Question

If y = y(x) is the solution of the differential equation, `(dy)/(dx) + 2ytanx = sinx, y(π/3)` = 0, then the maximum value of the function y (x) over R is equal to ______.

Options

• 8

• `1/2`

• `-15/4`

• `1/8`

Answer 1

If y = y(x) is the solution of the differential equation, `(dy)/(dx) + 2ytanx = sinx, y(π/3)` = 0, then the maximum value of the function y (x) over R is equal to `underlinebb(1/8)`.

Explanation:

Given differential equation is `(dy)/(dx) + 2ytanx = sinx`  ...(i)

Solution of this differential equation will be

y(I.F) = `intsinx(I.F)dx`  ...(ii)

I.F = `e^(int2tanxdx)`

I.F = `e^(2inttanxdx)`

∵ `inttanxdx` = In (secx) + c

I.F = `e^(2In secx)`

I.F = `e^(In sec^2x)`

I.F = sec²x  ...(iii)

Now using equation (ii) and (iii)

y(sec²x) = `int(sinx)(sec^2x)dx`

⇒ y(sec²x) = `int(sinx/cosx)(secx)dx`

⇒ y(sec²x) = `int(tanxsecx)dx`

⇒ y(sec²x) =  secx + C  ...(iv)

Given that `y(π/3)` = 0 i.e., when x = `π/3`, y = 0

⇒ `(0)(sec^2  π/3) = sec(π/3) + C` 

⇒ C = `-sec  π/3`

⇒ C = –2  ...(v)

Using equation (iv) and (v)

y(sec²x) = secx – 2

⇒ y = `(secx - 2)/(sec^2x)`

⇒ y = `secx/(sec^2x) - 2/(sec^2x)`

⇒ y = `1/secx - 2/(sec^2x)`

⇒ y = cosx – 2cos²x

⇒ y = `-2{cos^2x - 1/2 cosx}`

⇒ y = `-2{cos^2x - 1/2 cosx + (1/4)^2 - (1/4)^2}`

⇒ y = `1/8 - 2(cosx - 1/2)^2`

For y_max put cosx = `1/2`⇒ y_max = `1/8`