Your solution to the first part is excellent.
Now, given , note that: (i) If then as ; (ii) If then as . Thus the only value of which won’t give a runaway value with time is , i.e. must take the constant value with regard to time.
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.
Your solution to the first part is excellent.
Now, given , note that: (i) If then as ; (ii) If then as . Thus the only value of which won’t give a runaway value with time is , i.e. must take the constant value with regard to time.