Math Lab

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.

0.5
type a value
-22
1
type a value
-55
Live values
  • Half/double time1.3863
  • y(1)1.6487
  • y(5)12.1825
xy
  • y = y₀ e^(kx)

Formulas in this lab

  • Half/double time
    t1/2=ln2/kt_{1/2} = \ln 2 / |k|
  • y(1)
    y0eky_0 e^{k}
  • y(5)
    y0e5ky_0 e^{5k}
Tip: k > 0 → growth, k < 0 → decay, k = 0 → constant.

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.