Question

Solution of `("d"y)/("d"x) - y` = 1, y(0) = 1 is given by ______.

Options

• xy = – e^x

• xy = – e^-x 

• xy = – 1

• y = 2e^x – 1

Answer 1

Solution of `("d"y)/("d"x) - y` = 1, y(0) = 1 is given by y = 2e^x – 1.

Explanation:

The given differential equation is `("d"y)/("d"x) - y` = 1

Here, P = –1, Q = 1

∴ Integrating factor, I.F. = `"e"^(intPdx)`

= `"e"^(int -1"d"x)`

= `"e"^-x`

So, the solution is `y xx "I"."F". = int "Q" ."I"."F". "d"x + "c"`

⇒ `y xx "e"^-x = int 1."e"^-x  "d"x + "c"`

⇒ `y * "e"^-x = -"e"^-x + "c"`

Put x = 0, y = 1

⇒ `1. "e"^0 = - "e"^0 + "c"`

⇒ 1 = `-1 + "c"`

∴ c = 2

So the equation is `y * "e"^-x = -"e"^-x + 2`

⇒ y = `-1 + 2"e"^x`

= `2"e"^x - 1`.