## anonymous 5 years ago 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!

1. anonymous

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

2. anonymous

Is this how you started?

3. anonymous

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

4. anonymous

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

5. 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

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

What exactly have you been plugging in?

10. anonymous

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

11. anonymous

cubed root of (2/729-t)

12. anonymous

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

13. anonymous

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

14. anonymous

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

15. 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

16. anonymous

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

17. anonymous

you are correct a crapshoot it is

18. anonymous

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

19. anonymous

positive root only satisfies the boundary condition.

20. anonymous

21. anonymous

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

22. anonymous

and this checks out with wolfram alpha

23. anonymous

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

24. anonymous

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

25. anonymous

np

26. anonymous

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

27. anonymous

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

28. anonymous

but thank you

29. anonymous

you're welcome :)