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प्रश्न
The solution of `("d"y)/("d"x) + y = "e"^-x`, y(0) = 0 is ______.
विकल्प
y = ex(x – 1)
y = xe–x
y = xe–x + 1
y = (x + 1)e–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
∴ P = 1 and Q = e–x
∴ I.F = `"e"^(int 1."d"x)` = ex
So, the solution is `y xx "I"."F". = int "Q". "I"."F". "d"x + "c"`
⇒ `y . "e"^x = int"e"^-x . "e"^x "d"x + "c"`
⇒ `y . "e"^x = int 1."d"x + "c"`
⇒ `y . "e"^x + "c"`
Put x = 0, y = 0
We have 0 = 0 + c
∴ c = 0
So, the solution is `y "e"^x` = x
⇒ y = `x . "e"^-x`
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