Show that the differential equation `((x-y)dy)/(dx)=x+2y ,` is homogeneous and solve it.

Explanation

we have `(x-y)(dy)/(dx)=x+2y`
`(dy)/(dx) = (x+2y)/(x-y) = f(x,y)` say
`x = lambda x , y= lambday , f(lambda x, lambda y) = (lambdax+2lambday)/(lambda x-lambda y)= (x+2y)/(x-y) = f(x,y)`
Hence the differential equation is homogenous,
now
put `y = vx => (dy)/(dx)= v+ x(dv)/(dx)`
` :. v+ x(dv)/(dx)= (x+2vx)/(x-vx) = (1+2v)/(1-v)`
`=> x(dv)/(dx) = (1+2v-v+v^2)/(1-v) `
integrate on both side
` int(1-v)/((1+2v-v+v^2) dv = int dx/x `
Let `1-v= A d/(dv) [ (1+v+v^2)] +B => 1-v =A[2v+1]+B`
On equating the coefficients of like terms, we get `A= -1/(2) , B= 3/2`
`:. int((3)/(2(1+v+v^2)) - ((1)(2v+1))/(2(1+v+v^2)))= int dx/x`
`=> 3/(2) int 1/((v+1/(2) )+(sqrt(3)/2)^2) dv - 1/(2) log|1+v+v^2| = log|x|`
by solving the equation we get
`=> 3/(2) xx 2/(sqrt(3)) tan^-1 (( 2v+1)/(sqrt(3))) - 1/(2) log | (y^2+xy+x^2)/x^2 | = log|x| +c`