求解方程式dy/dx=(-2xy^3 - 2)/(3x^2y^2 + e^y)?

dy/dx = ( - 2xy³ - 2 ) / ( 3x²y² + e^y )
( - 2xy³ - 2 )dx = ( 3x²y² + e^y )dy
( 2xy³ + 2 )dx + ( 3x²y² + e^y )dy = 0

令 M = 2xy³ + 2 , N = 3x²y² + e^y
則 ∂M/∂y = ∂N/∂x = 6xy²
故原式為正合方程式( Exact ODE )
因此存在 Φ( x , y ) = c1 滿足原式, 使得 :
∂Φ/∂x = 2xy³ + 2 , ∂Φ/∂y = 3x²y² + e^y

Φ
= ∫ ( 2xy³ + 2 )dx + f(y)
= x²y³ + 2x + f(y)

∂Φ/∂y = 3x²y² + f ' (y) = 3x²y² + e^y
f ' (y) = e^y
f(y) = e^y + c2

Φ( x , y ) = x²y³ + 2x + f(y) = c1
x²y³ + 2x + e^y + c2 = c1
x²y³ + 2x + e^y = c ..... Ans