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

Nov 17, 2017

$y = C {e}^{- {x}^{3}} + \frac{1}{3}$

#### Explanation:

${e}^{{x}^{3}} \left(3 {x}^{2} y - {x}^{2}\right) \mathrm{dx} + {e}^{{x}^{3}} \mathrm{dy} = 0$ ..... [A]#

Suppose we have:

$M \left(x , y\right) \setminus \mathrm{dx} = N \left(x , y\right) \setminus \mathrm{dy}$

Then the DE is exact if ${M}_{y} - {N}_{x} = 0$

$M = {e}^{{x}^{3}} \left(3 {x}^{2} y - {x}^{2}\right) \implies {M}_{y} = 3 {x}^{2} {e}^{{x}^{3}}$
$N = {e}^{{x}^{3}} \implies {N}_{x} = 3 {x}^{2} {e}^{{x}^{3}}$

${M}_{y} - {N}_{x} = 0 \implies$ an exact DE

Then, our solution is given by:

${f}_{x} = M$ and ${f}_{y} = N$ and

Consider ${f}_{x} = M \implies \frac{\partial f}{\partial x} = {e}^{{x}^{3}} \left(3 {x}^{2} y - {x}^{2}\right)$

$\therefore f = \int \setminus {x}^{2} {e}^{{x}^{3}} \left(3 y - 1\right) \setminus \partial x$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{3} {e}^{{x}^{3}} \left(3 y - 1\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = {e}^{{x}^{3}} y - \frac{1}{3} {e}^{{x}^{3}}$

Consider ${f}_{y} = N \implies \frac{\partial f}{\partial y} = {e}^{{x}^{3}}$

$\therefore f = \int \setminus {e}^{{x}^{3}} \setminus \partial y$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = {e}^{{x}^{3}} y$

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

$f \left(x , y\right) = {e}^{{x}^{3}} y - \frac{1}{3} {e}^{{x}^{3}} = C$

And now we re-arrange to form an implicit solution

${e}^{{x}^{3}} y = C + \frac{1}{3} {e}^{{x}^{3}}$
$\therefore y = C {e}^{- {x}^{3}} + \frac{1}{3}$