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Question
Solve the following homogeneous differential equation:
`(x - y) ("d"y)/("d"x) = x + 3y`
Solution
`("d"y)/("d"x) = (x + 3y)/((x - y))` ........(1)
It is a homogeneous differential equation same degree in x and y
Put y = vx and `("d"y)/("d"x) = "v" + x "dv"/("d"x)`
Equation (1)
⇒ `"v" + x "dv"/("d"x) = ((x + 3 "v"x))/((x - "v"x))`
`"v" + x "dv"/("d"x) = (x(1 + 3"v"))/(x(1 - "v"))`
`"v" + x "dv"/("d"x) = ((1 + 3"v"))/((1 - "v"))`
`x "dv"/("d"x) = ((1 + 3"v"))/((1 - "v")) - "v"`
`x "dv"/("d"x) = ((1 + 3"v") - "v"(1 - "v"))/((1 - "v"))`
`x "dv"/("d"x) = (1 + 3"v" - "v" + "v"^2)/((1 - "v"))`
`x "dv"/("d"x) = ("v"^2 + 2"v" + 1)/((1 - "v"))`
`x "dv"/("d"x) = ("v" + 1)^2/((1 - "v"))`
`((1 - "v"))/("v" + 1)^2 = "dv" 1/x "d"x`
⇒ `(1 - "v" - 1 + 1)/("v" + 1)^2 "dv" = 1/x "d"x`
`(2 - "v" - 1)/("v" + 1)^2 "dv" = 1/x "d"x`
⇒ `(2 - ("v" + 1))/("v" + 1)^2 "dv" = 1/x "d"x`
`[2/("v" + 1)^2 - (("v" + 1))/("v" + 1)^2] "dv" = 1/x "d"x`
`[2("v" + 1)^2 - 1/(("v" + 1))] "dv" = 1/x "d"x`
Integrating on both sides
`int (2("v" + 1)^-2 - 1/(("v" + 1))) "dv" = int 1/x "d"x`
`2 int ("v" + 1)^-2 "dv" - int 1/(("v" + 1)) "dv" = int 1/x "d"x`
`2[("v" + 1)^(-2 + 1)/(-2 + 1)] - log ("v" + 1) = log x + log "c"`
`(-2)/(("v" + 1)) = log x + log "c" + log ("v" + 1)`
`(-2)/((y/x + 1)) = log "c" x("v" + 1)`
`(-2)/((x + y)) = log "c" ((x + y)/x)`
`((-2x)/(x + y)) = log "c" [x + y]`
⇒ `"e"^(- (2x)/(x + y)) = "c"(x + y)`
`1/"c" "e"^(-(2x)/(x + y)) = (x + y)` .....`{"Let" 1/"c" "be" "k"}`
⇒ `"ke"^(-(2x)/(x + y)) = (x + y)`
⇒ x + y = `"ke"^(-(2x)/(x + y))`
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