4.323325
1.353349
4.999996
-0.676676
0.5
0.676673
13.53
%
4.323325
1.353349
4.999996
-0.676676
0.5
0.676673
13.53
%
The Linear DE Calculator solves the first-order linear ordinary differential equation $$\frac{dy}{dx} + Py = Q$$ where $$P$$ and $$Q$$ are constants. This equation is one of the most fundamental and practically useful ODEs in mathematics, with direct applications in mixing problems, RC circuits, pharmacokinetics, and thermal equilibrium.
Unlike the simpler equation $$dy/dx = ky$$ (which has no forcing term), this equation includes a constant forcing term $$Q$$ that drives the system toward a steady state $$y_{ss} = Q/P$$. The general solution is:
$$y(x) = \frac{Q}{P} + Ce^{-Px}$$
where $$C = y_0 - Q/P$$ is determined by the initial condition. The solution consists of two parts: the steady-state (or particular) solution $$Q/P$$ and the transient (or homogeneous) solution $$Ce^{-Px}$$ that decays over time when $$P > 0$$.
When $$P > 0$$, the transient part decays exponentially, and the solution approaches the steady state $$Q/P$$ as $$x \to \infty$$. The time constant $$\tau = 1/P$$ determines how fast this approach occurs: after one time constant, about 63.2% of the adjustment has occurred; after five time constants, the system is within 0.7% of steady state.
This equation models many real-world processes. In an RC circuit, the voltage across the capacitor satisfies $$RC \cdot dV/dt + V = V_s$$ where $$V_s$$ is the source voltage. In mixing problems, the concentration of salt in a tank with inflow and outflow satisfies this form. In pharmacokinetics, drug concentration with continuous infusion follows $$dC/dt + kC = R/V$$ where $$R$$ is the infusion rate and $$V$$ is the volume of distribution.
Enter the coefficient $$P$$, the constant $$Q$$, and the initial condition $$y(0)$$, then specify the point $$x$$ at which to evaluate. The calculator returns the solution value, steady state, transient component, rate of change, time constant, and percentage of steady state reached.
The equation $$\frac{dy}{dx} + Py = Q$$ (with constant $$P$$ and $$Q$$) is solved using the integrating factor method.
Step 1: The integrating factor is $$\mu(x) = e^{Px}$$.
Step 2: Multiply both sides: $$\frac{d}{dx}[e^{Px} y] = Qe^{Px}$$
Step 3: Integrate: $$e^{Px} y = \frac{Q}{P}e^{Px} + C$$
Step 4: Divide by $$e^{Px}$$:
$$y(x) = \frac{Q}{P} + Ce^{-Px}$$
Step 5: Apply initial condition $$y(0) = y_0$$:
$$y_0 = \frac{Q}{P} + C \implies C = y_0 - \frac{Q}{P}$$
The steady state is $$y_{ss} = Q/P$$. The transient is $$Ce^{-Px}$$. The time constant is $$\tau = 1/|P|$$.
When $$P = 0$$, the equation reduces to $$dy/dx = Q$$, giving the linear solution $$y = y_0 + Qx$$.
y(x) is the total solution at position/time $$x$$, combining steady-state and transient components.
Steady State (Q/P) is the long-term equilibrium value that the system approaches when $$P > 0$$. For an RC circuit, this is the source voltage; for a mixing tank, the inflow concentration.
Transient Part shows how far the current value is from steady state. When $$P > 0$$, this decays exponentially toward zero.
Time Constant (1/|P|) is the characteristic time scale. After 1 time constant, ~63% of the transition is complete. After 3 time constants, ~95%. After 5, ~99.3%.
% of Steady State Reached indicates how close the system is to equilibrium at the given $$x$$. Values near 100% mean the transient has essentially died out.
Inputs
Results
Steady state = Q/P = 10/2 = 5. Time constant = 1/2 = 0.5. At x = 1 (two time constants): y(1) = 5 + (0-5)e^(-2) = 5 - 5·0.1353 = 4.323. About 86.5% of steady state has been reached, consistent with being two time constants into the process.
Inputs
Results
Starting with y₀ = 10, steady state = Q/P = 3/0.5 = 6. The system decays from 10 toward 6. At x = 4 (two time constants): y(4) = 6 + 4·e^(-2) = 6 + 0.541 ≈ 6.54. The transient part (0.54) shows we're still slightly above equilibrium.
When P < 0, the transient term grows exponentially instead of decaying. The solution diverges from the steady state Q/P rather than approaching it. This represents an unstable system. The percentage of steady state reached is not meaningful in this case.
The integrating factor μ(x) = e^(∫P dx) = e^(Px) transforms the left side of the equation into an exact derivative d/dx[μy]. This allows direct integration of both sides. For constant P and Q, the method yields the closed-form solution y = Q/P + Ce^(-Px).
The time constant τ = 1/P and the half-life t₁/₂ = ln(2)/P are related by t₁/₂ = τ·ln(2) ≈ 0.693·τ. After one time constant, 63.2% of the transition is complete. After one half-life, 50% is complete.
Yes, the general form dy/dx + P(x)y = Q(x) can be solved with a variable integrating factor μ(x) = e^(∫P(x)dx). However, the integral may not have a closed form. This calculator handles the constant-coefficient case, which covers most practical applications.
Q/P is the equilibrium or steady-state value — the value where dy/dx = 0. At this point, the forcing term Q exactly balances the dissipation term Py, so there is no further change. In circuit terms, Q represents the driving voltage/P the resistance, and Q/P is Ohm's law.
The time constant τ = 1/P is the fundamental measure of system speed. Engineers design RC filters, control systems, and thermal management based on the time constant. The rule of thumb '5τ to steady state' (99.3% complete) is used across all engineering disciplines.
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