Question

Solve the differential equation given by

(3x + 2y + y^2)dx + (x+4xy+5y^2)dy=0

Answer 1

$$\dfrac{\partial Q}{\partial x}=1+4y, \dfrac{\partial P}{\partial y}=2+2y$$ $$\dfrac{\partial Q}{\partial x}\not=\dfrac{\partial P}{\partial y}$$ $$\mu=x+y^2$$ $$(x+y^2)(3x + 2y + y^2)dx$$ $$+ (x+y^2)(x+4xy+5y^2)dy=0$$ $$P=3x^2+2xy+xy^2+3xy^2+2y^3+y^4$$ $$\dfrac{\partial P}{\partial y}=2x+8xy+6y^2+4y^3$$ $$Q=x^2+4x^2y+5xy^2+xy^2+4xy^3+5y^4$$ $$\dfrac{\partial Q}{\partial x}=2x+8xy+6y^2+4y^3$$ $$\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$$ $$\dfrac{\partial u}{\partial x}=3x^2+2xy+4xy^2+2y^3+y^4$$

Integrate

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

Then

$$\dfrac{\partial u}{\partial y}=x^2+4x^2y+6xy^2+4xy^3+\varphi'(y)$$ $$=Q=x^2+4x^2y+6xy^2+4xy^3+5y^4$$ $$\varphi'(y)=5y^4$$ $$\varphi(y)=y^5+C$$

The original differential equation has the following solution

$$x^3+x^2y+2x^2y^2+2xy^3+xy^4-y^5=C$$