(i) Problems based on R.H.S = eax + xn + cos ax

I (i) problems based on R.H.S = e^ax + xⁿ + cos ax

Example 5.1.58 Solve (D² - 4D + 4) y = e^2x + cos 4x + x²

Solution: Given: (D² - 4D + 4) y = e^2x + cos 4x + x²

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

 (m − 2)² = 0

m = 2, 2

Example 5.1.59 Solve (D²+1)² = x⁴ + sinh x

Solution: Given. (D² + I)² = x⁴ +  [ e^x – e^-x /2 ]

((D² + 1)² = x⁴  + 1/2 e^x – 1/2 e^-x

The auxiliary equation is (m² + 1)² = 0

m² + 1 = 0, m² + 1 = 0

m² = -1, m² = -1

m = ± i, m² = ± i

C.F = C₁+C₂) cosx + (C₃ + C₄x) sinx

P.I₁ =1/(D₂ + 1)₂ x⁴= [1/(1+ D²)² x⁴

⁼ [1 + D²]^-2 x⁴

=  [1 - 2D² + 3D⁴] x⁴ [Omitting D5 and higher powers]

P.I₁ = x⁴ - 24x2 + 72

P.I₃ = -1/8 e^x

y = C.F + P.I₁ + P.I₂+ P.I₃

y = (C₁ + C₂x) cosx + (C₃ + C₄x) sinx + x⁴ – 24x² + 72 + 1/8 e^x – 1/8 e^-x

Example 5.1.60 Solve: (D⁴+ D³ + D²) y = 12x2 + 2 cos 2x cosx

Solution: Given: (D⁴ + D³ + D²) y = 12x² + [cos (2 + 1)x + cos (2-1) x]

i.e., [D⁴ + D³ + D²ly = 12x² + cos 3x + cos x

The auxiliary equation is m⁴ + m³ + m² = 0

m² [m² + m + 1] = 0

m² = 0 , m² + m + 1 = 0

m = 0, 0 and m = -1/2 ± √3i/2

P.I₁ = 1/D⁴ + D³ + D² 12 x²

= 12/D² 1/ 1 + D+D² ]x²

P.I₁ = x⁴ - 4x³

P.I₂  = 1/D⁴ + D³ + D² cos 3x

P.I₂  =R.P. 1/D⁴ + D³ + D² e^i3x  [Replace D by (i 3)]

= R.P 1/(i 3)⁴ + (i 3)³ + (i 3)² e^i3x

 = RP 1/81 -27i – 9 e^i3x

= R.P 1/72 - 27i e^i3x

= 8/657 cos 3x – 1/219 sin 3x

P.I₃ = R.P 1/ D⁴ + D³ + D² e^ix

 = R.P 1/i⁴ +i³ + i² e^ix [Replace D by i]

= R.P 1/1 – i – 1 e^ix

=R.P 1/-i e^ix

= R.P i [cos x + i sin x]

= -sin x

P.I = P.I₁+ P.I₂ + P.I₃

= = x⁴ - 4x³ + 1/657 (8 cos 3x - 3 sin 3x) - sin x

The general solution is

y = C.F + P.I