dy/dx = k y, Solution y = y₀ e^(kx)
Differential Equations · Class XII
Pick growth/decay rate k and initial value y₀; the exponential curve responds.
- Half/double time1.3863
- y(1)1.6487
- y(5)12.1825
- y = y₀ e^(kx)
Formulas in this lab
- Half/double time
- y(1)
- y(5)
Frequently asked questions
▶What does $\frac{dy}{dx} = ky$ mean?
It is the simplest differential equation: the rate of change of $y$ is proportional to $y$ itself. The solution is $y = y_0 e^{kx}$, where $y_0 = y(0)$. Positive $k$ gives exponential growth; negative $k$ gives exponential decay.
▶How do I use the exponential ODE lab?
Slide rate $k$ and initial value $y_0$. The lab plots $y = y_0 e^{kx}$ and reports the half-life or doubling time $t_{1/2} = \ln 2/|k|$. Try $k = 0.5$, $y_0 = 1$: $y(1) \approx 1.65$ and doubling time $\approx 1.39$.
▶Where does exponential growth/decay appear in real life?
Radioactive decay (carbon-14 dating), bank compound interest, population growth (small populations), and cooling of hot tea (Newton's law of cooling, with a small twist) all follow $dy/dx = ky$. The lab's slider gives quick intuition for half-life calculations.
▶Common JEE pitfall: solving $dy/dx = ky$ by separation
Separate as $\frac{dy}{y} = k\,dx$, integrate to $\ln|y| = kx + C$, exponentiate to $y = Ae^{kx}$. Students sometimes forget the absolute-value in $\ln|y|$ or skip the constant. The lab solves the result instantly, but learn the steps for the long-answer marks.