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प्रश्न
The solution of `("d"y)/("d"x) + y = "e"^-x`, y(0) = 0 is ______.
विकल्प
y = `"e"^x (x - 1)`
y = xex
y = `x"e"^-x + 1`
y = xe–x
उत्तर
The solution of `("d"y)/("d"x) + y = "e"^-x`, y(0) = 0 is y = xe–x .
Explanation:
The given differential equation is `("d"y)/("d"x) + y = "e"^-x`
Since, it is a linear differential equation then P = 1 and Q = `"e"^-x`
Integrating factor I.F. = `"e"^(int Pdx)`
= `'e"^(int 1. "d"x)`
= ex
∴ Solution is `y xx "I"."F". = int "Q" xx "I"."F". "d"x + "c"`
⇒ `y xx "e"^x = int"e"^-x xx "e"^x"d"x + "c"`
⇒ `y xx "e"^x = int "e"^0 "d"x + "c"`
⇒ `y xx "e"^x = int 1."d"x + "c"`
⇒ `y xx "e"^x = x + "c"`
Put y = 0 and x = 0
∴ 0 = 0 + c
∴ c = 0
∴ Equation is `y xx "e"^x` = x
So y = `x"e"^-x`.
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