Consider the expression that arises from a first–order differential equation with

MATHEMATICS · Single correct · AP · Calculus AB · Differential Equations · Differential Equations

Problem

Consider the expression y(x)=y0+∫axf(t) dty(x)=y_0+\displaystyle\int_{a}^{x} f(t)\,dt that arises from a first–order differential equation dydx=f(x)\dfrac{dy}{dx}=f(x) with y(a)=y0y(a)=y_0. Which of the following best explains why this expression represents a unique curve passing through the point (a,y0)(a,y_0)? (a) The Fundamental Theorem of Calculus states that all antiderivatives differ by a constant, so different values of that constant produce the same curve. (b) Because f(t)f(t) is continuous near aa, the existence–uniqueness theorem for differential equations guarantees exactly one solution curve through (a,y0)(a,y_0). (c) The definite integral fixes the constant of integration, so y(a)=y0y(a)=y_0 is satisfied without introducing any additional unknowns. (d) The mean value theorem guarantees that f(t)f(t) attains its average value on [a,x][a,x], forcing the graph through (a,y0)(a,y_0).

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