18.01 PROBLEM SET 4 – SOLUTIONS

2. (a)

(g(x) · e^{−x})′ = g′(x) · e^{−x} + g(x) · e^{−x} · (−1) = 0.

Hence g(x) · e^{−x} = c_{1}. (b)

(h(x) · e^{x})′ = h′(x) · e^{x} + h(x) · e^{x} = 0.

Hence h(x) · e^{x} = c_{2}. (c)

(f(x) + f′(x))′ = f′(x) + f′′(x) = f(x) + f′(x),

so by part (a) f(x) + f′(x) = c_{1}. · e^{x}. Similarly

(f(x) − f′(x))′ = f′(x) − f′′(x) = − (f(x) − f′(x)) ,

so by part (b) f(x) − f′(x) = c_{2}. · e^{−x}.

(d) By (c) we have

f(x) = \frac{c_{1}}{2} · e^{x} + \frac{c_{2}}{2} · e^{−x}.

(10 points)

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