In Example 5 we saw that y Icirc brvbar 1 (x) (25 x 2 ) and y 2 (x) (25 x 2 ) are solutions of dy dx x y on the interval ( 5, 5) Explain why the peicewise defined function IV25 , 5

Question: In Example 5 we saw that y = Î¦1 (x) = (25 x 2 ) and y = 2 (x) = -(25 x 2 )

In Example 5 we saw that y = Î¦1 (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,

IV25 - , 5

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