Question

The solution of `("d"y)/("d"x) + y = "e"^-x`, y(0) = 0 is ______.

Options

• y = `"e"^x (x - 1)`

• y = xe^x

• y = `x"e"^-x + 1`

• y = xe^–x 

Answer 1

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)`

= e^x

∴ 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`.