Question

Find an integrating factor of the form yⁿ for the equation

(y²+2xy)dx-x²dy=0. Hence solve the equation

Answer 1

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

$$P(x,y)=y^2+2xy, \dfrac{\partial P}{\partial y}=2y+2x$$

$$Q(x,y)=-x^2, \dfrac{\partial Q}{\partial x}=-2x$$

$$\dfrac{\partial P}{\partial y}=2y+2x\not=-2x=\dfrac{\partial Q}{\partial x}$$

$$\dfrac{\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}}{P}=\dfrac{-2x-2y-2x}{y^2+2xy}=-\dfrac{2}{y}$$

An integrating factor is therefore given by $\mu(y)=\dfrac{1}{y^2}.$

Multiplying through by $\mu$ leads to the equation

$$(1+2\dfrac{x}{y})dx-\dfrac{x^2}{y^2}dy=0$$

which is exact.

$$M(x,y)=1+2\dfrac{x}{y}, \dfrac{\partial M}{\partial y}=-2\dfrac{x}{y^2}$$

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

$$\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}$$

There exists a function $u$ which satisfies

$$\dfrac{\partial u}{\partial x}=1+2\dfrac{x}{y}$$

$$\dfrac{\partial u}{\partial y}=-\dfrac{x^2}{y^2}$$

$$u(x, y)=\int(1+2\dfrac{x}{y})dx+\varphi(y)$$

$$=x+\dfrac{x^2}{y}+\varphi(y)$$

$$\dfrac{\partial u}{\partial y}=-\dfrac{x^2}{y^2}+\varphi'(y)=-\dfrac{x^2}{y^2}$$

$$=>\varphi'(y)=0=>\varphi(y)=C_1$$

The solution to the original equation is 

$$x+\dfrac{x^2}{y}=C$$

Or

$$y=\dfrac{x^2}{C-x}$$