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For the differential equation, find the general solution: dydx =4-y2 (-2<y<2) - Mathematics

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Question

For the differential equation, find the general solution:

`dy/dx = sqrt(4-y^2)      (-2 < y < 2)`

Sum
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Solution

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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Chapter 9: Differential Equations - Exercise 9.4 [Page 395]

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NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 9 Differential Equations
Exercise 9.4 | Q 2 | Page 395
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