I was wondering if someone could check my diff eq I had to solve with an initial condition:
dy/dt + 2y = 1, y(0) = 5/2
MY WORK:
dy/dt = 1 - 2y => dy = (1 - 2y) dt => dt = 1/(1 - 2y) dy
Take integral of both sides:
t = -1/2 * ln|2y - 1| + C
Using the initial condition:
0 = -1/2 * ln|2*0 - 1| + C
0 = 0 + C
Thus C = 0.
And therefore the solution is t = -1/2 * ln|2y - 1|.
Anyone see any mistakes?
Thanks for your help Jhevon.
Luckily for me, my prof. doesn't care if its in the form y(t) = something.
So here goes my solution:
dy/dt + 2y = 1
dy/dt = 1 - 2y => dy = (1 - 2y)dt => dt = 1/(1 - 2y) dy
Take integral of both sides:
t = -1/2 * ln|2y - 1| + C
Using the intial condition:
0 = -1/2 * ln|2*(5/2) - 1| + C
0 = -ln(2) + C
=> C = ln(2)
And thus, the solution is:
t = -1/2 * ln|2y - 1| + ln(2)