"We are asked to solve the differential equation:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (x - y ) {dy}/{dx} \ = \ x + 2 y.
"We rearrange a little:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad {dy}/{dx} \ = \ { x + 2 y } / { x - y }
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad {dy}/{dx} \ = \ { 1 + 2 ( y/x ) } / { 1 - ( y/x ) } \quad. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ (I)
"[While I may not need to mention this, this differential"
"equation is what is called a homogeneous differential equation."
"I'll discuss this after the work here. Knowing that the"
"differential equation is of that type, it tells us that the"
"following substitution will help.]"
"Use the substitution:" \qquad \qquad \qquad v = y/x. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ (II)
"Now we compute:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ y = vx.
\qquad \qquad :. \qquad \qquad \qquad \qquad \qquad \qquad \qquad {dy}/{dx} = v + x {dv}/{dx}. \qquad \qquad \qquad \qquad \qquad \qquad \qquad (III)
"Substituting (II) and (III) into (I):"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ v + x {dv}/{dx} \ = \ { 1 + 2 v } / { 1 - v }
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad x {dv}/{dx} \ = \ { 1 + 2 v } / { 1 - v } - v
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ { 1 + 2 v - v (1 - v) }/ { 1 - v }
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \= \ { 1 + v + v^2 }/ { 1 - v }.
"Thus, we have:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ x {dv}/{dx} \ = \ { 1 + v + v^2 }/ { 1 - v }.
"[Not necessary to note -- this is called a separable differential"
"equation.]"
\qquad \qquad \qquad \qquad \ { 1 - v } / { 1 + v + v^2 } {dv}/{dx} \ = \ 1/x
\qquad \qquad \qquad \ { 4 ( 1 - v ) } / { 4 + 4 v + 4 v^2 } {dv}/{dx} \ = \ 1 /x
\qquad \qquad \qquad \ { -4 ( v - 1 ) } / { 4 + 4 v + 4 v^2 } {dv}/{dx} \ = \ 1 /x
\qquad - 1/2{ 8 v - 8 } / { 4 v^2 + 4 v + 4 } {dv}/{dx} \ = \ 1 /x
\qquad - 1/2{ 8 v + 4 - 12 } / { 4 v^2 + 4 v + 4 }{dv}/{dx} \ = \ 1 /x
\qquad \qquad \qquad \ { 8 v + 4 - 12 } / { 4 v^2 + 4 v + 4 } {dv}/{dx} \ = \ -2/x
\qquad \qquad \qquad [ { 8 v + 4 } / { 4 v^2 + 4 v + 4 } - { 12 } /{ 4 v^2 + 4 v + 4 } ] {dv}/{dx} \ = \ -2/x
\qquad \qquad \ \ [ { 8 v + 4 } / { 4 v^2 + 4 v + 4 } - { 12 } / { ( 2 x + 1 )^2 + 3 } ] {dv}/{dx} \ = \ -2/x
\ [ { 8 v + 4 } / { 4 v^2 + 4 v + 4 } \ - \ 6 cdot ( { 2 } /{ ( 2 x + 1 )^2 + ( \sqrt{3} )^2 } ) ] {dv}/{dx} \ = \ -2/x
int \ [ { 8 v + 4 } / { 4 v^2 + 4 v + 4 } - 6 cdot ( { 2 } /{ ( 2 x + 1 )^2 + ( \sqrt{3} )^2 } ) ] dv
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ int -2/x dx
\quad \ ln| 4 v^2 + 4 v + 4 | - 6/\sqrt{ 3 } arctan( ( 2 v + 1 )/\sqrt{ 3 } ) \ = \ -2 ln|x| + C
\qquad \ ln| 4 v^2 + 4 v + 4 | + 2 ln|x| \ = \ 2 sqrt{ 3 } arctan( ( 2 v + 1 )/\sqrt{ 3 } )+ C
\qquad \ ln( | x^2 | cdot | 4 v^2 + 4 v + 4 | ) \ = \ 2 sqrt{ 3 } arctan( ( 2 v + 1 )/\sqrt{ 3 } )+ C.
"Now substitute back:" \qquad v = y/x:"
\qquad \quad ln| x^2 ( 4 (y/x)^2 + 4 (y/x) + 4 ) |
\qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ 2 sqrt{ 3 } arctan( ( 2 (y/x) + 1 )/\sqrt{ 3 } )+ C
\qquad \qquad ln| 4 y^2 + 4 x y + 4 x^2 | \ = \ 2 sqrt{ 3 } arctan( { x + 2 y }/ { \sqrt{ 3 } x } )+ C.
"This is our solution."
"Solution:"
\qquad \qquad ln| 4 y^2 + 4 x y + 4 x^2 | \ = \ 2 sqrt{ 3 } arctan( { x + 2 y }/ { \sqrt{ 3 } x } )+ C.