Question

Let y = y(x) be the solution of the differential equation `(dy)/(dx) + (sqrt(2)y)/(2cos^4x - cos2x) = xe^(tan^-1(sqrt(2)cost2x)), 0 < x < π/2` with `y(π/4) = π^2/32`. If `y(π/3) = π^2/18e^(-tan^-1(α))`, then the value of 3α² is equal to ______.

Options

• 1

• 2

• 3

• 4

Answer 1

Let y = y(x) be the solution of the differential equation `(dy)/(dx) + (sqrt(2)y)/(2cos^4x - cos2x) = xe^(tan^-1(sqrt(2)cost2x)), 0 < x < π/2` with `y(π/4) = π^2/32`. If `y(π/3) = π^2/18e^(-tan^-1(α))`, then the value of 3α² is equal to 2.

Explanation:

Given: `(dy)/(dx) + (sqrt(2)y)/(2cos^4x - cos2x)`

y = `xe^([tan^-1 (sqrt(5)cot2x)]); x ∈ (0, π/2)`

It is linear differential equation.

Comparing above differential equation with `(dy)/(dx) + py` = Q,

we get, p = `sqrt(2)/(2cos^4x - cos2x)`

and Q = `xe^(tan^-1(sqrt(5)cot2x)`

Now, I.F. = e

So, `intp.dx = intsqrt(2)/(2cos^4x - cos2x)dx`

⇒ `intp.dx = intsqrt(2)/(1/2(2cos^4x)^2 - cos2x)dx`

⇒ `intp.dx = intsqrt(2)/(1/2(1 + cos2x)^2 - cos2x)dx`

⇒ `intp.dx = int(2sqrt(2))/(1 + cos^2 2x)dx`

⇒ `intp.dx = int(2sqrt(2)sec^2 2x)/(2 + tan^2 2x)dx`

Let t = tan2x

⇒ dt = 2sec²2x dx

⇒ `intp.dx = sqrt(2) int(dt)/((sqrt(2)) + t^2)`

⇒ `intp.dx = sqrt(2), 1/sqrt(2)tan^-1(t/sqrt(2))`

⇒ `intp.dx = tan^-1((tan2x)/sqrt(2))`

∴ IF = e

So, solution of the given differential equation is given by y (IF) = `intQ.(I.F.).dx`

⇒ `ye^(tan^(-1^(((tan2x)/sqrt(2))))) = intxe^(tan-1(sqrt(2)cot2x)).e^(tan^(-1^(((tan2x)/sqrt(2))))dx` ...(i)

Now, `tan^-1(sqrt(2)cot2x) + tan^-1((tan2x)/sqrt(2))`

= `tan^-1((sqrt(2)cot2x + tan  (2x)/sqrt(2))/(1 - 1))` = `π/2`

From equation (i),

`ye^(tan^-1((tan2x)/sqrt(2))) = inte^(π/2)xdx`

⇒ `ye^(tan^-1((tan2x)/sqrt(2))) = e^(π/2).x^2/2 + C` ...(ii)

∵ `y(π/4) = π^2/32`

∴ `π^2/32e^(π/2) = e^(π/2).π^2/32 + C`

⇒ C = 0

Put x = `π/3` in equation (ii), we get

`y(π/3).e^(tan^-1)(-sqrt(3/2)) = e^(π/2).π^2/18`

⇒ `π^2/18e^(tan^-1(α)).e^(tan^-1(-sqrt(3/2))) = e^(π/2).π^2/18`

⇒ `e^(tan^-1(-α) + tan^-1(-sqrt(3/2))) = e^(π/2)`

⇒ `tan^-1(-α) + tan^-1(-sqrt(3/2)) = π/2`

⇒ `αsqrt(3/2)` = 1

⇒ α² = `2/3`

⇒ 3α² = 2