# Help with initial value problem

1. Jan 13, 2009

### bengaltiger14

Solve: dy/dt = 4y-2, y(0)=3

I first seperate variables: dy/(4y-2) = dt

I then integrate both sides to get: ln|4y-2|=t + c

I next set 4y-2=Ce^t

Do I just plug 3 in for y to get C=10 or am I going about this problem the wrong way??

2. Jan 13, 2009

### Staff: Mentor

You're close, but made an error.
Your antiderivate for 1/(4y - 2) is not ln|4y - 2|.

Factor 4 from the two terms in the denominator, and then multiply both sides by 4, getting

dy/(y - 1/2) = 4dt

Then continue from there.

3. Jan 13, 2009

### bengaltiger14

Ok, so it should be: ln|y-1/2| = 4t+C

then: y-1/2 = Ce^4t

I am not really sure what to do here.

4. Jan 13, 2009

### bengaltiger14

Do I just set y=3 and t=0 to get C = 2.5?

5. Jan 13, 2009

### NoMoreExams

You seem to be using C in 2 places which is fine as long as you recognize that they are different things.

ln|y-1/2| = 4t+C => y - 1/2 = Ae^(4t) where A = e^C

6. Jan 13, 2009

### bengaltiger14

so,
y-1/2=e^C*e^(4t)

Then C still equals 2.5 right?

7. Jan 13, 2009

### NoMoreExams

Well if t = 0 and y = 3 then 2.5 = e^C*e^(0) => e^C = 2.5 i.e. C = ln(2.5)... how are you getting C = 2.5?

The point IS C = ln(2.5) is just some constant once again, so I can call that constant. If I defined A = e^C = e^(ln(2.5)) = 2.5 then I would get what you started with. You need to understand that you can't just say C = e^C and say that that equals 2.5, realize which constant you are talking about.

8. Jan 14, 2009

### HallsofIvy

Staff Emeritus
If y- 1/2= Ce4t and you know that y(0)= 3, then 3-1/2= Ce0= C.
Yes, C= 5/2= 2.5.

y(t)= 1/2+ 2.5 e4t which can also be written
$$y(t)= \frac{1+ 5e^{4t}}{2}$$