What is the general solution of the differential equation ? #e^(x^3) (3x^2 y- x^2) dx + e^(x^3) dy =0 #

1 Answer
Nov 17, 2017

# y = Ce^(-x^3) + 1/3#

Explanation:

#e^(x^3) (3x^2 y- x^2) dx + e^(x^3) dy =0 # ..... [A]#

Suppose we have:

# M(x,y) \ dx = N(x,y) \ dy #

Then the DE is exact if #M_y-N_x=0#

# M = e^(x^3) (3x^2 y- x^2) => M_y = 3x^2e^(x^3) #
# N = e^(x^3) => N_x = 3x^2e^(x^3) #

# M_y - N_x = 0 => # an exact DE

Then, our solution is given by:

#f_x = M# and #f_y = N# and

Consider #f_x = M => (partial f)/(partial x) = e^(x^3) (3x^2 y- x^2) #

# :. f = int \ x^2e^(x^3) (3y- 1) \ partial x#
# \ \ \ \ \ \ \ = 1/3e^(x^3) (3y- 1) #
# \ \ \ \ \ \ \ = e^(x^3) y- 1/3e^(x^3) #

Consider #f_y = N => (partial f)/(partial y) = e^(x^3) #

# :. f = int \ e^(x^3) \ partial y#
# \ \ \ \ \ \ \ = e^(x^3) y #

If we combine the common components of both integrals we can form the general solution:

# f(x,y) = e^(x^3) y- 1/3e^(x^3) = C #

And now we re-arrange to form an implicit solution

# e^(x^3) y = C + 1/3e^(x^3)#
# :. y = Ce^(-x^3) + 1/3#