44.816891
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13.498588
44.816891
200.855369
44.816891
13.445067
4.481689
34.816891
3.333333
time
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The First Order DE Calculator solves the fundamental first-order linear ordinary differential equation $$\frac{dy}{dx} = ky$$ with initial condition $$y(0) = y_0$$. This is the simplest and most important differential equation in all of mathematics, appearing in virtually every branch of science from physics and chemistry to biology and economics.
The equation states that the rate of change of a quantity is proportional to its current value. When $$k > 0$$, this produces exponential growth — the quantity increases faster and faster over time. When $$k < 0$$, the result is exponential decay — the quantity diminishes, approaching zero asymptotically but never quite reaching it.
The solution $$y(t) = y_0 e^{kt}$$ is one of the most fundamental results in mathematics. It can be derived by separation of variables: rearranging to $$\frac{dy}{y} = k\,dt$$, integrating both sides gives $$\ln|y| = kt + C$$, and applying the initial condition yields $$y = y_0 e^{kt}$$.
This equation models an extraordinary range of phenomena. Radioactive decay follows this law with $$k = -\frac{\ln 2}{t_{1/2}}$$ where $$t_{1/2}$$ is the half-life. Compound interest with continuous compounding gives $$A = Pe^{rt}$$ — the same equation with financial variables. Population growth in unlimited environments, bacterial reproduction, atmospheric pressure variation with altitude, and capacitor discharge in electrical circuits all follow this model.
The calculator evaluates the solution at your specified time $$t$$ and also provides values at $$t = 1, 5, 10$$ for comparison, along with the instantaneous rate of change $$\frac{dy}{dt}$$ at time $$t$$. For positive $$k$$, the doubling time is computed; for negative $$k$$, the half-life is given. These characteristic times are independent of the initial value and depend only on the rate constant.
Enter your initial value $$y_0$$, rate constant $$k$$, and the time at which you wish to evaluate the solution. Positive $$k$$ models growth; negative $$k$$ models decay. The magnitude of $$k$$ determines how rapidly the function changes — larger $$|k|$$ means faster growth or decay.
The first-order ODE $$\frac{dy}{dt} = ky$$ is solved by separation of variables.
Step 1: Separate the variables:
$$\frac{dy}{y} = k\,dt$$
Step 2: Integrate both sides:
$$\ln|y| = kt + C$$
Step 3: Exponentiate:
$$y = Ae^{kt}, \quad A = e^C$$
Step 4: Apply the initial condition $$y(0) = y_0$$:
$$y_0 = Ae^0 = A$$
Therefore the unique solution is:
$$y(t) = y_0 e^{kt}$$
The doubling time (for $$k > 0$$) is found by setting $$y(t_d) = 2y_0$$:
$$2y_0 = y_0 e^{kt_d} \implies t_d = \frac{\ln 2}{k}$$
The half-life (for $$k < 0$$) is found by setting $$y(t_h) = \frac{y_0}{2}$$:
$$t_h = \frac{\ln 2}{|k|}$$
The instantaneous rate $$\frac{dy}{dt} = ky(t) = ky_0 e^{kt}$$ gives the speed of change at any moment.
y(t) is the value of the function at time $$t$$. For growth models this increases without bound; for decay models it approaches zero.
dy/dt at t is the instantaneous rate of change. For growth ($$k > 0$$), this is positive and increasing; for decay ($$k < 0$$), it is negative with decreasing magnitude.
Doubling Time (shown when $$k > 0$$) is the time required for the quantity to double. It depends only on $$k$$, not on $$y_0$$. A smaller doubling time means faster growth.
Half-Life (shown when $$k < 0$$) is the time for the quantity to decrease by half. Again, this depends only on $$|k|$$.
The values y(1), y(5), y(10) provide a trajectory snapshot, helping you visualize how quickly the function changes over different time horizons.
Inputs
Results
Starting with 10 bacteria and k = 0.3 per hour, after 5 hours: y(5) = 10·e^(0.3×5) = 10·e^1.5 ≈ 44.82. The population doubles every ln(2)/0.3 ≈ 2.31 hours. By t = 10, the population reaches about 201.
Inputs
Results
Starting with 500 g of a radioactive substance with k = -0.1 per year, after 7 years: y(7) = 500·e^(-0.7) ≈ 248.35 g. The half-life is ln(2)/0.1 ≈ 6.93 years, matching the near-halving observed at t = 7.
The rate constant k determines the speed and direction of change. When k > 0, the quantity grows exponentially; when k < 0, it decays. The magnitude |k| controls the speed: larger values mean faster change. Units of k are inverse time (e.g., per second, per year).
If you know the half-life t₁/₂, then k = -ln(2)/t₁/₂ for decay, or equivalently |k| = ln(2)/t₁/₂ ≈ 0.6931/t₁/₂. For example, carbon-14 has a half-life of 5,730 years, so k = -0.6931/5730 ≈ -0.000121 per year.
Exponential growth assumes unlimited resources. In reality, populations face constraints like food, space, and predation. The exponential model is accurate only for the early phase of growth. For long-term modeling, the logistic equation (which includes a carrying capacity) is more appropriate.
Mathematically, the formula y(t) = y₀e^(kt) works with negative y₀. Physically, negative initial values may represent quantities like charge deficit or temperature below a reference. The exponential function preserves the sign of y₀.
When k = 0, dy/dt = 0, meaning the function is constant: y(t) = y₀ for all t. There is no growth, no decay — the quantity remains unchanged. The doubling time and half-life are both undefined (infinite).
The Differential Equation Solver handles four different equation types (exponential, cooling, logistic). This First Order DE Calculator focuses exclusively on dy/dt = ky with detailed output including trajectory values at t = 1, 5, and 10, making it ideal for studying the pure exponential model in depth.
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