How do you calculate the particular integral of linear differential equations? Learn this and much more by watching this video. Here, we learn how the inverse differential operator is used to find the particular integral of trigonometric, exponential, polynomial and inverse hyperbolic functions. Problems are explained with the relevant formulae. This is useful for graduate students and engineering students learning Mathematics. For more videos, visit my page https://www.mathmadeeasy.co/about-4 Subscribe to my channel for more videos.

1. UCjmCXXIjd03JQad8-rUzs0Q
Graduate mathematics

2. Inverse Differential
Operator
Engineering mathematics
Undergraduate mathematics

3. What is an inverse differential operator?
Why do we need it ?

4. 𝐷 =
𝑑
𝑑𝑥
1
𝐷
𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑡ℎ𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑜𝑝𝑒𝑟𝑎𝑡𝑜𝑟
It is used to find the particular solution of a linear differential equation.
We learn how to calculate the particular integral for polynomial, hyperbolic , exponential
and trigonometric functions.

5. 1
𝐷
𝑥2 = 𝑥2 𝑑𝑥 =
𝑥3
3
𝑒 𝑎𝑥
𝑓(𝐷)
=
𝑒 𝑎𝑥
𝑓(𝑎)
𝑖𝑓 𝑓(𝑎) ≠ 0
= 𝑥
𝑒 𝑎𝑥
𝑓′(𝑎)
𝑖𝑓 𝑓′
(𝑎) ≠ 0
= 𝑥2
𝑒 𝑎𝑥
𝑓′′(𝑎)
𝑤ℎ𝑒𝑟𝑒 𝑓′′(𝑎) ≠ 0
𝑎𝑙𝑠𝑜 ℎ𝑜𝑙𝑑𝑠 𝑓𝑜𝑟 𝑒 𝑎𝑥+𝑏

6. 𝑠𝑖𝑛𝑎𝑥
𝑓(𝐷2)
=
sin 𝑎𝑥
𝑓(−𝑎2)
where f(−𝑎2
) ≠ 0
𝑐𝑜𝑠𝑎𝑥
𝑓(𝐷2)
=
𝑐𝑜𝑠𝑎𝑥
𝑓 −𝑎2
𝑖𝑓 𝑓(−𝑎2
) ≠ 0
𝑒 𝑎𝑥 𝑉
𝑓(𝐷)
= 𝑒 𝑎𝑥
𝑉
𝑓(𝐷 + 𝑎) 𝑐𝑜𝑠ℎ𝑥 =
𝑒 𝑥 + 𝑒−𝑥
2
sin ℎ𝑥 =
𝑒 𝑥 − 𝑒−𝑥
2

7. Do you want to excel in Maths?

8. Q 1 Solve
𝑑2 𝑦
𝑑𝑥2 − 4𝑦 = cosh 2𝑥 − 1 + 3 𝑥

9. (𝐷2
−4)𝑦 = 0
𝑚2 − 4 = 0 𝑎𝑢xillary equation
(m+2)(m-2)=0
𝑦𝑐 = 𝑐1 𝑒−2𝑥
+ 𝑐2 𝑒2𝑥

10. 𝑃𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑦 𝑃 =
1
𝐷2 − 4
𝑒2𝑥−1
+ 𝑒− 2𝑥−1
2
+
3 𝑥
𝐷2 − 4
=
1
2
𝑒2𝑥−1
𝐷2−4
+
1
2
𝑒−(2𝑥−1)
𝐷2−4
+
𝑒 𝑙𝑜𝑔3 𝑥
𝐷2−4
𝑒 𝑎𝑥
𝑓(𝐷)
=
𝑒 𝑎𝑥
𝑓(𝑎)
𝑖𝑓 𝑓(𝑎) ≠ 0
= 𝑥
𝑒 𝑎𝑥
𝑓′(𝑎)
𝑖𝑓 𝑓′
(𝑎) ≠ 0
=
𝑥
2
𝑒2𝑥−1
2𝐷
+
𝑥
2
𝑒− 2𝑥−1
2𝐷
+
3 𝑋
(𝑙𝑜𝑔3)2 − 4

