Question

Consider ODE

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

Find genaral solution

Answer 1

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

$$P=3x^2y^2+x^2, \dfrac{\partial P}{\partial y}=6x^2y$$

$$Q=2x^3y+y^2, \dfrac{\partial Q}{\partial x}=6x^2y$$

$$\dfrac{\partial P}{\partial y}=6x^2y=\dfrac{\partial Q}{\partial x}$$

$$\dfrac{\partial u}{\partial x}=P(x, y), \dfrac{\partial u}{\partial y}=Q(x,y)$$

$$u(x,y)=\int(3x^2y^2+x^2)dx+\varphi(y)$$

$$=x^3y^2+\dfrac{1}{3}x^3+\varphi(y)$$

$$\dfrac{\partial u}{\partial y}=2x^3y+\varphi'(y)=2x^3y+y^2$$

$$\varphi'(y)=y^2$$

$$\varphi(y)=\dfrac{1}{3}y^3+C_1$$

The general solution of the differential equation

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

is given by

$$x^3y^2+\dfrac{1}{3}x^3+\dfrac{1}{3}y^3=C$$