Answer to Question #241122 in Differential Equations for Jojo

Question

xy dx + (2x ^ 2 + 3y ^ 2 - 20) * d * y = 0 .

Answer 1

$$M(x,y)=xy$$

$$M_y=x$$

$$N(x,y)=2x^2+3y^2-20$$

$$N_x=4x$$

$$M_y=x≠4x=N_x$$

The equation is not exact.

$$xy^4dx+y^3(2x^2+3y^2-20)dy=0$$

$$M(x,y)=xy^4$$

$$M_y=4xy^3$$

$$N(x,y)=y^3(2x^2+3y^2-20)$$

$$N_x=4xy^3$$

$$M_y=4xy^3=N_x$$

The equation is exact.

There exists a function $f$ for which

$$∂f/∂x=M(x,y)$$

We can find $f$  by integrating $M(x,y)$ with respect to $x$ while holding $y$ constant:

$$f(x, y)=\int M(x,y)dx+g(y)$$

$$f(x, y)=\int xy^4dx+g(y)=\dfrac{1}{2}x^2y^4+g(y)$$

Taking the partial derivative of the last expression with respect to $y$ and setting the result equal to $N(x,y)$ gives

$$2x^2y^3+g'(y)=2x^2y^3+3y^5-20y^3$$

$$g'(y)=3y^5-20y^3$$

Integrate with respect to $y$

$$g(y)=\int(3y^5-20y^3)dy=\dfrac{1}{2}y^6-5y^4-C$$

Hence

$$f(x, y)=\dfrac{1}{2}x^2y^4+\dfrac{1}{2}y^6-5y^4-C$$

The solution of the equation in implicit form is

$$\dfrac{1}{2}x^2y^4+\dfrac{1}{2}y^6-5y^4=C$$