Question

Let y = y(x) be the solution of the differential equation, `(x^2 + 1)^2 ("dy")/("d"x) + 2x(x^2 + 1)"y"` = 1, such that y(0) = 0. If `sqrt("ay")(1) = π/32` then the value of  'a' is ______.

पर्याय

• `1/2`

• 1

• `1/16`

• `1/4`

Answer 1

Let y = y(x) be the solution of the differential equation, `(x^2 + 1)^2 ("dy")/("d"x) + 2x(x^2 + 1)"y"` = 1, such that y(0) = 0. If `sqrt("ay")(1) = π/32` then the value of  'a' is `underlinebb(1/16)`.

Explanation:

`(x^2 + 1)^2 ("dy")/("d"x) + 2x(x^2 + 1)"y"` = 1

`("dy")/("d"x) + (2xy)/(x^2 + 1) = 1/(x^2 + 1)^2`

From linear differential equation

P = `(2x)/(x^2 + 1)` Q = `1/(x^2 + 1)^2`

`"ye"^(int(2x)/(x^2 + 1)"d"x) = int1/(x^2 + 1)^2 "e"^(int(2x)/(x^2 + 1)"d"x) "d"x + "C"`

`"ye"^(log(x^2 + 1)) = int1/(x^2 + 1)^2 "e"^(log(x^2 + 1)) "d"x + "C"`

y(x² + 1) = `int1/(x^2 + 1) "d"x + "C"`

y(x² + 1) = tan^–1(x) + C

y(0) = 0  ...(Given)

⇒ C = 0

y(x² + 1) = tan^–1x

`sqrt("a")"y"(1) = π/32` then a = ?

Put x = 1  ...(i)

y(1){1 + 1} = tan^–11

y(1) = `π/8`

Substituting in equation (i)

`sqrt("a") xx π/8 = π/32`

a = `1/16`