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प्रश्न
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
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उत्तर
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.
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