Question

Solve the following differential equation? `(2x+y+3)dy/dx=x+2y+1`

Answer 1

`y+x+4/3 = A(y-x-2)^3`

Explanation:

We have:

`(2x+y+3)dy/dx = x+2y+1`

`:. dy/dx = (x+2y+1)/(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 +1 =0 ), (2x +y + 3=0) :} => { ( x=-5/3 ), (y=1/3) :}`

As a result we perform two linear transformations:

Let `{ (u=x+5/3 ), (v=y-1/3) :} <=> { ( x=u-5/3 ), (y=v+1/3) :} => { ( (dx)/(du)=1 ), ((dy)/(dv)=1) :}`

And if we substitute into the DE [A] we get

`dy/dx = ((u-5/3)+2(v+1/3)+1)/(2(u-5/3)+(v+1/3)+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+5/3 ), (v=y-1/3) :}`

Giving us:

`(y-1/3)+(x+5/3) = A((y-1/3)-(x+5/3))^3`

`:. y+x+4/3 = A(y-x-2)^3`

This is the General Solution, in implicit form.

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Validation of Solutions:

Taking the solution:

`y+x+4/3 = A(y-x-2)^3`

We have via Implicit Differentiation:

`dy/dx+1 = 3A(y-x-2)^2(dy/dx-1)`

`:. dy/dx+1 = 3A(y-x-2)^2dy/dx-3A(y-x-2)^2`

`:. (3A(y-x-2)^2-1)dy/dx=3A(y-x-2)^2 +1`

`:. dy/dx = (3A(y-x-2)^2 +1)/(3A(y-x-2)^2-1)`

`\ \ \ \ \ \ \ \ \ \ = (3A(y-x-2)^2 +1)/(3A(y-x-2)^2-1) * (y-x-2)/(y-x-2)`

`\ \ \ \ \ \ \ \ \ \ = (3A(y-x-2)^3 + (y-x-2))/(3A(y-x-2)^3-(y-x-2))`

`\ \ \ \ \ \ \ \ \ \ = (3(y+x+4/3) + (y-x-2))/(3(y+x+4/3)-(y-x-2))`

`\ \ \ \ \ \ \ \ \ \ = (3y+3x+4 + y-x-2)/(3y+3x+4-y+x+2)`

`\ \ \ \ \ \ \ \ \ \ = (4y+2x+2)/(2y+4x+6)`

`\ \ \ \ \ \ \ \ \ \ = (2y+x+1)/(y+2x+3) \ \ \` QED