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

2 Answers
Apr 7, 2018

Answer:

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

Explanation:

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.

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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#

Apr 7, 2018

Answer:

See below.

Explanation:

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#

so the differential equation after the change of coordinates reads

#(dv)/(du) = (u+2v)/(2u+v)#

now making #v = lambda u rArr (dv)/(du) = (dlambda)/(du) u + lambda #

we have after that new transformation

#(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.