For the following initial-value problems, show that the given equation implicitly defines a solution. Approximate y(2) using
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a. y' = − y3 + y/((3y2 + 1)t) , 1≤ t ≤ 2, y(1) = 1; y3t + yt = 2
b. y' = − (y cos t + 2tey)/(sin t + t2ey + 2) , 1≤ t ≤ 2, y(1) = 0; y sin t + t2ey + 2y = 1
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a Differentiating y 3 t yt 2 gives 3y 2 yt y 3 yt y 0 Solv...View the full answer
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