Question

Solve the following differential equation:

`(1 + 3"e"^(y/x))"d"y + 3"e"^(y/x)(1 - y/x)"d"x` = 0, given that y = 0 when x = 1

Answer 1

The given differential equation may be written as

`("d"y)/("d"x) = (-3"e"^(y/x)(1 - y/x))/((1 + 3"e"^(y/x))` ......(1)

This is a homogeneous differential equation,

Putting y = vx

⇒ `("d"y)/("d"x) = "v"(1) + x "dv"/("d"x)`

(1) ⇒ `"v" + x "dv"/("d"x) = (-3"e"^(y/x)(1 - y/x))/(1 + 3"e"^(y / x))`

= `(- 3"e"^((vx)/x) (1 - "v"))/(1 + 3"e"^((vx)/x)`

= `(- 3"e"^"v"(1 - "v"))/(1 + 3"e"^((vx)/x)`

`x "dv"/("d"x) = (- 3"e"^"v" + 3"e"^"v" "v")/((1 + 3"e"^"v")) - "v"`

= `(- 3"e"^"v" + 3"e"^"v" "v" - "v"(1 + 3"e"^"v"))/(1 + 3"e"^"v")`

= `(- 3"e"^"v" + 3"e"^"v" "v" - "v" - 3"e"^"v" "v")/(1 + 3"e"^"v")`

`x "dv"/("d"x) = (- 3"e"^"v" - "v")/(1 + 3"e"^"v")`

`((1 + 3"e"^"v"))/(- 3"e"^"v" - "v") "dv" = ("d"x)/x`

`- ((1 + 3"e"^"v"))/(("v" + 3"e"^"v")) "dv" = ("d"x)/x`

`- int ((1 + 3"e"^"v"))/("v" + 3"e"^"v") "dv" - int ("d"x)/x = log ("c")`

`- int ((1 + 3"e"^"v"))/("v" + 3"e"^"v") "dv" + int ("d"x)/x = log ("c")`

`log("v" + 3"e"^"v") + log(x) = log("c")`

`log ("v" + 3"e"^"v")x = log "c"`

`x("v" + 3"e"^"v") = "c"`

`x(y/x + 3"e"^(y/x))` = c   .........`(∵ "v" = y/x)`

`(xy)/x + 3x"e"^(y/x)` = c

`y + 3x"e"^(y/x)` = c

Given that y = 0 when x = 1

0 + 3(1) e° = c

3 = c

∴ `y + 3x"e"^(y/x)` =  3 is a required solution.