Question

Find the general solution of the following differential equation. Primes denote derivatives with respect to x. 9 x (x+5y )y prime equals 9 y (x -5 y )

Answer 1

`(x)/(5y) -lnxy = K`

Explanation:

Given the differential equation

`9x(x+5y )y' = 9y(x -5y )\\x(x+5y )(dy)/(dx) = y(x -5y)\\(dy)/(dx) = (y(x -5y))/(x(x+5y )) .\\let\ y = vx\\(dy)/(dx) =v+ x(dv)/(dx) \\v+ x(dv)/(dx) = (vx(x -5vx))/(x(x+5vx )) \\v+ x(dv)/(dx) = (v(1 -5v))/((1+5v )) \\x(dv)/(dx) = (v(1 -5v))/((1+5v )) - v\\x(dv)/(dx) = (v(1 -5v)-v(1+5v))/((1+5v )) \\x(dv)/(dx) = (-10v^(2) )/(1+5v) \\(dx)/(x) = ( 1+5v)/(-10v^(2))dv\\`

`\int\limits(dx)/(x) = (-1)/(10) \int\limits v^(-2) dv - (1)/(2)\int\limits (dv)/(v) \\lnx + C = (1)/(10v)-(1)/(2)lnv\\ since\ y =vx\\lnx+C = (1)/(10((y)/(x) ))-(1)/(2)ln((y)/(x) )\\lnx+C = (x)/(10y)-(1)/(2)ln((y)/(x) )`

`2lnx +2C = (x)/(5y) - ln (y)/(x) \\(x)/(5y) - ln (y)/(x) -2lnx = 2C\\(x)/(5y) -(ln (y)/(x) +2lnx) = 2C\\(x)/(5y) -(ln (y)/(x) +lnx^(2) ) = 2C\\\\(x)/(5y) -lnxy = 2C\\(x)/(5y) -lnxy = K\ where \ K = 2C`