Solve dy/dt = ky with an Initial Value: Separate, Integrate and Check

For constant k, the solutions of dy/dt = ky are y(t) = Ce^(kt). An initial value y(t₀) = y₀ gives y(t) = y₀e^[k(t − t₀)], including the zero solution when y₀ = 0.

A rate proportional to the current amount

The equation says the instantaneous change is k times the amount currently present. If t is measured in hours, k has units per hour. Positive k produces growth for positive y, negative k produces decay, and k = 0 gives a constant solution.

This is a mathematical model under stated assumptions. It is not a guarantee that a real population, balance or physical quantity will follow the formula indefinitely. In particular, continuous proportional growth differs from a once-per-period percentage update.

Worked example

Solve dy/dt = 0.3y with y(0) = 40. For a nonzero solution, separate variables: dy/y = 0.3 dt. Integrate to get ln|y| = 0.3t + C₁. Exponentiating and absorbing the constant and sign gives y = Ce^(0.3t).

Use the initial value: 40 = Ce⁰, so C = 40. The solution is y(t) = 40e^(0.3t). At t = 2, it equals 40e^0.6, approximately 72.885. Differentiate the solution to obtain 12e^(0.3t), which is exactly 0.3 times y(t), and check y(0) = 40 separately.

Do not lose the zero solution

Dividing by y assumes y ≠ 0. Check y = 0 directly before or after separation: its derivative is zero and ky is zero, so it also solves the equation. The final family Ce^(kt) includes it when C = 0.

If an initial condition is supplied at t₀ rather than zero, the shifted form y₀e^[k(t − t₀)] keeps the meaning clear. For example, y(2) = 5 with k = −0.4 gives y(t) = 5e^[−0.4(t − 2)]. Substituting t = 2 returns 5 exactly.

Continuous rate versus finite percentage

With k = 0.3 per time unit, the factor after one unit is e^0.3, not 1.3. To model an exact one-unit multiplier of 1.3 using this continuous equation, use k = ln 1.3. The numerical label “30 percent” must therefore be interpreted carefully.

Common mistake

Do not stop after fitting the initial value. A proposed function must satisfy both the differential equation and that value. Also include the constant of integration, and keep the sign of k when calculating decay. A negative rate does not make a positive solution negative; it makes its exponential factor decrease over time.

Equation aur initial value dono check karo; y se divide karte waqt zero solution yaad rakho.

Three practice questions with worked answers

1. Solve y′ = 2y with y(0) = 3.

   The family is Ce^(2t). The initial value gives C = 3, so y = 3e^(2t). Its derivative 6e^(2t) equals 2y.

2. Solve y′ = −y with y(0) = 8.

   Use y = Ce^(−t), then C = 8. Thus y = 8e^(−t), which stays positive and decreases as t increases.

3. Solve y′ = 5y with y(1) = 0.

   The initial condition forces C = 0 in Ce^(5t). Therefore y(t) = 0 for every t; check that both sides of the equation are zero.

Related study guides

• Differentiation Formulas Class 12: Complete List with Examples
• Integration Formulas Class 12: Complete List with Methods and Examples

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References and authorship

Original examples and practice questions, prepared with AI assistance and editorial arithmetic review. Related definitions and methods: OpenStax: Separable Equations. No textbook or exam-board affiliation is claimed.