In Example 5 we saw that y = Φ1 (x) = (25 x 2 ) and y

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In Example 5 we saw that y = Φ1 (x) = ˆš(25 €“ x2) and y =2(x) = -ˆš(25 €“ x2) are solutions of dy/dx = -x/y on the interval (-5, 5). Explain why the peicewise-defined function

IV25 - х, —5 <х<0 0 <x< 5 y = 1-V25 - х,

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