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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`.
Solution
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.
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