Solve the differential equation (dy)/(dx)=(x+2y+3)/(2x+y+3)dydx=x+2y+32x+y+3 ?

2 Answers
Nov 29, 2017

See below.

Explanation:

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)

Nov 29, 2017

(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