Question

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

Answer 1

`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`

Answer 2

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.