First Order, Non-Linear DE - Not Seperable

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Hello, I'm studying for a test and this is a question on a practice test...

cos(x)+y^2+(2yx-1)y'=0


I can't separate the variables (it's not homogeneous, either), this isn't exact and bernoulli won't work...

dy/dx=-cos(x)/(2yx-1)-y^2/(2yx-1)

I changed the equation so it would look like this but I can't simplify it any more than that and I can't just take the integral of it here...If anyone could give me some help with this problem it would be much appreciated. Thanks!
 
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Looks like an exact equation, if it's on your test you should know how to solve it. That said, it doesn't look like the solution is anything obvious so you'll have to go through the usual P = df/dx and Q = df/dy
 
Oh you're right, thanks!
 
Hi !

cos(x)+y²+2yxy'-y'=0
cos(x)+(x y² -y)'=0
 
Thread 'Direction Fields and Isoclines'

Thread 'Direction Fields and Isoclines'

I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
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