Question

Find the solution of the differential equation p2+y2 = 1​

Answer 1

find the solution of differential equation p2+y2=1

Answer 2

SOLUTION

TO DETERMINE

The solution of the differential equation

$\sf{ {p}^{2} + {y}^{2} = 1 \: }$

EVALUATION

$\sf{ {p}^{2} + {y}^{2} = 1 \: }$

$\implies \sf{ {p}^{2} = 1 - {y}^{2} \: }$

$\implies \sf{ {p} = \pm \: \sqrt{ 1 - {y}^{2} } \: }$

$\displaystyle \implies \sf{ \frac{dy}{dx} = \pm \: \sqrt{ 1 - {y}^{2} } \: }$

$\displaystyle \implies \sf{ \frac{dy}{ \sqrt{1 - {y}^{2} } } = \pm \: dx \: }$

On integration

$\displaystyle \implies \sf{ \int\frac{dy}{ \sqrt{1 - {y}^{2} } } = \pm \int dx \: }$

$\displaystyle \implies \sf{ { \sin}^{ - 1}y = \pm \: x + c \: }$

Where c is integration constant

FINAL ANSWER

The required solution is

$\boxed{ \displaystyle \: \: \sf{ { \sin}^{ - 1}y = \pm \: x + c \: } \: \: }$

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