(b) Problems based on P.I.= 1/f(D) eax ⇒ Replace D by a

1. (b) PROBLEMS BASED ON P.I.= 1/f(D) e^ax ⇒Replace D by a

 

Example 5.1.10 (a) Solve (D² - 4D +13) y = e^2x.

Solution: Given: (D² - 4D + 13) y = e^2x

The auxiliary equation is m² - 4m + 13 = 0

 

Example 5.1.10(b) Solve: y'' + 7y' - 8y = e^2x

Solution: Given: [D² + 7D - 8] y = e^2x

The auxiliary equation is m² + 7m – 8 = 0 ⇒ m = 1, m = −8

C.F = A e^x + Be^-8x

P.I = (1/ D² + 7D – 8) e^2x = (1 / 4 + 14 – 8) e^2x

[Replace D by 2]

= 1 / 10 e^2x

y = C.F + P.I

y = A e^x + Be^-8x + 1/10 e^2x

 

Example 5.1.10(c) Find the particular integral of (D² + 4D+8) y = e^2x

[A.U M/J 2016 R-8]

Solution: P.I = 1 / D² + 4D + 8 e^2x

= (1 / 4 + 8 + 8) e^2x

[Replace D by 2]

= 1/ 20 e^2x

 

Example 5.1.11 (a) Solve (D - 2)² y = e^2x.

Solution: Given: (D - 2)² y = e^2x

The auxiliary equation is (m - 2)² = 0

i.e., m = 2, 2.

 

Example 5.1.11 (b) Find the particular integral of (D² - 4D + 4) y = 2^x

[A.U N/D 2012]

Solution:

Given: (D² - 4D + 4) y = 2^x

i.e., (D - 2)² y = 2^x = e^log2x = e^(log2x)

P.I = 1 / (D – 2)² e^log2x = 1 / (log2 – 2)² e^log2x

[Replace D by log 2]

 

Example 5.1.12 Solve (D² + 7D + 12) y = 14 e^-3x

[A.U M/J 2016 R-8]

Solution: Given: (D² + 7D + 12) y = 14e^-3x

The auxiliary equation is m² + 7m + 12 = 0

⇒ m = -3, m = -4

 

Example 5.1.13. Solve (4D2 - 4 D + 1) y = 4.

Solution: Given: (4 D² - 4D + 1)y = 4

The auxiliary equation is 4m² - 4m + 1 = 0

m = 1/2, 1/2

 

Example 5.1.14 Solve (D² - 4) y = 1

[A.U N/D 2013 R-8]

Solution: Given: (D² - 4) y = e^0x

The auxiliary equation is m² - 4 = 0  ⇒ m = ±2

 

Example 5.1.15. Solve (D² - 4) y = e^2x + e^-4x.

Solution: Given: (D² - 4) y = e^2x + e ^̄4x

The auxiliary equation is m² – 4 = 0

 

Example 5.1.16. Solve d²y / dx² + 4 dy/dx + 5y = -2 cosh x

 [AU, April 2002]

Solution:

The auxiliary equation is m2 + 4m +5 = 0

 

Example 5.1.17. Find the P.I of (D²-1) y = (e^x + 1)²

Solution: Given: (D² - 1)y = (e^x + 1)²

(D² - 1)y = (e^x)² + 1 + 2^ex

 

Example 5.1.18. Solve (D³ - 3D² + 4D - 2) y = e^x.

Solution: Given: (D³ - 3D² + 4D - 2)y = e^x

The auxiliary equation is m³ - 3m² + 4m − 2 = 0

⇒ m 1, m = 1 ± i

C.F. Ae^x + e^x [B cos x + C sin x]

P.I. = (1 / D³ - 3D² + 4D – 2) e^x

= (1 / 1 – 3 + 4 – 2) e^x [Replace D by 1]

= 1/0 e^x  [Ordinary rule fails]

= x (1 / (1 / D³ - 3D² + 4D – 2) e^x

= x ( 1 /  3 – 6 + 4)e^x = xe^x

Hence, the solution is

y = C.F. + P.I

y = Ae^x + e^x [B cos x + C sin x] + x e^x

 

Example 5.1.19. Solve (D5 - D) y = 12 e^x.

Solution: Given: (D⁵ - D)y = 12^ex

The auxiliary equation is m⁵ – m = 0