1. ## Differential Equations

I think I'm understanding better, but still not quite there yet. Please point out my mistakes (if any lol).

Find the general solution of the given differential:

dy/dt = [te^y]/(2t-1)

d/e^y (te^y) = d/dt 2t-1

integral d/e^y (te^y) d/dy = integral d/dy (2t-1) d/dt

integral e-y = 2 integral (t-1) dt
ln -[y] = 2/(t-1) + C

2. Originally Posted by startingover
I think I'm understanding better, but still not quite there yet. Please point out my mistakes (if any lol).

Find the general solution of the given differential:

dy/dt = [te^y]/(2t-1)

d/e^y (te^y) = d/dt 2t-1

integral d/e^y (te^y) d/dy = integral d/dy (2t-1) d/dt

integral e-y = 2 integral (t-1) dt
ln -[y] = 2/(t-1) + C
The mistake is that you are dividing both sides by $te^{y}$ you shoudl inside divide by $e^y$.

So you end up with,
$e^y \cdot \frac{dy}{dx} = \frac{t}{2t-1}$

3. ## here is the solution

Here is the solution