Help with a simple initial value problem

[bengaltiger14]

Homework Statement

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?

 

[rock.freak667]

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

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

Separation of variables. Now integrate.

 

[HallsofIvy]

Homework Statement

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?

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.