# Help with initial value problem

1. Jan 30, 2012

### cameuth

OK, so clearly I am missing something, because I know this is supposed to be a simple problem. It reads:

solve the following initial value problem:
dy/dt=-y+5
y(0)=y_naught

my process is as follows:
dy/(5-y)=dt
integrate
ln(5-y)=t+C
exponential both sides
5-y=(e^t)(e^c)
y=5-(e^t)(e^c)

solve for constant:
y_naught=5-e^c
e^c=5-y_naught

y=5-(5-y_naught)e^t

My book disagrees with this answer slightly, can anyone see where I've stumbled in the process?

2. Jan 30, 2012

### tiny-tim

hi cameuth!
nooo

try dy/(y-5) = -dt

3. Jan 30, 2012

### cameuth

I'm sorry... but can you explain why that is. Intuitively what I see is
dy/dt=5-y
divide by (5-y) multiply by dt
dy/5-y=dt

I'm sorry again that I'm such a beginner at this, I just don't understand why you did what you did. I see that it gets me the right answer your way, but not why we go that path.

4. Jan 30, 2012

### Char. Limit

Because the integral of 1/(5-y) isn't ln(5-y). Chain rule.

5. Jan 31, 2012

### tiny-tim

(just got up :zzz: …)

yes … you missed out a minus

6. Jan 31, 2012

### cameuth

sorry for the late reply. I got the answer guys, and y'all were a ton of help. Seriously, thanks. It's been awhile since six months since I've done cal 3 and so my integrating has some rust to knock off.