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प्रश्न
For the differential equation, find the general solution:
`dy/dx = sqrt(4-y^2) (-2 < y < 2)`
उत्तर
We have,
`dy/dx = sqrt(4 - y^2) (-2 < y < 2)`
⇒ `intdy/ sqrt(4 - y^2) = intdx`
On integrating
`int dy/ sqrt(2^2 - y^2) = int dx`
`sin^-1 (y/2)` = x + C
`y/2` = sin (x + C)
y = 2 sin (x + C)
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