Question

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

Answer 1

`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`