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

Answer:

$\frac{x}{5y} -lnxy = K$

Step-by-step explanation:

Given the differential equation

$9x(x+5y )y' = 9y(x -5y )\\x(x+5y )\frac{dy}{dx} = y(x -5y)\\\frac{dy}{dx} = \frac{y(x -5y)}{x(x+5y )} .\\let\ y = vx\\\frac{dy}{dx} =v+ x\frac{dv}{dx} \\v+ x\frac{dv}{dx} = \frac{vx(x -5vx)}{x(x+5vx )} \\v+ x\frac{dv}{dx} = \frac{v(1 -5v)}{(1+5v )} \\x\frac{dv}{dx} = \frac{v(1 -5v)}{(1+5v )} - v\\x\frac{dv}{dx} = \frac{v(1 -5v)-v(1+5v)}{(1+5v )} \\x\frac{dv}{dx} = \frac{-10v^{2} }{1+5v} \\\frac{dx}{x} = \frac{ 1+5v}{-10v^{2}}dv$

$\int\limits\frac{dx}{x} = \frac{-1}{10} \int\limits v^{-2} dv - \frac{1}{2}\int\limits \frac{dv}{v} \\lnx + C = \frac{1}{10v}-\frac{1}{2}lnv\\ since\ y =vx\\lnx+C = \frac{1}{10(\frac{y}{x} )}-\frac{1}{2}ln(\frac{y}{x} )\\lnx+C = \frac{x}{10y}-\frac{1}{2}ln(\frac{y}{x} )$

$2lnx +2C = \frac{x}{5y} - ln \frac{y}{x} \\\frac{x}{5y} - ln \frac{y}{x} -2lnx = 2C\\\frac{x}{5y} -(ln \frac{y}{x} +2lnx) = 2C\\\frac{x}{5y} -(ln \frac{y}{x} +lnx^{2} ) = 2C\\\\\frac{x}{5y} -lnxy = 2C\\\frac{x}{5y} -lnxy = K\ where \ K = 2C$