Question

Find the solution of the differential equation that satisfies the given initial condition.

Answer 1

\[\frac{ dP }{ dt }=5\sqrt{Pt}, P(1) = 2\]

Answer 2

\[\frac{ 1 }{ \sqrt{P} }\frac{ dP }{ dt }= 5\sqrt{t}*dt\]\[\int\limits_{}^{}P^{-1/2}dP = 5\int\limits_{}^{}t^{1/2}dt\]\[2P^{1/2} = 5[\frac{2 }{ 3 }t^{3/2}] + C\]
Let C/2 = K
\[P = (\frac{ 5 }{ 3 }t^{3/2}+K)^2 = \frac{ 25 }{ 9 }t^3+\frac{ 10K }{ 3 }t^{3/2}+K^2\]

Given condition P(1) = 2,
\[\frac{ 25 }{ 9 }(1)^3+\frac{ 10K }{ 3 }(1)^{3/2}+K^2=2 => \frac{ 25 }{ 9 }+\frac{ 30k }{ 9 }+\frac{ 9 }{ 9 }K^2=2\]\[9k^2+30k+25=18 => 9k^2+30k+7 = 0\]
\[K = \frac{ -30 \pm \sqrt{900-4(9)(7)} }{ 2(9) } = \frac{ -30 \pm \sqrt{648} }{ 18 }= \frac{ -30 \pm 25 \sqrt{23} }{ 18 }\]
\[P = ( \frac{ 5 }{ 3 }t^{3/2}-\frac{ 30 + 25 \sqrt{23} }{ 18 })^2\]

My answer is wrong, what do?

Answer 3

Why did you square both sides omg\[2\sqrt 2 = \frac{10}3+C\]

Answer 4

don't you need to square both sides to get rid of P^1/2?

Answer 5

You got two values for \(C\) when there is only actually one. Squaring both sides always leads to extraneous solutions.

Answer 6

how do you know it's the positive one and not the negative one? lol you have a non-linear equation here so you need to make sure you IVP fits one solution and not the other.

Answer 7

for the positive and negative, I just figure whatever the output condition is: like if P(1) = -2, then it is negative and if P(1) = 2 then it's positive

Answer 8

You don't square the sides at all. Just use\[2P^{1/2} = \frac{10}3t^{3/2}\]

Answer 9

+ c

Answer 10

then divide by 2 both sides?

Answer 11

you can solve for the constant at any time lol... then after you've done so, you can get isolat ethe dependent variable

Answer 12

\[P^{1/2}=\frac{ 5 }{ 3 }t^{3/2}+\frac{ C }{ 2 }\]
Then use the initial condition P(1) = 2 now?

Answer 13

Yeah, sure.

Answer 14

\[\sqrt{2}=\frac{ 5 }{ 3 }+\frac{ C }{ 2 } => \sqrt{2}-\frac{ 5 }{ 3 }= \frac{ C }{ 2 } = > C = 2\sqrt{2}-\frac{ 10 }{ 3 }\]

Answer 15

\[P = (\frac{ 5 }{ 3 }t^{3/2}+\frac{ 2\sqrt{2}-\frac{ 10 }{ 3 } }{ 2 })^2\]

Like that?

Answer 16

OK yeah that works

Answer 17

The website isn't accepting this as an answer, maybe I have to simplify it?

Answer 18

Source of the question

Answer 19

Show's you're multiplying by 2 rather than squaring it

Answer 20

It's not being multiplied by 2, it is actually raised to the 2nd power. It is difficult to see since the parenthesis are large making the exponent look like a multiplication

Answer 21

Maybe try simplifying \[\frac{ 2\sqrt{2}-\frac{ 10 }{ 3 } }{ 2 } = \sqrt{2}-\frac{ 5 }{ 3 }\]

Answer 22

Yep that was it, had to simplify it.

Answer 23

That's silly, but glad it worked out!

Answer 24

Haha yea, Thanks all!