Answer to Question #211002 in Differential Equations for Syed

Question

Solve

solve (3x^2+2xy^2)dx+(2x^2y)dy=0 where y(2)=-3

Answer 1

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

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

$$Q=2x^2y, \dfrac{\partial Q}{\partial x}=4xy$$

$$\dfrac{\partial P}{\partial y}=4xy=\dfrac{\partial Q}{\partial x}$$

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

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

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

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

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

$$\varphi(y)=C_1$$

The general solution of the differential equation

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

is given by

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

 y(2)=-3

$$(2)^3+(2)^2(-3)^2=C$$

$$C=44$$

The solution of the IVP is

$$x^3+x^2y^2=44$$