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A first-order linear differential equation relates a function to its first rate of change and models many relaxation and growth processes.
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Formula
For time-dependent quantities, a has units of inverse time and r has units of y per time.
| Symbol | Meaning | Unit |
|---|---|---|
| Unknown state function | unit of the state variable | |
| Constant decay or restoring coefficient | 1/time | |
| External forcing or source | unit of y / time | |
| Integration constant fixed by initial data | unit of y |
y\prime + a y = r(t)The linear equation y′+a y=r(t) is turned into a product derivative using the integrating factor e^{at}. For constant r, the solution is the sum of the steady value r/a and an exponentially decaying initial term. If a>0, coefficient a sets the time constant 1/a.
For a linear equation y′+a(t)y=r(t) with continuous coefficients; here a is constant.
Multiplying by the integrating factor e^{at} gives (e^{at}y)′=e^{at}r(t).
Solve an initial-value problem with constant forcing
For a linear equation y′+a(t)y=r(t) with continuous coefficients; here a is constant.
y=3+Ce^(−2t), C=−2; therefore y=3−2e^(−2t).
Solve y′+2y=6 with y(0)=1.
The solution is y=3+Ce^(−2t); y(0)=1 gives C=−2.
Thus y(t)=3−2e^(−2t).
Check: y′=4e^(−2t), so y′+2y=4e^(−2t)+6−4e^(−2t)=6 and y(0)=1.
Forgetting the homogeneous solution or initial condition.
The full solution includes a steady part and a homogeneous exponential term; the initial condition fixes its coefficient.
Formula cluster
The integrating factor turns the derivative-plus-state expression into a product derivative.
Solve y′+2y=6 with y(0)=1. The solution is y=3+Ce^(−2t); y(0)=1 gives C=−2. Thus y(t)=3−2e^(−2t). Check: y′=4e^(−2t), so y′+2y=4e^(−2t)+6−4e^(−2t)=6 and y(0)=1.
Cooling, discharge, drug concentration, population models with inflow or outflow and simple control systems.
Here is how to work through a typical First-order linear differential equation (y′ + a y = r(t)) task step by step:
Task
y=3+Ce^(−2t), C=−2; therefore y=3−2e^(−2t).
Solution path
Solve y′+2y=6 with y(0)=1.
The solution is y=3+Ce^(−2t); y(0)=1 gives C=−2.
Thus y(t)=3−2e^(−2t).
Check: y′=4e^(−2t), so y′+2y=4e^(−2t)+6−4e^(−2t)=6 and y(0)=1.
For y′+a(t)y=r(t), the integrating factor is μ(t)=e^{∫a(t)dt}. The left side becomes (μy)′, so it can be integrated directly rather than guessed. For constant a, μ=e^{at}. Multiplication by e^(at) works because the product rule reproduces the terms in the equation: (e^(at)y)′=e^(at)(y′+ay). Then only the right-hand side must be integrated. The method extends to variable coefficients using μ=e^(∫a(t)dt). The initial value in this example is specified at t=0. For a start at t=t0, shift the initial condition accordingly; the same integration step determines the unique solution.
The differential equation determines the trajectory only up to an integration constant. A value such as y(0)=y0 selects one curve from that family. In a physical model, it represents the measured initial state. For a first-order linear ODE, one initial value fixes the integration constant when coefficients are continuous. Different initial values give different solutions of the same equation. In models, stating the initial time and state clearly avoids ambiguity. The solution is unique because the equation is linear and its coefficients are continuous here. With discontinuous inputs, solve piecewise and match the state continuously at jump points.
For a>0, the homogeneous part decays as e^(−at). After t=1/a, the remaining fraction is e^(−1)≈36.8 %. A larger a means faster approach to the steady value r/a. The time constant describes an e-folding, not the full duration of a process. After three time constants, about five percent of the decaying component remains; after five, less than one percent. If a is negative, the homogeneous component grows exponentially and the system is unstable. A negative coefficient a reverses the stability conclusion. Check its sign in the model before claiming exponential approach to a steady state.
The general solution is y=e^(−at)(C+∫e^(at)r(t)dt). Multiply r(t) by the integrating factor, integrate and divide back. The initial value determines C. For constant r this reduces to r/a+Ce^(−at). For a time-varying input, the method is unchanged, but a steady value r/a does not generally apply. The integral solution accumulates past inputs weighted by the exponential system response. It shows that the current state depends on the full input history. The initial condition sets the integration constant; the forcing term alone does not. Every proposed solution must satisfy both the differential equation and the initial condition.
Differentiate y(t) and substitute both y′ and y into the original equation. The equation must hold for all t, not just initially. Then check the initial condition separately. Direct substitution often catches sign errors in the exponential term. A useful check also examines t→∞. For a>0 and constant r, the solution must approach r/a, since y′ tends to zero and ay=r remains. If the limit is wrong, the sign or particular solution is often mistaken. For a>0, the initial component eventually vanishes while constant input sets the final value. This limit is a quick plausibility check on the calculation.
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Which formula describes First-order linear differential equation?
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How do you rearrange First-order linear differential equation for Solve an initial-value problem with constant forcing?
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Which common mistake happens with First-order linear differential equation?
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