relation
Note: 𝑙, 𝑚, 𝑛 may be constant values or functions of 𝑥. 𝑦, 𝑧
 𝑑𝑥±𝑑𝑦
Exact forms:1) = log 𝑥 ± 𝑦
𝑥±𝑦
2) 𝑥𝑑𝑦 + 𝑦𝑑𝑥 = 𝑑 𝑥𝑦 = 𝑥𝑦
𝑑𝑥 𝑑𝑦 𝑑𝑧
Example 1: = =
𝑦 −𝑥 2𝑥−3𝑦
𝑑𝑥 𝑑𝑦
= ⇒ −𝑥𝑑𝑥 = 𝑦𝑑𝑦 ⇒ −𝑥𝑑𝑥 = 𝑦𝑑𝑦
𝑦 −𝑥
−𝑥 2 𝑦 2
= + 𝑐 ⇒ 𝑥 2 + 𝑦 2 = 𝑐1
2 2
3𝑑𝑥+2𝑑𝑦+𝑑𝑧
Consider combination we get 0 in the 𝐷𝑟 .
3𝑦−2𝑥+2𝑥−3𝑦
So we get 3𝑑𝑥 + 2𝑑𝑦 + 𝑑𝑧 = 0

3𝑑𝑥 + 2𝑑𝑦 + 𝑑𝑧 = 0 ⇒ 3𝑥 + 2𝑦 + 𝑧 = 𝑐2
 𝑑𝑥 𝑑𝑦 𝑑𝑧
Example 2: = =
𝑧 𝑥+𝑦 𝑧 𝑥−𝑦 𝑥 2 +𝑦 2
𝑑𝑥 𝑑𝑦 𝑑𝑥 𝑑𝑦
= ⇒ =
𝑧 𝑥+𝑦 𝑧 𝑥−𝑦 𝑥+𝑦 𝑥−𝑦
𝑥 − 𝑦 𝑑𝑥 = 𝑥 + 𝑦 𝑑𝑦 ⇒ 𝑥𝑑𝑥 − 𝑦𝑑𝑥 = 𝑥𝑑𝑦 + 𝑦𝑑𝑦
𝑥𝑑𝑥 − 𝑦𝑑𝑥 − 𝑥𝑑𝑦 − 𝑦𝑑𝑦 = 0

𝑥𝑑𝑥 − 𝑦𝑑𝑥 + 𝑥𝑑𝑦 − 𝑦𝑑𝑦 = 0

𝑥2 𝑦2
− 𝑥𝑦 − = 𝑐1
2 2
Choose multipliers 𝑥, −𝑦, −𝑧
𝑥𝑑𝑥 − 𝑦𝑑𝑦 − 𝑧𝑑𝑧
𝑥 2 𝑧 + 𝑥𝑦𝑧 − 𝑥𝑦𝑧 + 𝑦 2 𝑧 − 𝑥 2 𝑧 − 𝑦 2 𝑧
⇒ 𝑥𝑑𝑥 − 𝑦𝑑𝑦 − 𝑧𝑑𝑧 = 0 ⇒ 𝑥 2 − 𝑦 2 − 𝑧 2 = 𝑐2