Question

General solution of `("d"y)/("d"x) + y` = sinx is ______.

Answer 1

General solution of `("d"y)/("d"x) + y` = sinx is y = `((sinx - cosx)/2) + "c"."e"^-x`.

Explanation:

The given differential equation is `("d"y)/("d"x) + y`  = sinx

Since, it it a linear differential equation

∴ P = 1 and Q = sinx

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

= `"e"^(int1."d"x)`

= e^x

∴ Solution is `y xx "i"."F". = int "Q" xx "I"."F". "D"x + "C"`

⇒ `y . "e"^x = int sin x . "e"6x "d"x + "c"`  ....(1)

Let I = `int sin_"I"x . "e"_"II"^x "d"x`

I = `sin x . int "e"^x  "d"x - int ("D"(sinx) . int"e"^x "d"x)"d"x`

I = `sinx . "e"^x - int cos_"I"x . "e"_"II"^x  "d"x`

I = `sinx . "e"^x - [cosx . int "e"^x "d"x - int ("D"(cosx) int"e"^x "d"x)"d"x]`

I = `sin x . "e"6x - [cosx . "e"^x - int - sin x . "e"^x "d"x]`

I = `sin x . "e"^x - cos x . "e"^x - int sin x . "e"^x "d"x`

I = `sin x . "e"^x - cos x . "e"^x - "I"`

⇒ I + I = `"e"^x (sin x - cos x)`

⇒ 2I = `"e"^x (sinx - cosx)`

∴ I = `"e"^x/2 (sinx - cosx)`

From equation (1) we get

`y . "e"^x = "e"^x/2 (sinx - cosx) + "c"`

y = `((sinx - cosx)/2) + "c" . "e"^-x`