Mathematics

Mathematics

Problem

For a non–zero constant kk, the differential equation dydx=k y\dfrac{dy}{dx}=k\,y arises frequently in models of exponential change. a. (i) Show that ln⁡∣y∣=kx+C1\ln|y|=k x+C_1, where C1C_1 is an arbitrary constant. b. (ii) Hence obtain the general solution y=Cekxy=C e^{k x}, clearly identifying the arbitrary constant CC. c. (iii) Prove, by examining the sign of dydx\dfrac{dy}{dx}, that every solution curve is increasing when k>0k>0 and decreasing when k<0k<0. State what happens when k=0k=0.
Intuitive explanation

This equation captures the essence of exponential growth or decay: the rate of change of yy is always proportional to yy itself. The key insight is that solutions have the form y=Cekxy = Ce^{kx}—the function grows (or shrinks) at a rate determined entirely by its current size. Solving it requires separating variables to isolate yy and xx on opposite sides, then integrating both sides.

Intuitive explanation

This equation captures the essence of exponential growth or decay: the rate of change of yy is always proportional to yy itself. The key insight is that solutions have the form y=Cekxy = Ce^{kx}—the function grows (or shrinks) at a rate determined entirely by its current size. Solving it requires separating variables to isolate yy and xx on opposite sides, then integrating both sides.

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