Thread: 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 (0(2n+1)2), 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

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 (0(2n+1)2), 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 to find the value of that constant.

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)