Ordinary differential equations
導入済み| English | 意味 |
|---|---|
| ordinary differential equation/ˈɔːdɪnəri ˌdɪfəˈrenʃl ɪˈkweɪʒn/ | An equation that relates a function of one variable to its derivatives. |
| homogeneous/ˌhɒməˈdʒiːniəs/ | 同次 |
| particular/pəˈtɪkjʊlə/ | 特解 |
| initial condition/ɪˈnɪʃl kənˈdɪʃn/ | 初期条件 |
The switch closes; the voltage slides
- At $t = 0$ a switch closes and a capacitor voltage slides to its new value. Course 1's RC response was one example of a pattern this lesson makes general.
- Circuit laws written with $\dfrac{d}{dt}$ in them are differential equations. Solve one method once, and every first-order circuit follows.
The first-order linear ODE
- An ordinary differential equation 常微分方程 links a quantity to its rate of change. The first-order linear form with constant coefficients is $\dfrac{dy}{dt} + ay = f(t)$.
- $a$ belongs to the circuit (for an RC circuit, $a = \dfrac{1}{RC}$); $f(t)$ is the input that drives it.
Homogeneous plus particular
- The solution has two parts. The homogeneous 齐次 part $Ce^{-at}$ solves the equation with $f = 0$; the particular 特解 part is any one solution forced by $f$.
- The initial condition 初始条件 fixes $C$ and picks one curve from the family. The particular part is found first and is never adjusted to fit the start.
Put the steps for solving dy/dt + ay = f(t) in order.
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C can only be found once the particular part is in place, because y(0) includes both parts.
Why the shape is exponential
- Write the equation as $\dfrac{dy}{dt} = -a(y - y_{\infty})$ and separate: $\dfrac{dy}{y - y_{\infty}} = -a\,dt$.
- Integrating gives $\ln(y - y_{\infty}) = -at + \text{const}$, so $y - y_{\infty} = Ce^{-at}$. The gap to the final value shrinks by the same percentage in every equal time step.
dy/dt + 4y = 0 with y(0) = 3. What is y at t = 0.25? Give two decimal places.
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$y = 3e^{-4t}$, so $y(0.25) = 3e^{-1} \approx 1.10$. With no input, only the homogeneous part remains.
A constant input
- $\dfrac{dy}{dt} + 2y = 10$, $y(0) = 1$. Homogeneous: $Ce^{-2t}$. A constant input forces a constant: $y_{p} = \dfrac{10}{2} = 5$.
- So $y = 5 + Ce^{-2t}$, and $y(0) = 5 + C = 1$ gives $C = -4$: $y = 5 - 4e^{-2t}$.
Check both ends. At $t = 0$, $y = 1$. As $t \to \infty$, $y \to 5$ and $\dfrac{dy}{dt} \to 0$, so the equation reads $0 + 2 \times 5 = 10$.
y = 5 − 4e^(−2t) solves dy/dt + 2y = 10. What value does y approach for large t?
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With $\dfrac{dy}{dt}$ gone, $2y = 10$ and $y = 5$. The exponential dies away to reveal the steady state.
dy/dt + 5y = 20 with y(0) = 1. Writing y = 4 + Ce^(−5t), what is C?
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The particular part is $\dfrac{20}{5} = 4$. Then $y(0) = 4 + C = 1$, so $C = -3$ and $y = 4 - 3e^{-5t}$.
A decaying input
- $\dfrac{dy}{dt} + 3y = 6e^{-t}$, $y(0) = 2$. Guess a particular part with the input's shape: $y_{p} = ke^{-t}$ gives $-k + 3k = 6$, so $k = 3$.
- Then $y = 3e^{-t} + Ce^{-3t}$, and $y(0) = 3 + C = 2$ gives $C = -1$: $y = 3e^{-t} - e^{-3t}$.
Check at $t = 0.5$: $y = 3(0.6065) - 0.2231 = 1.596$ and $\dfrac{dy}{dt} = -1.150$, so the left side is $-1.150 + 3(1.596) = 3.639$, and $6e^{-0.5} = 3.639$. The response mixes the input's exponential ($-1$) and the circuit's own ($-3$).
For dy/dt + 3y = 6e^(−t), y(0) = 2, the guess y_p = ke^(−t) gives k =
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Substituting gives $2k = 6$, so $k = 3$; then $y(0) = 3 + C = 2$ gives $C = -1$. The initial condition fixes $C$, never $k$.
Three common slips
- Forgetting the particular part. The homogeneous part alone always decays to zero.
- Using the initial condition to choose $k$. Only the homogeneous constant $C$ is free.
- Checking only $t = 0$. Also check the large-$t$ limit, where $\dfrac{dy}{dt} \approx 0$.
The initial condition is used to choose the constant k in the particular solution.
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The particular part is fixed by the input alone. The initial condition sets the homogeneous constant C.
Recap
- $\dfrac{dy}{dt} + ay = f(t)$: solution = homogeneous $Ce^{-at}$ + particular.
- Find the particular part from the input's shape; then the initial condition fixes $C$.
- Check the start value and the steady state.