a) Solve dy/dt - y = 1

b) What should be the value of y(0) = yo so that the solution to part a is finite even when t -> infinity

This is how I tried:

a) dy/dt - y = 1

dy/dt = y+1---------------(1)

Case 1: y+1 = 0

Or, y = -1

Case 2: y+1 not= 0

Divide both sides of (1) by y+1

1/(y+1) dy/dt = 1

dy/(y+1) = dt

Integrating,

ln |y+1| = t + C

|y+1| = e^(t+C)

|y+1| = B e^t (where B = e^C; B > 0 because e^C > 0 for any value of C)

y+1 = B e^t, or y+1 = -B e^t

y+1 = A e^t (where A = +-B; A > 0 or A < 0)

y = A e^t - 1

The solution in case 1 also can be written as y = A e^t - 1 if A = 0

Therefore, the generic solution is

y = A e^t - 1 (where A is any real number as constant)

Is this correct?

b) Let dy/dt = f(y)

f(y) = y+1

f(y) = 0 at y = -1

When y < -1, then f(y) < 0

When y > -1, then f(y) > 0

Therefore f(y) moves away from y=-1

Therefore the solution is unstable for any value other than y = -1 irrespective of y = -1

When A = 0, then y = -1

The solution remains finite even when t -> infinity

Therefore the required value of y0 is y0 = -1

Is this correct? But I am not sure of the solution to part b.