# Separating Variables

1. Sep 2, 2009

### KevinL

I would just like to have my first two questions checked, and on the last one I'm not even quite sure how to start it.

Find the general solution of the differential equation.

1) dy/dt = t^4*y

dy/y = t^4 dt

Integrate both sides:

ln|y| = (t^5)/5 +c

Raise both sides by e to get rid of ln:

y = +/- ce^(t^5/5)

Is y = 0 also a solution (equilibrium)?

2) dy/dt = 2-y

dy/(2-y) = 1 dt

integrate both sides:

-ln|2-y| = t+c
ln|2-y| = -t+c

Raise both sides by e to get rid of ln:

|2-y| = ce^-t
y=ce^-t +2

Is y=2 also a solution (equilibrium)?

3) dy/dt = 1/(ty + t + y + 1)

Im not even sure how to get this in the form dy/dt = f(t)g(y)

2. Sep 2, 2009

### rock.freak667

This is good.

so if y=0 is a solution, then 0=cet5/5

is there any value of t that satisfies this?

This is good as well. Similar, with the one above, 2=ce-t+2. Is there any t to satisfy this?

EDIT: even though it is correct, you should use a different letter for the constant c, just so it isn't technically wrong.

ty+t+y+t ≡ ty+y +t+1 ≡ (t+1)y+(t+1)

see it now?

Last edited: Sep 2, 2009
3. Sep 2, 2009

### KevinL

t=0. The way I thought of it originally was that we had the integral dy/y, so if y=0 you obviously would get problems. Is this a fair way to think of it?

t=0. Similarly, when you have dy/(2-y), y=2 would give you problems. Again, is this a fair way to find the solution y=2?

Ah, of course. So...

(y+1) dy = 1/(t=1) dt

Integrate both sides:

y^2/2 + y = ln|t+1|

e^(y^2/2)*e^y -1 = t

Now we're stuck (or at least im stuck! . The HW instructions mentioned that this could happen, but to get as "far as you can". Does this suffice for an answer?

4. Sep 2, 2009

### rock.freak667

Say we had ex=2 and we wanted to find x. We'd take ln of both sides

lnex=ln2 => x*lne=ln2 =>x=ln2

so now for ex=0, take ln lnex=ln(0). We have a problem now, ln(0) does not exist. So there is no x that satisfies ex=0. Better now?

5. Sep 2, 2009

### KevinL

Makes sense now. Thank you for the help :)