What is the general solution of the differential equation #dy/dx = (x+2y-3)/(2x+y-3)#?

Question

Emilia Aguilar · Accepted Answer

# y+x-2 = A(y-x)^3 #

We have: # dy/dx=(x+2y-3)/(2x+y-3) # ..... [A] Our standard toolkit for DE's cannot be used. However we can perform a transformation to remove the constants from the linear numerator and denominator. Consider the simultaneous equations # { ( x + 2y -3 =0 ), (2x +y - 3=0) :} => { ( x=1 ), (y=1) :} # As a result we perform two linear transformations: Let # { (u=x-1 ), (v=y-1) :} <=> { ( x=u+1 ), (y=v+1) :} => { ( (dx)/(du)=1 ), ((dy)/(dv)=1) :}# And if we substitute into the DE [A] we get # dy/dx = ((u+1)+2(v+1)-3)/(2(u+1)+(v+1)-3) # # \ \ \ \ \ \ = (u+1+2v+2-3)/(2u+2+v+1-3) # # \ \ \ \ \ \ = (u+2v)/(2u+v) # And utilising the chain rule we have: # (dy)/(dx) = (dy)/(dv) (dv)/(du) (du)/(dx) => (dy)/(dx) = (dv)/(du) # Thus we have a transformed equation # (dv)/(du) = (u+2v)/(2u+v) # ..... [B] This is now in a form that we can handle using a substitution of the form #v=wu# which if we differentiate wrt #u# using the product gives us: # (dv)/(du) = (w)(d/(du)u) + (d/(du)w)(u) = w + u(dw)/(du) # Using this substitution into our modified DE [B] we get: # \ \ \ \ \ w + u(dw)/(du) = (u+2wu)/(2u+wu) # # :. w + u(dw)/(du) = (u+2wu)/(2u+wu) # # :. u(dw)/(du) = (u+2wu)/(2u+wu) - w # # :. u(dw)/(du) = ( (u+2wu) - w(2u+wu) ) / (2u+wu) # # :. u(dw)/(du) = ( u+2wu - 2uw-w^2u ) / (2u+wu) # # :. u(dw)/(du) = ( u(1-w^2) ) / (u(2+w)) # # :. u(dw)/(du) = (1-w^2) / (2+w) # This is now a separable DE, so we can rearrange and separate the variables to get: # int \ (2+w)/(1-w^2) \ dw = int \ 1/u \ du # # :. int \ (2)/(1-w^2)+(w)/(1-w^2) \ dw = int \ 1/u \ du # # :. int \ (2)/((1+w)(1-w))+(w)/(1-w^2) \ dw = int \ 1/u \ du # And utilising a Partial Fraction decomposition: # int \ 1/(w+1)-1/(w-1)+(w)/(1-w^2) \ dw = int \ 1/u \ du # Which is now readily integrable (giving: # ln |w+1| - ln|w-1| - 1/2ln|w^2-1| = ln| u| + lnC # This is now an algebraic problem, and we get: # ln |w+1|/|w-1| - ln sqrt(w^2-1) = ln |Cu| # # :. ln( |w+1|/( |w-1|sqrt(w^2-1)) ) = ln |Cu| # # :. |w+1|/( |w-1|sqrt(w^2-1)) = |Cu| # And squaring we get: # (w+1)^2/( (w-1)^2(w^2-1)) = C^2u^2 # # :. (w+1)^2/( (w-1)^2(w+1)(w-1)) = Au^2 # # :. (w+1)/( (w-1)^3) = Au^2 # # :. (w+1) = Au^2 (w-1)^3# Then restoring the earlier #w# substitution, using #w=v/u# we get: # v/u+1 = Au^2 (v/u-1)^3# # :. (v+u)/u = Au^2 ((v-u)/u)^3# # :. (v+u)/u = Au^2 (v-u)^3/u^3# # :. v+u = A(v-u)^3# Finally, we restore the earlier substitutions for #u# and #v#, using: # { (u=x-1 ), (v=y-1) :} # Giving us: # (y-1)+(x-1) = A((y-1)-(x-1))^3# # :. y+x-2 = A(y-x)^3 # This is the General Solution, in implicit form. ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||| Validation of Solutions: Taking the solution: # y+x-2 = A(y-x)^3 # We have via Implicit Differentiation: # dy/dx+1 = 3A(y-x)^2(dy/dx-1) # # :. dy/dx+1 = 3A(y-x)^2dy/dx-3A(y-x)^2 # # :. (3A(y-x)^2-1)dy/dx=3A(y-x)^2 +1# # :. dy/dx = (3A(y-x)^2 +1)/(3A(y-x)^2-1) # # \ \ \ \ \ \ \ \ \ \ = (3A(y-x)^2 +1)/(3A(y-x)^2-1) * (y-x)/(y-x) # # \ \ \ \ \ \ \ \ \ \ = (3A(y-x)^3 + (y-x))/(3A(y-x)^3-(y-x)) # # \ \ \ \ \ \ \ \ \ \ = (3(y+x-2) + (y-x))/(3(y+x-2)-(y-x)) # # \ \ \ \ \ \ \ \ \ \ = (3y+3x-6 + y-x)/(3y+3x-6-y+x) # # \ \ \ \ \ \ \ \ \ \ = (4y+2x-6)/(2y+4x-6) # # \ \ \ \ \ \ \ \ \ \ = (2y+x-3)/(y+2x-3) \ \ \ QED#

Charles Anctil · Answer

See below.

Calling #{(u=x+2y-3),(v=2x+y-3):}# we have #{(x=1/3(2v-u+3)),(y=1/3(2u-v+3)):}# and also #{(dx=1/3(2dv-du)),(dy=1/3(2du-dv)):}# and also #dy/dx = (2du-dv)/(2dv-du) = (2-(dv)/(du))/(2(dv)/(du)-1) = u/v# thus following the coordinate shift, the differential equation reads #(dv)/(du) = (u+2v)/(2u+v)# now making #v = lambda u rArr (dv)/(du) = (dlambda)/(du) u + lambda # After that recent change, we have #(du)/u = (lambda+2)/(1-lambda^2) dlambda# giving #lnu = 1/2(ln(1+lambda)-3ln(1-lambda))+C_0# then #u^2 = C_1^2(1+lambda)/(1-lambda)^3# or #u^2(1-v/u)^3=C_1^2(1+v/u)# or #(u-v)^3/u=C_1^2(v+u)/u# or #(u-v)^3=C_1^2(u+v)# but #u+v = 3(x+y-2)# and #u-v = y-x# so finally #(x+y-2)^3+C_2(y-x)=0# is the general solution in implicit form.

Axel Alvarez · Answer

undefined To find the general solution of the given differential equation $ \frac{dy}{dx} = \frac{x + 2y - 3}{2x + y - 3} $, follow these steps:

1. Rearrange the equation to isolate $ dy $ on one side and $ dx $ on the other side.

2. Integrate both sides with respect to $ x $.

3. Solve the resulting integral equation for $ y $ to obtain the general solution.

4. If initial conditions are given, use them to find the particular solution.

5. Express the solution in its simplest form if possible.