# Homework Help: Help with a simple initial value problem

1. Jan 8, 2009

### bengaltiger14

1. The problem statement, all variables and given/known data

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

This is how I work it. I integrate both sides and get y= 2y^2 - 2y + C

I then solve for C: 3=2(0)^2 - 2(0) + C

C = 3

Is this the correct way to solve for C?

2. Jan 8, 2009

### rock.freak667

$$\frac{dy}{dt}=4y-2$$

$$\frac{1}{4y-2}dy= 1 dt$$

Separation of variables. Now integrate.

3. Jan 8, 2009

### HallsofIvy

It's not solving for C that is the problem! You have integrated on the left side with respect to t and on the right with respect to y. You can't do that. In order to integrate with respect to t you have to "multiply" by dt: (dy/dt) dt= dy, but then the right side is (4y- 2)dt which is NOT the same as (4y- 2)dy.

What you can do, as rock.freak667 suggested is "separate" the equation so you have only y on one side and t on the other:
dy= (4y-2)dt so dy/(4y-2)= dt. NOW you can integrate the left side with respect to y (because of the "dy") and the right side with respect to t (because of the "dt"). After you have done that you can set y= 3, t= 0 to find C, the constant of integration.