Initial value problem - differential equations

[Cocoleia]

Homework Statement

I am given (y^2 + y sin x cos y) dx + (xy + y cos x sin y) dy = 0, y(0) = π/2 .
I need to solve this

Homework Equations

The Attempt at a Solution

At this point they still aren't exact, so I gave up. I can't figure out what the problem is. Is it possible that I have to continue trying to make them exact?

I know that the answer is: xy-cosxcosy=0

 

[Mark44]

Homework Statement

I am given (y^2 + y sin x cos y) dx + (xy + y cos x sin y) dy = 0, y(0) = π/2 .
I need to solve this

Homework Equations

The Attempt at a Solution

View attachment 110019

At this point they still aren't exact, so I gave up. I can't figure out what the problem is. Is it possible that I have to continue trying to make them exact?

I know that the answer is: xy-cosxcosy=0

Divide both sides by y, to get $(y + \sin(x)\cos(y))dx + (x + \cos(x)\sin(y)) dy = 0$.
Now check for exactness. Note that since $y(0) = \pi/2$, it's reasonable to assume that $y \ne 0$.

 

[Cocoleia]

Divide both sides by y, to get $(y + \sin(x)\cos(y))dx + (x + \cos(x)\sin(y)) dy = 0$.
Now check for exactness. Note that since $y(0) = \pi/2$, it's reasonable to assume that $y \ne 0$.

Can we always do this if we have a common factor like y in this case?
Thank you, I spent a very long time trying to figure this out

 

[Mark44]

Can we always do this if we have a common factor like y in this case?

I wouldn't say you can always do it. By getting rid of a factor of y, we are possibly dividing by zero. Since the initial condition is that $y(0) = \pi/2$, if necessary, we can restrict y to some interval that includes $\pi/2$ but doesn't include zero.