VITEEE 2015 Paper

We can write given differential equation as,
(D2 - 1) x = k ... (i)
where , D =

d

dy

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 =

1

D2−1

k
= k .

1

D2−1

e0.y
= K .

1

02−1

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)