Question

1/xy² +y4 is an integrating factor for the differential equation (x²y + y²)dx + (y³ - x³)dy = 0

x1/

Answer 1

Solution.

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

If $\frac{y^2}{x}+y^4$ is an integrating factor for the differential equation, then

$$(\frac{y^2}{x}+y^4 )(x^ 2 y + y^ 2 ) dx +(\frac{y^2}{x}+y^4 ) (y^ 3 - x^ 3 ) dy = 0.$$

So,

$F(x,y)=\int (\frac{y^2}{x}+y^4 )(x^ 2 y + y^ 2 ) dx=\int(xy^3+x^2y^5+\frac{y^4}{x}+y^6)dx=\newline =y^3\frac{x^2}{2}+y^5\frac{x^3}{3}+y^4\ln{x}+y^6x+g(y).$

$F_y'(x,y)=\frac{3}{2}x^2y^2+\frac{5}{3}x^3y^4+4\ln{x}y^3+6xy^5+g'(y),$

$\frac{3}{2}x^2y^2+\frac{5}{3}x^3y^4+4\ln{x}y^3+6xy^5+g'(y)=\frac{y^5}{x}+y^7-x^2y^2-x^3y^4.$

From here

$g'(y)=\frac{y^5}{x}+y^7-\frac{5}{2}x^2y^2-\frac{8}{3}x^3y^4-4\ln{x}y^3-6xy^5.$

$g(y)=\frac{y^6}{6x}+\frac{y^8}{8}-\frac{5x^2y^3}{6}-\frac{8x^3y^5}{15}-\ln{x}y^4-xy^6+C,$

where $C$ is some constant.

Answer.

$y^3\frac{x^2}{2}+y^5\frac{x^3}{3}+y^4\ln{x}+y^6x+\frac{y^6}{6x}+\frac{y^8}{8}-\frac{5x^2y^3}{6}-\frac{8x^3y^5}{15}-\ln{x}y^4-xy^6+C=\newline =-\frac{1}{3}x^2y^3-\frac{1}{5}x^3y^5+\frac{1}{6}\frac{y^6}{x}+\frac{1}{8}y^8+C.$