Question

solve the initial value problem:
y' = 2x cos^2(y), (3,0)

Answer 1

how do i find the derivative of cos^2(y)?

Answer 2

antiderivative***

Answer 3

\[\frac{dy}{dx}=2xcos^2(y)\]

Answer 4

ok u use trig identity

Answer 5

i will show u

Answer 6

\[\sin^2(x)+\cos^2(x)=1\]
\[-\sin^2(x)+\cos^2(x)=\cos(2x)\]
by adding both equation yields to
\[2\cos^2(x)=1+\cos(2x)\]
\[\therefore \cos2(x)=\frac{1}{2}[1+\cos(2x)]\]

Answer 7

\[\frac{1}{2}\int\limits\limits [1+\cos(2y)]dy= \frac{1}{2}y+\frac{1}{4}\sin(2y)\]

Answer 8

+c

Answer 9

are u there?

Answer 10

Woops, this problem is actually a bit easier than that. We don't need to know the anti-derivative of cos^2x.

Notice that to solve this IVP you'll need to divide the cos^2 to the other side, so it'll turn into sec^2x

Answer 11

\[\large \frac{dy}{dx}= 2x \cos^2y \qquad \rightarrow \qquad \frac{1}{\cos^2y}\cdot \frac{dy}{dx}=2x\]
\[\large \sec^2y\cdot\frac{dy}{dx}=2x \qquad \rightarrow \qquad \sec^2y\;dy=2x\;dx\]Confused by any of those steps?
This is the process you would want to take BEFORE you integrate.

Answer 12

sorry about that, i just got back home. and ooooh okay i understand that.

Answer 13

so now im at tan(y) = 2x + c
i guess now i just plug in my point?

Answer 14

tan(y)=x^2+c right? Don't forget to integrate the X side :)

Answer 15

haha oops thats what i meant ;p

Answer 16

Mmmmm yah try plugging in the point. You should be able to solve for C if you do that.

Answer 17

tan y = x^2 - 9

Answer 18

Yay good job!
If they want you to solve explicitly for y, then you have another tricky step to do.

Answer 19

i was going to ask. i dont know how to get y by itself. lol

Answer 20

\[\large \tan(a)=b \qquad \qquad \rightarrow \qquad \qquad a=\arctan(b)\]
The inverse function will swap the... pieces :p
If you wanted to be more thorough about it, you can say that you're applying the arctangent to both sides, and the arctan will "undo" the tangent, leaving you with y on the left.

Answer 21

okay so it'll just be the y = arctan(x^2 - 9) ?

Answer 22

yay good job \c:/

Answer 23

thank you zep! definitely my most helpful person on here c:

Answer 24

You seem to be getting these a lot faster now! That's good to see. No potato math today! c: