Question

help please: find particular solution:

cos^2(y^2)cos(x)dx-2sin(x)ydy=0

y(pi/2)=sqr((pi/2))

Answer 1

To confirm:
\[\cos^2 y^2 . \cos x. dx - 2 \sin x . y . dy = 0\]

Answer 2

is that right?

Answer 3

you there?

Answer 4

Ok ... I see how to solve it. Let me know when you're back.

Answer 5

Im here, is right

Answer 6

Ok, the first thing you want to get rid of are the y^2 terms. So substitute u = y^2 and this will simplfy the equation somewhat. Write down what you get in terms of u and x.

Answer 7

Yes we have the same thing. Now follow the first step I just gave you.

Answer 8

and with 2sinxy?

Answer 9

If u = y^2, then du = 2y dy.

Answer 10

So
cos^2(y^2)cos(x)dx-2sin(x)ydy=0
becomes

cos^2 u . cos x . dx - sin x du = 0

Now you should have no problem in theory solving that equation because it is separable.

Answer 11

ok

Answer 12

nicely done jj. Medal for you.

Answer 13

cos^2u/du=sinx/cosx ·dx

Answer 14

doesn't look right.

Answer 15

How I do separate rightly?

Answer 16

cos^2 u . cos x . dx - sin x du = 0
=> cos^2 u . cos x . dx = sin x du
=> cos x/sin x . dx = sec^2 u du

Answer 17

ok, them integrate?

Answer 18

yes

Answer 19

log(sin(X)) =tan(u) du ?

Answer 20

log(sin(X)) =tan(u) du ?

Answer 21

ln(sin x) = tan u + C, yes

Answer 22

Well, first I'd solve for y(x) and then find the particular solution.

Answer 23

and now to find the particular solution ?

Answer 24

undo the u and replace the value, right?

Answer 25

solve for y, yes

Answer 26

thank u very much 4 your help!