Answer to Question #271010 in Differential Equations for Aysu

Question #271010

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


1
Expert's answer
2021-11-29T16:23:00-0500
MyNxN=6y2y2xy=4x\dfrac{M_y-N_x}{N}=\dfrac{-6y-2y}{2xy}=-\dfrac{4}{x}

Integrating factor


μ=μ(x)=e(4/x)dx=x4\mu=\mu(x)=e^{\int(-4/x)dx}=x^{-4}

x4(x23y2)dx+2x4xydy=0x^{-4}(x^2-3y^2)dx+2x^{-4}xydy=0

(x23x4y2)dx+2x3ydy=0(x^{-2}-3x^{-4}y^2)dx+2x^{-3}ydy=0

My=6x4y\dfrac{\partial M}{\partial y}=-6x^{-4}y

Nx=6x4y\dfrac{\partial N}{\partial x}=-6x^{-4}y

The differential equation


(x23x4y2)dx+2x3ydy=0(x^{-2}-3x^{-4}y^2)dx+2x^{-3}ydy=0

is the equation in exact differentials.

Integrating factor


μ=μ(x)=x4\mu=\mu(x)=x^{-4}

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