Answer to Question #271010 in Differential Equations for Aysu

Question

Find the integrating factor to transform the given differential equation into the equation in exact differentials: (x²-3y²)dx+2xydy=0 μ=μ(x)

Answer 1

$$\dfrac{M_y-N_x}{N}=\dfrac{-6y-2y}{2xy}=-\dfrac{4}{x}$$

Integrating factor

$$\mu=\mu(x)=e^{\int(-4/x)dx}=x^{-4}$$

$$x^{-4}(x^2-3y^2)dx+2x^{-4}xydy=0$$

$$(x^{-2}-3x^{-4}y^2)dx+2x^{-3}ydy=0$$

$$\dfrac{\partial M}{\partial y}=-6x^{-4}y$$

$$\dfrac{\partial N}{\partial x}=-6x^{-4}y$$

The differential equation

$$(x^{-2}-3x^{-4}y^2)dx+2x^{-3}ydy=0$$

is the equation in exact differentials.

Integrating factor

$$\mu=\mu(x)=x^{-4}$$