anonymous
  • 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!
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
This is a separable differential equation. You just need to rearrange:\[6p^{-4}dp=dt\]integrate and solve for your constant.
anonymous
  • anonymous
Is this how you started?
anonymous
  • 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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anonymous
  • anonymous
So\[p(0)=9 \rightarrow 9^{-3}=c-\frac{0}{2}=c\]
anonymous
  • 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
  • anonymous
I get\[p(t)=\sqrt[3]{\frac{2}{c-t}}\]
anonymous
  • anonymous
\[p(0)=9 \rightarrow 9=\sqrt[3]{\frac{2}{c}} \rightarrow c=\frac{2}{9^3}\]
anonymous
  • anonymous
yep thats what I've been plugging into wiley plus to no avail
anonymous
  • anonymous
What exactly have you been plugging in?
dumbcow
  • dumbcow
i get a constant of -2/243 -> p(t) = cubed root (-486/(243t - 2))
anonymous
  • anonymous
cubed root of (2/729-t)
anonymous
  • anonymous
Maybe you need to use \[c=-\frac{2}{9^3}\]for a solution\[p(t)=\sqrt[3]{\frac{-2}{t+c}}\]
anonymous
  • anonymous
\[p(t)=9\frac{\sqrt[3]{-2}}{\sqrt[3]{9^3t-2}}\]
anonymous
  • anonymous
These are just different ways of saying the same thing. These online submission things are such a crapshoot.
anonymous
  • 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
  • anonymous
substituting gives me c=1/12 so u(t)=square root (2/3t+1/12)?
anonymous
  • anonymous
you are correct a crapshoot it is
anonymous
  • anonymous
\[u= \frac{6}{\sqrt{1-24t}}\]
anonymous
  • anonymous
positive root only satisfies the boundary condition.
anonymous
  • anonymous
I had c=-1/24
anonymous
  • anonymous
once you integrated did you get 3/2 u^2=t+C
anonymous
  • anonymous
and this checks out with wolfram alpha
anonymous
  • anonymous
\[-\frac{3}{2}u^{-2}=t+c\]after integration.
anonymous
  • anonymous
aha i see what you did, nice job! Thank you for the help!
anonymous
  • anonymous
np
anonymous
  • anonymous
So, does that mean your other question is defunct now?
anonymous
  • anonymous
well i have an understanding of the problem but all of my attepmts are used
anonymous
  • anonymous
but thank you
anonymous
  • anonymous
you're welcome :)

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