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Calculus Help Please Fan + Medal
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@ganeshie8
\[\frac{ dy }{ \sqrt{1-y^2} }=2x dx\]
\[\sin^{-1}(y) =x^2+c\]
\[y=\sin(x^2+c)\]
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Is that correct?
c_1+x^2 = integral_1^(y(x)) 1\/(r(1-xi^2)) dxi
@ParthKohli
Yes, that's right.
How do I do part B??
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Can the sin(0+c)=3 for any value c?
Try graphing sin(x) for any value x.
There is no value for C.
So that initial point is not a point on the solution for y.
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