Question

Help please

Answer 1

Differential Equations.
\[\frac{ dp }{ dt }=2\sqrt{pt}\] with y(0) = −1 as the initial value.

Answer 2

Here is my work. I messed up somewhere. Could someone please check my work?

Answer 3

@random231

Answer 4

You can quickly check wolfram to make sure you have the right 'final' answer.

Answer 5

What happened to the 2?

Answer 6

Where did the two go?

Answer 7

Oh Sorry. One sec...

Answer 8

\[ \frac{dp}{dt} = 2\sqrt{pt}
\\ \, \\
\frac{dp}{dt} = 2\sqrt{p}\sqrt{t}
\\ \, \\
\frac{1}{\sqrt p} \frac{dp}{dt} = 2\sqrt{t}
\\ \, \\
\int p^{-1/2}~ \frac{dp}{dt} ~dt= 2\int t^{1/2}~dt
\\ \, \\
\int p^{-1/2}~ dp= 2\int t^{1/2}~dt
\\ \, \\
\frac{ p^{1/2}}{1/2} =2 \frac{t^{3/2}}{3/2} + C
\\ \, \\
2\sqrt p =2 \frac 2 3 t^{3/2}+ C
\\ \, \\
\sqrt p = \frac 2 3 t^{3/2}+ C^{*}
\\ \, \\
p = \left( \frac 2 3 t^{3/2}+ C \right)^2


\]

Answer 9

you can leave it in the square root form
to find C

Answer 10

In your directions you have a mixture of variables.
p , t, and y.

Answer 11

Differential Equations.
\[\frac{ dp }{ dt }=2\sqrt{pt}\] with y(0) = −1 as the initial value. <--- did you mean p(0) = -1 ?

Answer 12

Then would... \[c = \frac{ 3\sqrt{5}-2 }{ 3 }\]

Answer 13

Yes i did, sorry

Answer 14

@owen3 Oh wait. no i didnt. I looked at at a different problem...

Answer 15

p(1)=5

Answer 16

I was going to say, this gives you a non real solution.

Answer 17

Haha, sorry

Answer 18

\[
\sqrt p = \frac 2 3 t^{3/2}+ C^{*}

\\ \, \\
\sqrt 5 = \frac 2 3 1^{3/2}+ C^{}
\\ \, \\
C = \sqrt 5 - \frac 2 3
\]

Answer 19

\[ p = \left( \frac 2 3 t^{3/2}+ \sqrt 5 - \frac 2 3 \right)^2

\]

Answer 20

Is it possible they meant \[ \frac{dP}{dt} = 2 \sqrt{ P(t)} \]
also that should be capital \[P \]

Answer 21

oh okay. haha... thanks.

Answer 22

Yes.

Answer 23

Could you check something for me real quick though?

Answer 24

I watched this video tutorial on a problem of mine, but it wont accept the answer? Did i happen to input it incorrectly?

Answer 25

@owen3 See, i got it right :)

Answer 26

You want to solve
$$ \frac{dy}{dt} = ky( 1 - y) $$