differential equation

1. WELCOMETOMYPRESENTATION...
Name : Arup Sarkar
ID : 161-15-6721

2. QUESTION - 11
Solve the following nonhomogeneous LDE by
operator method :-
(𝐷2
- 5D + 4)y = sin 5x

3. SOLUTION
Given,
(𝐷2 - 5D + 4)y= sin 5x
Complementary Function :
Considering the Homogeneous case of the given DE .
∴ Auxilary Equation is,
𝑚2 - 5m + 4 = 0
𝑚2 - m - 4m + 4 = 0
(m – 1)(m – 4) = 0
∴ m = 1 , 4
∴ yc = C1 𝒆 𝒙 + C2 𝒆 𝟒𝒙

4. Particular Integral :
yp =
𝟏
𝑫 𝟐 −𝟓𝑫+𝟒
. sin 5x
=
𝟏
−𝟓 𝟐 −𝟓𝑫+𝟒
. sin 5x
=
𝟏
− 𝟐𝟏 −𝟓𝑫
. sin 5x
= ( - 1) . [
𝟏
𝟐𝟏 + 𝟓𝑫
] . sin 5x
= ( - 1) . [
(𝟐𝟏 − 𝟓𝑫 )
(𝟐𝟏 + 𝟓𝑫 )(𝟐𝟏 − 𝟓𝑫 )
] . sin 5x
= ( - 1) . [
(𝟐𝟏 − 𝟓𝑫 )
𝟐𝟏 𝟐 − 𝟓𝑫 𝟐 ] . sin 5x
= ( - 1) . [
(𝟐𝟏 − 𝟓𝑫 )
𝟒𝟒𝟏 −𝟐𝟓𝑫 𝟐 ] . sin 5x
= ( -
𝟏
𝟏𝟎𝟔𝟔
) . (𝟐𝟏 − 𝟓𝑫 ) . sin 5x
= ( -
𝟏
𝟏𝟎𝟔𝟔
) . [ 𝟐𝟏𝒔𝒊𝒏 𝟓𝒙 − 𝟓
𝒅
𝒅𝒙
𝒔𝒊𝒏 𝟓𝒙 ]
= ( -
𝟏
𝟏𝟎𝟔𝟔
) . (𝟐𝟏sin 5x − 𝟓. 𝟓 𝒄𝒐𝒔 𝟓𝒙)
= ( -
𝟏
𝟏𝟎𝟔𝟔
) . (𝟐𝟏sin 5x − 𝟐𝟓𝒄𝒐𝒔 𝟓𝒙)

5. ∴ The Complete Solution,
y = yc + yp
= C1 𝒆 𝒙 + C2 𝒆 𝟒𝒙
+ [( -
𝟏
𝟏𝟎𝟔𝟔
) . (𝟐𝟏sin 5x −
𝟐𝟓𝒄𝒐𝒔 𝟓𝒙)]

6. Thank you...