Solve the differential equation #(dy)/(dx)=(x+2y+3)/(2x+y+3)# ?

#dy/dx = (x+2y+3)/(2x+y+3) rArr (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)#

#(dy)/(dx)=(x+2y+3)/(2x+y+3)#

#(2x+y+3)*dy=(x+2y+3)*dx#

Now I solved #2x+y+3=0# and #x+2y+3=0# equation system. From them, #x=y=-1#

Hence I used #x=m-1#, #y=p-1#, #dx=dm# and #dy=dp# substitution, this differential equation became

#(2m+p)*dp=(m+2p)*dm#

I used #m=p*z# and #dm=pdz+zdp# transformation, it became

#(2pz+p)*dp=(pz+2p)(pdz+zdp)=0#

#(2z+1)*dp=(z+2)(pdz+zdp)=0#

#(2z+1)*dp=p(z+2)*dz+(z^2+2z)*dp#

#(z^2-1)*dp+p(z+2)*dz=0#

#(dp)/p+((z+2)*dz)/(z^2-1)=0#

#(2dp)/p+((2z+4)*dz)/[(z+1)(z-1)]=0#

#(2dp)/p+(3dz)/(z-1)-(dz)/(z+1)=0#

#2Lnp+3Ln(z-1)-Ln(z+1)=LnC#

#Ln[p^2*(z-1)^3/(z+1)]=LnC#

#p^2*(z-1)^3/(z+1)=C#

After using #m=p*z# and #z=m/p# inverse transformation,

#p^2*(m/p-1)^3/(m/p+1)=C#

#(m-p)^3/(m+p)=C#

After using #x=m-1#, #y=p-1#, #m=x+1# and #p=y+1# inverse transformation, I found

#(x-y)^3/(x+y+2)=C#