Answer to Question #207870 in Differential Equations for Dany

Question

y dx-4(x+y⁶)dy=0

Answer 1

Solution

Writing DE in the form P(x,y)dx + Q(x,y)dy = 0 we’ll get P(x,y) = y, Q(x,y) = -4(x+y⁶)

∂P/∂y=1 ≠ ∂Q/∂x=-4

Let’s find function  M(x,y), called an integrating factor, for which equality ∂(M(x,y)P(x,y))/∂y=∂(M(x,y)Q(x,y))/∂x is satisfied.

-[∂P/∂y - ∂Q/∂x]/P=-5/y

So M(x,y) may be found from equation

dM/M = -[∂P/∂y - ∂Q/∂x]dy/P=-5dy/y

ln|M| = -5ln|y|+lnC

M = C/y⁵  

Multiplying DE by M(x,y) with arbitrary constant C=1

dx/y⁴-4(x/y⁵+y)dy = 0

Let’s find function u(x,y) for which du = u_xdx+u_ydy = 0, where u_x = 1/y⁴ and u_y = -4(x/y⁵+y)

u(x,y) = ∫dx/y⁴ + f(y)  = x/y⁴ + f(y)  , where f(y) – arbitrary function.

u_y = -4 x/y⁵ +f’(y) = -4(x/y⁵+y) => f’(y) = -4y => f(y) = -2y² - C

Substituting this into expression for u(x,y) we’ll get

u(x,y) =  x/y⁴ - 2y² - C

Therefore general solution of DE is x/y⁴ - 2y² = C