Differential Equations: Find all solutions of these equations

1. Jan 21, 2010

ballajr

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

1) y' + ycos(t) = 0
2) y' - 2yt = 0
3) y'tan(t) + ysec2(t) = 0

4) Let p(x) be an arbitrary polynomial function. Note that the following is a polynomial
solution to y" = p(x) + y:

y = -p - p" - p(4) - p(6) - ... - p(n)

where n is the smallest positive even integer such that p(n+1) is the zero polynomial.
Prove that the following differential equation has a polynomial solution:

y''' = p(x) + 2y

2. Relevant equations

3. The attempt at a solution

1) First, I tried moving the ycos(t) to the right, but I don't know where to go from there.
y' = ycos(t)

2) I moved 2ty to the right. Integrated it.
y' = 2ty

3) This one is just a beast. I was never good at trigonometry. I moved the ysec^2t over to the right and canceled out the trig functions to get down to this:
y' = ysec(t)

I don't know where to go from here.

4) I don't even know how to go about doing this problem.

2. Jan 21, 2010

sutupidmath

All three of them can be solved by separation of variables. That is if we have a diff. eq of the form:

$$\frac{dy}{dt}=H(y,t)$$

we say it is a separable diff. eq. if $$H(y,t)=\frac{f(t)}{g(y)}$$.

Try to do simmilar stuff with your particular functions. You've got the first step right for the 1st: moving ycos(t) to the other side(be careful of the sign though). what next? also good job with the 2nd.

3. Jan 21, 2010

ballajr

Oh okay. There's going to be more than one solution for all three of those right? My teacher likes to trick us like that. The second one i haven't solved yet. The fourth one is a bit confusing. I need some help with that.