Question

will award medal, find the solution to the differential equaiton. 3(du/dt)=u^2 where u(0)=5

Answer 1

have trouble separating the variables?

Answer 2

try dividing u^2 on both sides and multiply dt on both sides

Answer 3

\[\int\limits1/u^2du=\int\limits1/3dt\]

Answer 4

is this how you would set up your integration

Answer 5

\[3 du =u^2 dt \\ 3 \frac{1}{u^2} du=dt \\ \text{ or } \frac{1}{u^2} du=\frac{1}{3} dt \]
2nd line or 3rd line works
so in other words what you have is good

Answer 6

so to solve for the equation you have |u^2|=(e^(t/3))e^c where (e^c=b)

Answer 7

where did all e^ come from?

Answer 8

you don't have 1/u du on the left

Answer 9

when you integrate you have ln|u^2|, and to get rid of the ln you need to raise both sides under e

Answer 10

\[\int\limits_{}^{}u^{-2} du= \frac{u^{-2+1}}{-2+1}+C\]

Answer 11

\[\int\limits_{}^{}u^{-n} du=\frac{u^{-n+1}}{-n+1}+C ; n \neq 1 \\ \text{ when n=1 we have } \int\limits_{}^{}u^{-1} du=\ln|u|+C\]

Answer 12

ohhh right

Answer 13

=-1/u

Answer 14

-1/u=t/3+c

Answer 15

yep yep :)

Answer 16

c=-1/5

Answer 17

u(0)=5
gives
-1/5=0/3+c
yes c=-1/5
you are right

Answer 18

so it asks the question in terms of u=...

Answer 19

so the first step I would make is writing the right hand side as one fraction by combining the fractions

Answer 20

then just flip both sides

Answer 21

u=-(15/5t-3)

Answer 22

hell yea thanks

Answer 23

np