Ordinary differential equations
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| homogeneous/ˌhɒməˈdʒiːniəs/ | homogeneous |
| particular/pəˈtɪkjʊlə/ | particular solution |
| initial condition/ɪˈnɪʃl kənˈdɪʃn/ | initial condition |
ODE + Laplace: two routes, one answer
- Classical: homogeneous $Ce^{-at}$ plus a particular part, then the initial condition selects from the family; check both endpoints (start value, steady state).
- Laplace: $L{dy/dt} = sY - y(0)$ makes it algebra; partial fractions then invert. Poles predict the waveform before you invert.
$dy/dt + 2y = 10, y(0) = 1$ gives $y = 5 - 4e^{-2t}$ by BOTH routes; the poles $s = 0, -2$ say "constant plus decay" before any inversion.
{exercise:uni-ee2x-laplace}
Y(s) has poles at s = 0 and s = −2. The response shape is —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
A pole at 0 is a constant term; a real pole at −a is e^{−at}. Poles read the waveform before any inversion — complex pole pairs would mean oscillation.
For dy/dt + 3y = 6e^{-t}, y(0) = 2, the particular guess y_p = k e^{-t} gives k = —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
Substituting: -k + 3k = 2k = 6, so k = 3 and y = 3e^{-t} + Ce^{-3t}; y(0) = 3 + C = 2 gives C = -1. The initial condition fixes the HOMOGENEOUS constant, never the particular k.
The solution y = 5 - 4e^{-2t} must satisfy the large-t limit of dy/dt + 2y = 10. There, y approaches —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
With dy/dt gone, 2y = 10 so y = 5 - the steady state is the input divided by the coefficient, and the exponential dies to reveal it. Endpoint checks at BOTH ends are free.