Question

(3x^2+2xy+y^3)dx + (x^2+y^2)dy=0

Answer 1

$\begin{aligned} (3x^2y + 2xy + y^3)dx &+ (x^2 + y^2) dy = 0\\ \\ M=3x^2y + 2xy + y^3 &; \quad N = x^2 + y^2\\ M_y = 3x^2 + 2x + 3y^2 &; \quad N_x = 2x\\ \\ \therefore \dfrac{M_y-N_x}{N} = 3\\ \\ Thus,\ \mu(x)= e^{\int{3dx}} &= e^{3x} \end{aligned}$Employing $\mu(x)$ into the general equation, we have;

$\begin{aligned} e^{3x}(3x^2y + 2xy + y^3)dx &+ e^{3x}(x^2 + y^2) dy = 0\\ \end{aligned}$which, by construction, must be exact. So we seek a function $\psi$ such that;

$\frac{\partial\psi}{\partial x} =e ^{3x}(3x ^2y + 2xy + y ^3 )\\ \frac{\partial \psi }{\partial y} = e ^{3x} (x ^2 + y ^2 )$

Integrating the first equation with respect to x and the second equation with respect to y, we get;

$\psi (x, y) = x^2ye^{3x} + \frac{1}{3}y^3e^{3x}+h_1(y)\\ \psi (x, y) = x^2ye^{3x} +\frac{1}{3}y^3e^{3x}+h_2(x)$

Comparing the two expressions above we observe that we must take a $h_1y = h_2x =$ C (a constant).

Thus, function $\psi$ satisfying the exact function must be of the form;

$\psi (x, y) = e ^{3x}x^2y + e ^{3x}y^3 + C$

Thus, the general solution to the equation is of the form;

$e ^{3x}x ^2y + e ^{3x }y^3 = C$