The equation is dy/dt=-ay+b where a and b are positve numbers. I have the solutions guide and it says that equation also equals dy/dt=-a(y-(b/a)) and then dy/(y-b/a)=-a. I can't seem to get the equation from the 1st form to the 2nd for some reason. I haven't taken calc in over a year so I'm a bit rusty.
No. As far as I can see, a and b have nothing to do with the initial conditions.
I hope you realise that when you integrate, an arbitrary constant of integration is introduced. The value of that constant can be determined if intitial conditions are given eg. y = 1 when t = 0.
By the way, if you're studying differential equations you better make it your urgent business to rust rid your calculus.
Note for dy/dt =- ay + b
y = b/a is an equiilbrium solution.
your initial conditions do depend on a and b in the sense that if y(0) < b/a
then the solution curve approaches the line y= b/a from below.
If y(0) > b/a then the solution curve decays to the line y =b/a from above.