Question

For the differential equation, find the general solution:

`dy/dx = sin^(-1) x`

Answer 1

We have `dy/dx = sin^-1 x`

⇒ dy = sin^-1 x dx                   ...(1)

Integrating (1) both sides, we get

`intdy = intsin^-1 x dx`

⇒ `y = sin^-1 x int 1 dx - int (d/dx (sin^-1 x) int 1 dx)  dx`

⇒ `y = x sin^-1 x - intx/sqrt(1 - x^2)  dx`

⇒ `y = x sin^-1 x + 1/2 int ((-2x) dx)/sqrt(1 - x^2)`

⇒ `y = x sin^-1 x + 1/2 int1/sqrtt  dt`      

[Putting 1 - x² = t ⇒ -2x dx = dt]

⇒ `y = x sin^-1 x + 1/2 (t^(1/2))/(1/2) + C`

⇒ `y = x sin^-1 x + sqrtt + C`

⇒ `y = x sin^-1 x + sqrt(1+ x^2) + C`

Which is the required solution.