Question

Find the general solution of the DE equation in comments

Answer 1

\[y'=(\frac{ \sin x }{ y })^2\]

Answer 2

Square the top and bottom. It's seperable

Answer 3

so i would have sin^2(x)/y^2?

Answer 4

Yup

Answer 5

then multiply both sides by y^2 correct?

Answer 6

Yup

Answer 7

how do i find the integral of sin^2(x) dx?

Answer 8

it wouldn't be just -cos^2(x) right

Answer 9

No. Remember sin^2(x) = (sin(x))^2 which you can't integrate directly. Differentiate -cos^2(x) to see you don't get the right answer.
Use the power reduction identity sin^2(x) = (1-cos(2x))/2

Answer 10

so i get \[\frac{ y^3 }{ 3 }=\frac{ 1 }{ 2 } \int\limits 1-\cos(2x)dx\]

Answer 11

what about that integral?

Answer 12

Yes. It would be nicer to put the parens in \[\frac{1}{2}\int\limits(1-\cos(2x))dx\]
You can split it across the subtraction \[\frac{1}{2}\int\limits(1-\cos(2x))dx = \frac{1}{2}\int\limits1\ dx - \frac{1}{2}\int\limits \cos(2x)dx\]

Answer 13

i get \[\frac{ y^3 }{ 3 }= \frac{ 1 }{ 2 }(x-\frac{ 1 }{ 2 }\sin(2x))+C\] is that correct?

Answer 14

Looks good

Answer 15

why does wolframalpha have the answer as \[\frac{ 1 }{ 2 }(x-\sin(x)\cos(x))+C\]

Answer 16

Because sin(2x)=sin(x+x)=sin(x)cos(x)+sin(x)cos(x)=2sin(x)cos(x)

Answer 17

so its the same answer either way?

Answer 18

Yes.

Answer 19

Alright. Thanks a lot!

Answer 20

Sure thing