Question

y' - y = 2te^(2t) ; y(0) = 1

first order differential equation..

i need to know how to solve it well basic steps i know it is following the general form dy/dt = p(t)y = g(t)

Answer 1

got it nevermind :)

Answer 2

There is a 'particular' solution, yp, and the homogeneous solution yh. The solution yh satisfies the equation y' - y = 0, and the general solution y = yp + C.yh satisfies the inhomogeneous equation of the problem

So find these two solutions, add them up, and then make them meet the initial condition.

Step 1: Homogeneous solution
if y' - y = 0, you can separate variables and dy/y = dx. Thus yh = ....

Step 2: Particular solution
There is a definite procedure for this that I know from more general theory, which I won't give you here and I'm guessing you haven't learnt it yet.

But if you had to guess, you would say it's something like,
yp(x) = Ate^(2t) + Be^(2t).
Put this into the equation and you'll find what A and B work.

Step 3: Solve the IVP
Write now the general solution
y(x) = yp(x) + C.yh(x)
and substitute the initial condition y(0) = 1 to solve for C.

Answer 3

thanks jamesj i think i get it now i was making a slight error in my initial value so my factor wasn't right