Question

Find the general solution of `("d"y)/("d"x) -3y = sin2x`

Answer 1

Given equation is `("d"y)/("d"x) -3y = sin2x`

Here, P = –3 and Q = sin2x

∴ Integrating factor I.F. = `"e"^(int Pdx)`

= `"e"^(int-3dx)`

= `"e"^(-3x)`

∴ Solution is `y xx "I"."F". = int "Q" . "I"."F". "d"x + "c"`

⇒ `y . "e"^(-3x) = int sin2x . "e"^(-3x) "d"x + "c"`

Let I = `int sin_"I" 2x . "e"_"II"^(-3x) "d"x`

⇒ I = `sin 2x . int "e"^(-3x)"d"x - int("D"(sin 2x) . int"e"^(-3x) "d"x)"d"x`

⇒ I = `sin 2x . "e"^(-3x)/(-3) - int 2 cos2x . "e"^(-3x)/(-3) "d"x`

⇒ I = `"e"^(-3x)/(-3) sin2x + 2/3 int cos_"I" 2x . "e"_"II"^(-3x) "d"x`

⇒ I = `"e"^(-3x)/(-3) sin 2x + 2/3 [cos 2x . int "e"^(-3x) "d"x - int["D" cos2x . int "e"^(-3x) "d"x]"d"x]`

⇒ I = `"e"^(-3x)/(-3) sin 2x + 2/3 [cos 2x . "e"^(-3x)/(-3) - 2sin 2x . "e"^(-3x)/(-3)]"d"x`

⇒ I = `"e"^(-3x)/(-3) sin 2x - 2/9 cos2x . "e"^(-3x) - 4/9 int sin 2x. "e"^(-3x) "d"x`

⇒ `"e"^(-3x)/(-3) sin2x - 2/9 "e"^(-3x) cos 2x - 4/9 "I"`

⇒ `"I" + 4/9 "I" = "e"^(-3x)/(-3) sin 2x - 2/9 "e"^(-3x) cos 2x`

⇒ `13/9 "I" = - 1/9 [3"e"^(-3x) sin2x + 2"e"^(-3x) cos2x]`

⇒ I = `- 1/13 "e"^(-3x) [3 sin 2x + 2 cos2x]`

∴ The equation becomes `y . "e"^(-3x) = - 1/13 "e"^(-3x) [3 sin 2x + 2 cos 2x] + "c"`

∴ y = `- 1/13 [3 sin 2x + 2 cos 2x] + "c" . "e"^(3x)`

Hence, the required solution is y = `-[(3sin2x + 2cos2x)/13] + "c" . "e"^(3x)`