Differential Equations

Just started a differential equation class. Any tips on how to get started solving the following;

y' = (2x*cos(x^2+1)-5y^2) / (10xy + cos(y))

Thanks!

Question

Differential Equations 

Just started a differential equation class. Any tips on how to get started solving the following;

y' = (2x*cos(x^2+1)-5y^2) / (10xy + cos(y)) 

Thanks!

anonymous · Answer

Have you discussed Exact differential equations already?

anonymous · Answer

dy/dx = (2x*cos(x^2 +1) - 5y^2)/(10xy + cos(y))
(10xy + cosy)dy - (2xcos*x^2 + 1) - 5y^2)dx = 0
this is in the form Ndy - Mdx = 0 and is an exact differential equation
dM/dy = -(-10y) = 10y
dN/dx = 10y

anonymous · Answer

Oh wow, we have... could not, for the life of me, see putting it into that form, haha. Thank you very much for the help!