Question

The differential equation having y = (cos^-1 x)² + P (sin^-1 x) + Q as its general solution, where P and Q are arbitrary constants, is 

Options

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

• `(1 - x^2) ("d"^2y)/("d"x^2) - x ("d"y)/("d"x)` = 2

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

• None of these

Answer 1

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

Explanation:

y = (cos^-1 x)² + P (sin^-1 x) + Q

⇒ y = (cos^-1 x)² + P `(pi/2 - cos^-1 x)` + Q   .....`[because sin^-1 x + cos^-1 x = pi/2]`

⇒ y = (cos^-1 x)² - P cos^-1 x + `(pi"P")/2` + Q    ....(i)

Differentiating w.r.t. x, we get

`"dy"/"dx" = (- 2 cos^-1 x)/sqrt(1 - x^2) + "P"/sqrt(1 - x^2)`

`=> (1 - x^2) ("dy"/"dx")^2`

= (P - 2 cos^-1 x)²

= 4(cos^-1 x)² - 4P cos^-1 x + P²

= 4[(cos^-1 x)² - P cos^-1 x] + P²

`= 4 ("y" - (pi"P")/2 - "Q") + "P"^2`   ...[From (i)]

`therefore (1 - x^2) ("dy"/"dx")^2 = 4y - 2pi"P" - 4"Q" + "P"^2`

Differentiating w.r.t. x, we get

`(1 - "x"^2)*2 "dy"/"dx" * ("d"^2y)/"dx"^2 - 2x ("dy"/"dx")^2 = 4 "dy"/"dx"`

⇒ `(1 - x^2) ("d"^2y)/("d"x^2) - x ("d"y)/("d"x)` = 2