# 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.