dy/dx = (x+2y+3)/(2x+y+3) rArr (2x+y+3) dy = (x+2y+3)dxdydx=x+2y+32x+y+3⇒(2x+y+3)dy=(x+2y+3)dx
Now we will do a series of variable transformations to get an amenable format in the differential equation
1) {(x+2y = u),(2x+y = v):} rArr {(dx = 1/3(-du+2dv)),(dy=1/3(2du-dv)):}
(v+3)(2du-dv) = (u+3) (-du+2dv)
Now making the transformation
2) {(U = u+3),(V = v+3):} rArr {(dU = du),(dV = dv):}
V(2dU-dV) = U (-dU+2dV) or
2VdU-VdV=-UdU+2UdV
Now making
3) {(eta = U^2),(xi = V^2):} rArr {(1/2(d eta)/sqrt eta = dU),(1/2(d xi) /sqrt xi= dV):}
2 sqrt(xi/eta)d eta - d xi = - d eta + 2 sqrt(eta/xi) d xi or
(2 sqrt(xi/eta)+1)d eta = (2 sqrt(eta/xi)+1)d xi
now introducing
4) eta = lambda xi rArr d eta = lambda d xi + xi d lambda
(2/sqrt lambda +1)(lambda d xi+xi d lambda) = (2 sqrtlambda+1)d xi
or grouping variables
(d xi)/xi + f(lambda)/(lambda f(lambda)-1) d lambda = 0
with f(lambda) = (2/sqrt lambda+1)/(2 sqrt lambda + 1)
After integration we obtain
log xi +3log(1-sqrt lambda)-log(1+sqrt lambda) = C_0 or
xi= (C_1(1+sqrt lambda))/(1-sqrt lambda)^3 or
V^2=(C_1 (1+U/V))/(1-U/V)^3 and finally
(2x+y+3)^2 = (C_1(1+(x+2y+3)/(2x+y+3)))/(1-(x+2y+3)/(2x+y+3))^3
An implicit form solution.
This can be reduced to
(x-y)^3=C_2(x+y+2)