Question

differentiate
dy/dx = xsqrt(y)cos^2(sqrt(y))

Answer 1

So you want to find y'' given y'
so we have
\[y'=x \sqrt{y} \cos^2(\sqrt{y})\]

Answer 2

no i think you need to find y

Answer 3

the directions said solve for y

Answer 4

so we want to solve the differential equation not differentiate ?

Answer 5

so this is a separation of variables problem

Answer 6

it is a separable diffeq...not too bad

Answer 7

\[\frac{1}{\sqrt{y} \cos^2(\sqrt{y})}dy=x dx\]

Answer 8

integrate both sides

Answer 9

don't forget to put +C on of the sides

Answer 10

one*

Answer 11

how would you integrwte the left side though? thats what confuses me

Answer 12

that one side is not that bad with a substitution u=sqrt{y}
and write 1/cos^2 as sec^2

Answer 13

\[u=\sqrt{y} => du=\frac{1}{2 \sqrt{y}} dy => 2 du=\frac{dy}{\sqrt{y}}\]

Answer 14

\[\int\limits_{}^{} 2 \sec^2(u) du\]

Answer 15

\[2 \tan(u)+C=2 \tan(\sqrt{y})+C\]

Answer 16

did i lose you?

Answer 17

no thank you so much that really helped

Answer 18

ok you still have to do the other side don't forget

Answer 19

wait so how would you format your final answer?

Answer 20

\[2 \tan(\sqrt{y})+C=\frac{x^2}{2}\]

Answer 21

I would leave it like this

Answer 22

I think Some teachers like you to solve it for y I think

Answer 23

yeah that would be a pain but thanks for all your help

Answer 24

I would leave it like that too.

Answer 25

Zarkon has a phd so listen to him.

Answer 26

okay that saves a lot of work

Answer 27

since the tangent is periodic when you take the inverse tangent you might not reclaim the y value you wanted.

Answer 28

right moving stuff around sometimes messes with domain right?

Answer 29

yes...best to leave it as it is

Answer 30

just like you aren't suppose to solve this x=x^2 by dividing x on both sides