Question

Find the solution to the differential equation, subject to the given initial condition 6*dp/dt=p^4, p(0)=9. it seems so simple but cant get the answer right
please Help!

Answer 1

This is a separable differential equation. You just need to rearrange:\[6p^{-4}dp=dt\]integrate and solve for your constant.

Answer 2

Is this how you started?

Answer 3

Anyway, integrate both sides:\[6\frac{p^{-3}}{-3}=t+c \rightarrow -2p^{-3}=t+c \rightarrow p^{-3}=c-\frac{t}{2}\]

Answer 4

So\[p(0)=9 \rightarrow 9^{-3}=c-\frac{0}{2}=c\]

Answer 5

yes and i have to write it out as p(t)= but i keep getting it wrong
so far i havep^3=(t+C)/-2

Answer 6

I get\[p(t)=\sqrt[3]{\frac{2}{c-t}}\]

Answer 7

\[p(0)=9 \rightarrow 9=\sqrt[3]{\frac{2}{c}} \rightarrow c=\frac{2}{9^3}\]

Answer 8

yep thats what I've been plugging into wiley plus to no avail

Answer 9

What exactly have you been plugging in?

Answer 10

i get a constant of -2/243
-> p(t) = cubed root (-486/(243t - 2))

Answer 11

cubed root of (2/729-t)

Answer 12

Maybe you need to use \[c=-\frac{2}{9^3}\]for a solution\[p(t)=\sqrt[3]{\frac{-2}{t+c}}\]

Answer 13

\[p(t)=9\frac{\sqrt[3]{-2}}{\sqrt[3]{9^3t-2}}\]

Answer 14

These are just different ways of saying the same thing. These online submission things are such a crapshoot.

Answer 15

i have new values now 3 *du/dt= u^3, u(0)=6 which then i get once integrated
u^-2= 2/3t+C

Answer 16

substituting gives me c=1/12 so u(t)=square root (2/3t+1/12)?

Answer 17

you are correct a crapshoot it is

Answer 18

\[u= \frac{6}{\sqrt{1-24t}}\]

Answer 19

positive root only satisfies the boundary condition.

Answer 20

I had c=-1/24

Answer 21

once you integrated did you get 3/2 u^2=t+C

Answer 22

and this checks out with wolfram alpha

Answer 23

\[-\frac{3}{2}u^{-2}=t+c\]after integration.

Answer 24

aha i see what you did, nice job! Thank you for the help!

Answer 25

np

Answer 26

So, does that mean your other question is defunct now?

Answer 27

well i have an understanding of the problem but all of my attepmts are used

Answer 28

but thank you

Answer 29

you're welcome :)