Answer to Question #178125 in Differential Equations for Nakul Kumar singh

Question #178125

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

x1/


1
Expert's answer
2021-04-25T07:37:55-0400

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\n=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\n=-\\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."


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