Question

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

Answer 1

See below.

Explanation:

`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)`

Answer 2

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

Explanation:

`(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`