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 = xe^–x + 1

• y = (x + 1)e^–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

∴ P = 1 and Q = e^–x

∴ I.F = `"e"^(int 1."d"x)` = e^x

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`