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Question

Show that

y=sin(tan1x+C)y = \sin \left( \tan ^ { - 1 } x + C \right)

is the general solution of

y=1y2/(1+x2).y ^ { \prime } = \sqrt { 1 - y ^ { 2 } } / \left( 1 + x ^ { 2 } \right).

Then use the addition formula for the sine function to show that the general solution may be written

y=(cosC)x+sinC1+x2y = \frac { ( \cos C ) x + \sin C } { \sqrt { 1 + x ^ { 2 } } }

Solution

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Step 1
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We are given the equation:

y=1y21+x2y'=\dfrac{\sqrt{1-y^2}}{1+x^2}

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