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

2 Answers
Apr 7, 2018

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

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.