Question

tg

Answer 1

+-arccos(0.25cosecx(c+4log(sinx)))

Answer 2

how did you get that? =)

Answer 3

first order equations should have a dx and dy.. and im confused

Answer 4

It does. y' = dy/dx.

Answer 5

Yep, 'exact'ly!

Answer 6

thank you.. and what do you think is the first step to answer this? im sorry.

Answer 7

Okay, well exact equations are special brand. They satisfy the following condition:\[\frac{dM}{dy}=\frac{dN}{dx}\]
When given an exact equation of the form:\[M(x,y) dx+N(x,y)dy=0\]
The general solution f(x,y) also satisfies the following equation:\[df = \frac{df}{dx}dx+\frac{df}{dy}dy\]
So you need to find your partials of your terms, and then compare those to the 'df' equation. Then you'll need to integrate to get the actual f(x,y).

Answer 8

Since it already told you this is exact, you do not need to find dm/dy and dn/dx, as that is obviously already true (and I checked it, it is).

Answer 9

how to check if a given equation is exact?

Answer 10

partially. You have the right terms assigned to M and N, but you didn't differentiate at all.

Answer 11

Here, you have:

M(x,y) = cosxcosy - cotx
N(x,y) = -sinxsiny

Answer 12

This gives:

dM/dy = -cosxsiny
dN/dx = -cosxsiny

They match, therefore an exact ODE

Answer 13

Your ODE is:
M(x,y)dx + N(x,y)dy = 0 --> (cosxcosy-cotx)dx + (-sinxsiny)dy = 0

Answer 14

yes..

Answer 15

So you are seeking a solution f(x,y), where it's differential df is as I said above a few posts ago.

Answer 16

so you compare directly:

M(x,y) = df/dx
N(x,y) = df/dy

Answer 17

yes the gen solution =)

Answer 18

df/dx = cosxcosy - cotx
df/dy = -sinxsiny

Answer 19

So you need to integrate the first with a constant of integration as a function of 'y' and then go on.

Answer 20

Basically, do this:\[f(x,y) = C(y)+\int\limits \frac{df}{dx}dx\]

Answer 21

c is constant, right?

Answer 22

yep! because your integral is indefinite

Answer 23

So what do you get for this:\[f(x,y)=C(y)+\int\limits [\cos(x)\cos(y)-\cot(x)]dx\]

Answer 24

i did this..