Answer to Question #270221 in Differential Equations for Haris

Question

solve the ODE (3x²+4xy-6)dy+(6xy+2y²-5)=0

Answer 1

 

$$(6xy+2y^2-5)dx +(3x^2+4xy-6) dy=0$$

 

$$M(x,y)=6xy+2y^2-5$$

$$=> {\partial M(x,y)\over \partial y}=M_y=6x+4y$$

$$N(x,y)=3x^2+4xy-6$$

$$=> {\partial N(x,y)\over \partial x}=N_x=6x+4y$$

Hence exact since $M_y=N_x$

There exist a function $u(x,y)$ such that

$${\partial u(x,y) \over \partial x}=M(x,y)=6xy+2y^2-5$$

$${\partial u(x,y) \over \partial y}=N(x,y)=3x^2+4xy-6$$

Hence

$$u(x, y)=\int(6xy+2y^2-5)dx+\varphi(y)$$

$$=3x^2y+2xy^2-5x+\varphi(y)$$

Differentiating with respect to $y,$ we substitute the function into the second equation:

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

$$=N(x,y)=3x^2+4xy-6$$

Then

$$\varphi'(y)=-6$$

Integrate

$$\varphi(y)=-\int 6dy=-6y-C$$

$$u(x, y)=3x^2y+2xy^2-5x-6y-C$$

The general solution is

$$3x^2y+2xy^2-5x-6y=C$$