Question

If y(t) is a solution of `(1 + "t")"dy"/"dt" - "t"y` = 1 and y(0) = – 1, then show that y(1) = `-1/2`.

Answer 1

Given equation is `(1 + "t")"dy"/"dt" - "t"y` = 1

⇒ `"dy"/"dt" - ("t"/(1 + "t")) y = 1/(1 + "t")`

Here, P = `(-"t")/(1 + "t")` and Q = `1/(1 + "t")`

∴ Integrating factor I.F. = `"e"^(intpdt)`

= `"e"^(int (-1)/(1 + "t") "dt")`

= `"e"^(-int (1 + "t" - 1)/(1 + "t") "dt")`

= `"e"^(-int(1 - 1/(1 + "t"))"dt")`

= `"e"^(-["t" - log(1 + "t")])`

= `"e"^(-"t" + log(1 + "t"))`

= `"e"^(-"t") * "e"^(log(1 + "t"))`

∴ I.F. = `"e"^(-"t") * (1 + "t")`

Required solution of the given differential equation is

y . I. F. = `int "Q" . "I"."F". "dt" + "c"`

⇒ `y * "e"^-"t" (1 + "t") = int 1/((1 + "t")) * "e"^-"t" * (1 + "t")  "dt" + "c"`

⇒ `y * "e"^-"t" (1 + "t") = int "e"^-"t"  "dt" + "c"`

⇒ `y * "e"^-"t" (1 + "t") = - "e"^-"t" + "c"`

Put t = 0 and y = –1  ....[∵ y(0) = –1]

⇒ `-1 * "e"^0 * 1 = -"e"^0 + "c"`

⇒ –1 = –1 + c

⇒ c = 0

So the equation becomes

`y"e"^-"t" (1 + "t") = -"e"^-"t"`

Now put t = 1

∴ `y * "e"^-1 (1 + 1) = -"e"^-1`

⇒ 2y = –1

⇒ y = `- 1/2`

Hence y(1) = `-1/2` is verified.