Answer:
Step-by-step explanation:
Given the differential equation
y^(4) + 8y' = 6
We want to write this in the form
L(y) = g(x)
Where L is a linear differential operator with constant coefficient.
A linear differential operator of the nth order is a linear combination of derivative operators up to n.
L = D^n + a_1D^(n-1) + a_2D^(n-2) + ... + a_n,
defined by
Ly = y^n + a_1y^(n-1) + a_2y^(n-2) + ... + a_(n-1)y' + a_ny
Where a_i are continuous functions of x.
Now, we have
y^(4) + 8y' = 6
Let d/dx = D
Then
D^4 y + 8Dy = 6
D(D³ + 8)y = 6
Consider
D(D³ + 8)y = 0
The auxiliary equation is
m(m³ + 8) = 0
m = 0
Or
m³ + 8 = 0
=> m³ = -8
=> m = -2
The complimentary solution is
y = C1 + (C2 + C3x + C4x²)e^(-2x)
The particular integral is
y_p = Ax
y' = A
y'' = y''' = y^(4) = 0
Using these
0 + 8A = 6
A = 6/8 = 3/4
So
y = C1 + (C2 + C3x + C4x²)e^(-2x) + 3/4