Question

can anyone help me with this differential eq please t dy/dt + 4y^2=0 t>0

Answer 1

rearrange the equation separate the variables.

Answer 2

is it ok if I leave the -4 on the right side

Answer 3

you mean on the right side?
sure, wy not...

Answer 4

After doing that I got -y^-1=-4lnt+c

Answer 5

So I know the negative would cross out but what about the C would it be -C?

Answer 6

negative would cross out (?), elaborate please...

Answer 7

uh so I have y= 1/4lnt +C. I'm not sure if the answer is that or y=1/4lnt-C

Answer 8

if you had y^-1 = -4 ln(t) + C then wouldn't y be over that ?

Answer 9

If i was solving for y for example

Answer 10

y^-1 = 4 ln(t) + C
raise both sides to -1 exponent,
y = 1 / ( -4 ln(t) + C )

Answer 11

oh I meant that there was a negative before y^-1

Answer 12

to be fair though, integral 1/t = |t| and you don't need t>0

Answer 13

and I have a typo

Answer 14

-y^-1 = 4 ln(t) + C
y^-1 = - 4 ln(t) + C

(+C or -C, you are still adding an arbitrary constant)

y = 1 / ( - 4 ln(t) + C )

Answer 15

gotcha thank you so much :)

Answer 16

this is your y function apparently

Answer 17

https://www.desmos.com/calculator/go2zsfuum4

Answer 18

** set of functions