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प्रश्न
For the differential equation, find the general solution:
`dy/dx = sin^(-1) x`
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उत्तर
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.
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