Question

Solve the differential equation dy/dt = 1/(2y+1)

Answer 1

(2y+1) dy = 1 dt

integrate now

Answer 2

I just started self studying DIFF EQ yesterday (taking the course once semester starts in 2 weeks) so I'll have a crack at it.
(2y+1) dy = 1 dt
y^2+ y = t
Is that it?

Answer 3

That was the easy part I need the whole thing y(t) =

Answer 4

dont forget the constant of integration

Answer 5

Here is another one dy/dt = y(1-y) do you know how to do that ? It says the answer is y(t) = ke^t/(ke^t+1)

Answer 6

ooo right, u(t)=e^t
multiply u(t) throughout
(e^t * y)'=te^t
integrate: ye^t=e^t (t-1)
y(t)=t-1+C
is that what you got for the 1st question as well?

Answer 7

No that's not what I got. I figured it out. The answer was (-1 +- sqrt(4t+c))/2

Answer 8

yah you need to complete the square for first equation

Answer 9

right

Answer 10

Is there any particular reason why you have to complete the square for the 1st?

Answer 11

cuz you want to solve the function, \(y(t)\) that satisfies the given differential equaiton

Answer 12

you have a quadratic equation

Answer 13

for 2nd question :-

\(\large \frac{dy}{dt} = y(1-y)\)

\(\large \frac{1}{y(1-y)} dy = dt\)

integrate now

(left side requires partial fractions)

Answer 14

can you split left side into 2 fractions?
ita quite straight forward