Question

HIGH SCHOOL DIFFERENTIAL EQUATIONS: http://gyazo.com/74ab054a4598bd38974685146682659e.png
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Answer 1

http://gyazo.com/74ab054a4598bd38974685146682659e.png

Answer 2

Please do not mass tag.

Answer 3

You won't understand the feeling of falling from heaven to hell xD

Answer 4

Do you know how I can start this problem? This teacher is insane.

Answer 5

You can start by this:
\[\begin{array}{rcl}
\frac{dy}{dt}&=&t\sqrt y\\
dy&=&t\sqrt ydt\\
\frac1{\sqrt y}dy&=&t\hspace{3pt}dt\\
\int\frac1{\sqrt y}dy&=&\int t\hspace{3pt}dt\\
\end{array}\]

Answer 6

y = (t^2/2+c)^2

Answer 7

How the hell did \(\int\dfrac1{\sqrt y}dy\) become \(y\)? lol

Answer 8

I integrated it to get 2sqrt(y) then brought the 2 over to the other side then squared it

Answer 9

oh sorry lol

Answer 10

I doubt that it's 2t^2 instead of t^2/2

Answer 11

What do you mean? Might want to check your math lol

Answer 12

oh yes sorry my fault

Answer 13

Oh wait! Is it (t^2/4+c)^2?

Answer 14

oh yes lol, so now you're better than me :)

Answer 15

Ok, continue on

Answer 16

I don't know how to do xP I'm only grade 9 xP

Answer 17

Kenny did a great job of putting this problem into appropriate math symbols. I'll start from there.
\[\int\limits_{-}^{-}\frac{ dy }{ \sqrt{y} }=\int\limits_{-}^{-}y ^{-1/2}dy\]

Answer 18

Integrating this requires the power rule. Are you, Aunt Anna, familiar with that? If so, would you try doing the integration yourself?

Answer 19

I said by doing separation of fariables you would get y = (t^2/4+c)^2

Answer 20

@TanteAnne, you are right the answer is
\[
y=\left(c+\frac{t^2}{4}\right)^2
\]

Answer 21

Continue

Answer 22

Continue

Answer 23

There's nothing to continue, Anna. eliassaab's result is in simplest form.

Answer 24

tHERE ARE STILL 4 MORE PARTS.