Question

Find the integrating factor to transform the given differential equation into the equation in exact differentials: (3x²y+y³)dx-(2x³+5y)dy=0, μ=μ(y)

Answer 1

$$(3x^2y+y^3)dx-(2x^3+5y)dy=0$$

$$\dfrac{M_y-N_x}{-M}=\dfrac{3x^2+3y^2+6x^2}{-3x^2y-y^3}=-\dfrac{3}{y^2}$$

Integrating factor

$$\mu=\mu(y)=e^{\int(-3/y^2)dy}=\dfrac{1}{y^3}$$

$$\dfrac{3x^2+y^2}{y^2}dx+\dfrac{-2x^3-5y}{y^3}dy=0$$

$$M(x,y)=\dfrac{3x^2+y^2}{y^2}$$

$$\dfrac{\partial M}{\partial y}=\dfrac{2yy^2-2y(3x^2+y^2)}{y^4}=-\dfrac{6x^2}{y^3}$$

$$N(x,y)=\dfrac{-2x^3-5y}{y^3}$$

$$\dfrac{\partial N}{\partial x}=-\dfrac{6x^2}{y^3}$$

$$\dfrac{\partial M}{\partial y}=-\dfrac{6x^2}{y^3}=\dfrac{\partial N}{\partial x}$$

Integrating factor $\mu=\mu(y)=\dfrac{1}{y^3}.$