11. =
𝑥
8
𝑥𝑒2𝑥−1-
𝑥
8
𝑒−(2𝑥−1)+
3 𝑥
(𝑙𝑜𝑔3)2−4
𝑦 = 𝑦𝑐 = 𝑐1 𝑒−2𝑥 + 𝑐2 𝑒2𝑥 +
𝑥
8
𝑥𝑒2𝑥−1
-
𝑥
8
𝑒−(2𝑥−1)
+
3 𝑥
(𝑙𝑜𝑔3)2−4

13. Q 2 Solve (𝐷2−2𝐷 + 4)𝑦 = 𝑒 𝑥 𝑐𝑜𝑠𝑥

14. 𝐴𝑢xillary equation 𝑚2 − 2𝑚 + 4 = 0
𝑟𝑜𝑜𝑡𝑠 𝑎𝑟𝑒
2± −12
2
=1±𝑖 3
𝑦𝑐 = 𝑒 𝑥
(𝑐1 𝑐𝑜𝑠 3𝑥 + 𝑐2 𝑠𝑖𝑛 3𝑥)
𝑦𝑝 =
𝑒 𝑥
𝑐𝑜𝑠𝑥
𝐷2 − 2𝐷 + 4
𝑒 𝑎𝑥
𝑉
𝑓(𝐷)
= 𝑒 𝑎𝑥
𝑉
𝑓(𝐷 + 𝑎)
Type equation here.

15. 𝑦𝑝 = 𝑒 𝑥
𝑐𝑜𝑠𝑥
(𝐷 + 1)2−2 𝐷 + 1 + 4
𝐷2
+ 2𝐷 + 1 − 2𝐷 − 2 + 4
= 𝑒 𝑋
𝑐𝑜𝑠𝑥
𝐷2 + 3
= 𝑒 𝑥
𝑐𝑜𝑠𝑥
−1 + 3
= 𝑒 𝑥
𝑐𝑜𝑠𝑥
2

16. 𝑦 = 𝑒 𝑥
𝑐1 𝑐𝑜𝑠 3𝑥 + 𝑐2 𝑠𝑖𝑛 3𝑥 +
𝑒 𝑥
𝑐𝑜𝑠𝑥
2

18. Q 3 Solve
𝑑3 𝑦
𝑑𝑥3 +
𝑑2 𝑦
𝑑𝑥2 + 4
𝑑𝑦
𝑑𝑥
+ 4𝑦 = 𝑥2
− 4𝑥 + 6

19. (𝐷3
+𝐷2
+ 4𝐷 + 4)𝑦 = 𝑥2
− 4𝑥+6
𝐴𝑢𝑥𝑖𝑙𝑙𝑎𝑟𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑚3
+ 𝑚2
+ 4𝑚 + 4 = 0
1 1 4 4 -1 is a root-1
1 0 4 0 𝑚 + 1 𝑚2
+ 4 = 0
-1 0 -4

20. 𝑚 = −1, ±2𝑖
𝑦𝑐 = 𝑐1 𝑒−𝑥 + 𝑐3 𝑐𝑜𝑠2𝑥 + 𝑐4 sin 2𝑥
𝑦𝑝 =
𝑥2 − 4𝑥 − 6
𝐷3 + 𝐷2 + 4𝐷 + 4

21. 𝑥2
− 4𝑥 − 64 + 4𝐷 + 𝐷2
+ 𝐷3
𝑥2
4
−
3
2
𝑥 -
1
8
4𝐷
𝑥2
4
= 2𝑥
𝐷2
𝑥2
4
=
𝐷2𝑥
4
=
1
2
𝑥2
+ 2𝑥 +
1
2
−6𝑥 −
13
2
4𝐷
−3
2
𝑥 = 4 −
3
2
= −6
𝑦𝑐 =
𝑥2
4
−
3
2
𝑥-
1
8
-6x -6
-
1
2
-
1
2
0

22. 𝑦 = 𝑐1 𝑒−𝑥 + 𝑐2 𝑐𝑜𝑠2𝑥 + 𝑐3 sin 2𝑥 +
𝑥2
4
−
3
2
𝑥 −
1
8

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