Answer to Question #214929 in Differential Equations for sandra

Question

prove that u(x,y)=x²y is an integrating factor of the equation

(3y+4xy²)dx+(2x+3x²y)dy=0

Hence solve the equation

Answer 1

$$x^2y(3y+4xy^2)dx+x^2y(2x+3x^2y)dy=0$$

$$P(x, y)=3x^2y^2+4x^3y^3, \dfrac{\partial P}{\partial y}=6x^2y+12x^3y^2$$

$$Q(x, y)=2x^3y+3x^4y^2, \dfrac{\partial Q}{\partial x}=6x^2y+12x^3y^2$$

$$\dfrac{\partial P}{\partial y}=6x^2y+12x^3y^2=\dfrac{\partial Q}{\partial x}$$

Then exists $u(x, y)$ such that

$$\dfrac{\partial u}{\partial x}=P(x, y)=3x^2y^2+4x^3y^3$$

$$\dfrac{\partial u}{\partial y}=Q(x, y)=2x^3y+3x^4y^2$$

$$u=\int (3x^2y^2+4x^3y^3)dx+\varphi(y)$$

$$=x^3y^2+x^4y^3+\varphi(y)$$

$$\dfrac{\partial u}{\partial y}=2x^3y+3x^4y^2+\varphi'(y)=2x^3y+3x^4y^2$$

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

The general solution of the exact differential equation is given by

$$x^3y^2+x^4y^3=C$$

where $C$ is an arbitrary real number.