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title: 工程數學 03/15+17
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###### tags: `工程數學 2021`

### 04/07

Nonhomogeneous Equations

$y'' + py' + gy = 0 - (H)$
$h''+py' + gy = r - (N)$

Relationship between soltions of (H) & (N)
a. THe difference of 2 soltions of (N) is a solution of (H)

$y_1'' + py_1' + gy_1 = r$ ~(1)
$y_2'' + py_2' + gy_2 = r$ ~(2)

$(1) - (2)$ $(y_1 - y_2)'' + p(y_1 - y_2) + g(y_1 - y_2) = 0$
$y_1 - y_2$ is solution of (H)
b. The sum of a solution of (H) and a soltions of (N) is also a soltion of (N)

$y_1'' + py_1' + gy_1 = r$ — (1)
$y_2'' + py_2' + gy_2 = 0$ — (1)
$(1) + (2)$
$(y_1 + y_2)'' + p(y_1 + y_2) + g(y_1 + y_2) = r$
$y_1 + y_2 is a soltion of (N)$

Def. A gerneal soltion of (N) is of the form
$y_N = y_H + y_p$
$y_H = c_1y_1 + c_2y_2$ is a GS of (H)
$y_p$ is any soltion of (N)
(Containing no arbitrary constant)

Ex. $y'' + 2y' + 101y = 10.4 e^x$
1. Solve $y'' + 2y' + 101 y = 0$
$λ^2 + 2λ + 101 = 0$
$y_H = e^{-x}( c_1 cos(10x) + c_2 sin(10x) )$
2. Find $y_p$
Guess $y_p = ce^x$
$ce^x + 2ce^x + 101ce^x = 10.4 e^x$
$→ c = 0.1$ $y_p = 0.1e^x$ $y=e^{-x}(c_1 \cos 10x + c_2 \sin 10x) + 0.1e^x$