Question

Solve the following:

If `("d"y)/("d"x) + 2 y tan x = sin x` and if y = 0 when x = `pi/3` express y in term of x.

Answer 1

`("d"y)/("d"x) + 2 y tan x = sin x`

It is of the form `("d"y)/("d"x) + "P"y` = Q

Here P = 2tan x

Q = sin x

`int "Pd"x = int 2 tan x  "d"x`

= `2 int tan x  "d"x`

= `2 log sec x`

= `log sec^2x`

I.F = `"e"^(int pdx)`

= `"e"^(log(sec^2x)`

= sec²x

The required solution is

y(I.F) = `int "Q"("I.F")  "d"x + "c"`

y(sec²x) = `int sin x (sec^2x)  "d"x + "c"`

y(sec²x) = `int sin x (1/(cos x)) sec x  "d"x + "c"`

y(sec²x) = `int (sinx/cosx) sec x "d"x + "c"`

y(sec²x) = `int tan x sec x  "d"x + "c"`

⇒ y(sec²x) = sec x + c  ........(1)

If y = 0

When x = `pi/3`

Then (1)

⇒ `0(sec^2 (pi/3)) = sec (pi/3) + "c"`

0 = 2 + c

⇒ c = – 2

∴ Equation (1)

⇒ y sec²x = sec x – 2