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

Question

Christopher Agresta · Accepted Answer

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

#e^(x^3) (3x^2 y- x^2) dx + e^(x^3) dy =0 # ..... [A]# Assume 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 Next, our resolution is provided 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 # The general solution can be formed by combining the elements of both integrals that are common: # f(x,y) = e^(x^3) y- 1/3e^(x^3) = C # We now rearrange in order to create an implicit solution. # e^(x^3) y = C + 1/3e^(x^3)# # :. y = Ce^(-x^3) + 1/3#

Ezekiel Alexander · Answer

undefined To find the general solution of the given differential equation $e^{x^3}(3x^2y - x^2)dx + e^{x^3}dy = 0$, we can separate the variables and integrate.

First, rewrite the equation as $e^{x^3}(3x^2y - x^2)dx = - e^{x^3}dy$.

Divide both sides by $e^{x^3}$ to isolate $dy$, yielding $\frac{3x^2y - x^2}{e^{x^3}}dx = -dy$.

Now, integrate both sides with respect to their respective variables. On the left side, integrate with respect to $x$, and on the right side, integrate with respect to $y$.

This yields $\int \frac{3x^2y - x^2}{e^{x^3}}dx = -\int dy$.

Solve the left integral, which might involve a substitution or another integration technique. After finding the antiderivative, set it equal to $-y + C$, where $C$ is the constant of integration.

This gives the general solution to the differential equation.

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