Question

Determine whether the given differential equation is exact. If it is exact, solve it.
(5x+4y) dx + (4x-8y^3) dy =0.

Answer 1

Question #16146 Determine whether the given differential equation is exact. If it is exact, solve it. $(5x + 4y)dx + (4x - 8y^3)dy = 0$.

Solution. We are to verify $\frac{\partial(5x + 4y)}{\partial y} = \frac{\partial(4x - 8y^3)}{\partial x}$. Hence the differential equation is exact.

To solve it, write $\frac{\partial U(x,y)}{\partial x} = 5x + 4y$, thus $U(x,y) = 5 / 2x^{2} + 4yx + \varphi (y)$. Next, $\frac{\partial U(x,y)}{\partial y} = 4x - 8y^3$ and $4x + \varphi '(y) = 4x - 8y^3$, so $\varphi '(y) = -8y^3$ or $\varphi (y) = -2y^4 +C$. To conclude the general solution $U(x,y) = C$ or equivalently $5 / 2x^{2} + 4yx - 2y^{4} = C$.

Solution. $5 / 2x^{2} + 4yx - 2y^{4} = C$.