- #1
Want2BeSmart
- 1
- 0
Homework Statement
(3(x^2)y + 2xy + y^3)dx + (x^2 + y^2)dy = 0
The Attempt at a Solution
This is from my notes, so I already have the answer. I just don't understand the very last step with the integrations.
(3(x^2)y + 2xy + y^3)dx + (x^2 + y^2)dy = 0
My = 3x^2 + 2x + 3y^2
Nx = 2x
(My-Nx)/N = (3x^2 + 3y^2)/(x^2 + y^2) = 3
dlnu/dx = 3, u = e^(3x)
Multiply by integrating factor u = e^(3x)
(3(x^2)y + 2xy + y^3)e^(3x)dx + (x^2 + y^2)e^(3x)dy = 0
and then the part I don't understand:
phi(x,y) = [∫ 0 to y (x^2 + y^2)(e^(3x))dy] + [∫ 0 to x (0 dx)]
= (x^2)y(e^(3x)) + (1/3)(y^3)(e^(3x))
Thanks