- anonymous

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!

- jamiebookeater

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- anonymous

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

- anonymous

Is this how you started?

- anonymous

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

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## More answers

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

What exactly have you been plugging in?

- dumbcow

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

- anonymous

cubed root of (2/729-t)

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

you are correct a crapshoot it is

- anonymous

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

- anonymous

positive root only satisfies the boundary condition.

- anonymous

I had c=-1/24

- anonymous

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

- anonymous

and this checks out with wolfram alpha

- anonymous

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

- anonymous

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

- anonymous

np

- anonymous

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

- anonymous

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

- anonymous

but thank you

- anonymous

you're welcome :)

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