Question

is
La solución general de la siguiente ecuación diferencial
(6x +y - 1)dx + (x + 4y)dy = 0
es:

Answer 1

M=6x+y-1

N=x+4y

$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}=1$

$F=\int (6x+y-1)dx$

$F=3x^2+yx-x+\psi(y)$

$</p><p>\frac{\partial F}{\partial y}=x+\psi^\prime (y)=x+4y$

$\psi \prime (y)=4y$

$\psi (y)=2y^2+C$

solución

$F=3x^2+xy-x+2y^2+C$