If dy/dx= x+3xy+3y, then y=?

Question

This is just a calculus question I am struggling with. All I need is the answer, but an explanation would be nice. Thanks.

Answer 1

dy/dx = x + 3xy + 3y = x + 3y(x +1)
dy/dx - 3(x +1) y = x

Then apply the Integrating Factor (IF) and continue
>> P(x) = - 3(x+1) and Q(x) = x

Answer 2

y = 3xy + x^3y + 1/2 x^2

Answer 3

ignore the y's on the right side of the equation, and integrate it like any polynomial
y = 1/2 x² + 3/2 x²y + 3xy + c
then solve for y
y (-3/2 x² - 3x + 1) = 1/2 x² + c
y (3x² + 6x - 2) = -x² + c
y = (-x² + c)/(3x² + 6x - 2)

Answer 4

Given dy/dx = x + 3xy + 3y
Find y.

dy/dx = x + 3xy + 3y

Integrating we have

y = ∫(x + 3xy + 3y)dx = x²/2 + 3x²y/2 + 3xy + C
y(1 - 3x²/2 - 3x) = x²/2 + C
y = (x²/2)/(1 - 3x²/2 - 3x) + C
y = x²/(2 - 3x² - 6x) + C
y = x²/(2 - 6x - 3x²) + C

Answer 5

y=1+3y+3x+3
y=4+3y+3x

Answer 6

I would probably integrate it.
y=x²/2+3x²y/2+3xy+c
y-3x²y-3xy=x²/2+c
y(1-3x²-3x)=x²/2+c
y=x² / 2(1-3x²-3x)+c