Exercises 1.3 9. (a) The function φ ( x ), being a solution to the given initial value problem, satisFes dφ dx = x − φ ( x ) ,φ (0) = 1 . (1.4) Thus d 2 φ dx 2 = d dx ± dφ dx ² = d dx ( x − φ ( x )) = 1 − dφ dx =1 − x + φ ( x ) , where we have used (1.4) substituting (twice) x − φ ( x )for dφ/dx . (b) ±irst we note that any solution to the given di²erential equation on an interval I is continuously diferentiable on I . Indeed, if y ( x ) is a solution on I ,then y 0( x )doesex ist on I ,andso y ( x ) is continuous on I because it is di²erentiable. This immediately implies that y0 ( x ) is continuous as the di²erence of two continuous functions, x and y ( x ). ±rom (1.4) we conclude that dφ dx ³ ³ ³ x =0 =[ x − φ ( x )] ³ ³ x =0 =0 − φ (0) = − 1 <0 and so the continuity of φ0 ( x ) implies that, for | x | small enough, φ0 ( x ) < 0. By the Monotonicity Test, negative derivative of a function results that the function itself is decreasing. When x increases from zero, as far as φ ( x ) >x , one has φ0 ( x ) < 0andso φ ( x ) decreases.
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