We can write given differential equation as,
(
D2 - 1) x = k ... (i)
where , D =
Its auxiliary equation is
m2 - 1 = 0, so that
m = 1 , - 1
Hence , CF =
C1ey+C2e−y where
C1,C2 are arbitrary constants
Now , also PI =
k
= k .
e0.y = K .
e0.y = - K
So, solution of eq. (i) is
x =
C1ey+C2e*−y−k ... (ii)
Given that x = 0, when y = 0
So , 0 =
C1+C2 - k (From (ii))
⇒
C1+C2 = k ... (iii)
Multiplying both sides of eq. (ii) by
e−y we get x .
e−y =
C1+C2e−2y−ke−y ... (iv)
Given that x → m when y + oo,m being a finite quantity. So, eq (iv) becomes
x × 0 =
C1+C2 × 0 - (k × 0)
⇒
C1 = 0 ... (v)
From eqs. (iv) and (v), we get
C1 = 0 and
C2 = k
Hence, eq. (ii) becomes
x = k (
e−y - 1)