# Slope Field with Solution Passing Through Point

• February 11th 2010, 07:36 PM
pcannons
Slope Field with Solution Passing Through Point
Hi,

This is for Calc II homework..

The figure below shows the slope field for the equation y' = sin (x) cos (y).

(If you want I can post the picture, but I don't think its needed)

And the question is

What is the equation of the solution passing through (0https://instruct.math.lsa.umich.edu/...144/char3B.png(2n+1)https://instruct.math.lsa.umich.edu/...144/char19.pnghttps://instruct.math.lsa.umich.edu/...144/char3D.png2), where n is any integer?
y=

Now, I know that the slope is 0 along any of these points but I just don't know what to do with the rest of the information.

Thanks for any help
• February 12th 2010, 03:50 AM
HallsofIvy
Quote:

Originally Posted by pcannons
Hi,

This is for Calc II homework..

The figure below shows the slope field for the equation y' = sin (x) cos (y).

(If you want I can post the picture, but I don't think its needed)

And the question is

What is the equation of the solution passing through (0https://instruct.math.lsa.umich.edu/...144/char3B.png(2n+1)https://instruct.math.lsa.umich.edu/...144/char19.pnghttps://instruct.math.lsa.umich.edu/...144/char3D.png2), where n is any integer?
y=

Now, I know that the slope is 0 along any of these points but I just don't know what to do with the rest of the information.

Thanks for any help

You need to integrate y'= sin(x)cos(x) (which should be easy). You will, of course, have a "constant of integration". Use the fact that $y(0)= \frac{(2n+1)\pi}{2}$ to find the value of that constant.
• February 12th 2010, 05:41 AM
Calculus26
y' = sin (x) cos (y).

has equiilibrium solutions of the form y= constant where cos(y) = 0

which is (2n+1)pi/2

the solution curves are y = (2n+1)pi/2 which are horizontal lines

In general for dy/dx = f(x)g(y) equillibrium solutions are precisely the zeroes of g(y)