Differential Equation Question

Question

Find the solution for the following exact differential equations.

 

2xy^2 + 4 = 2(3 - x^2y)y' ; y(-1) = 8

Answer 1

2xy² + 4 = 2(3 - x²y)y'

A brief solution is as following.

(2xy²+4)dx+(2x²y-6)dy=0  (M(x,y)dx+N(x,y)dy=0

If we denote F(x, y)=x²y²+4x-6y, then 

∂F/∂x=2xy²+4x = M(x,y)

∂F/∂y=2x²y-6 = N(x, y)

DF(x,y)=0

F(x,y)=C, C -constant

x²y²+4x-6y=C 

y(-1)=8, therefore, C=36-4-48=-16

So, the solution is x²y²+4x-6y+16=0.

If any questions, let me know.