# 初值問題

${\displaystyle y'=0.85y,\qquad y(0)=19}$
${\displaystyle {\dot {y}}+3y=6t+5,\qquad y(0)=3}$

## 定義

${\displaystyle y'(t)=f(t,\ y(t))\quad {\text{with}}\quad f:\mathbb {R} \times \mathbb {R} \to \mathbb {R} \,\!}$

${\displaystyle (t_{0},\ y_{0})\in \mathbb {R} \times \mathbb {R} \,\!}$

• 假若初值問題的一個解是函數 ${\displaystyle y\,\!}$ ，則 ${\displaystyle y\,\!}$ 是微分方程式 ${\displaystyle y'(t)=f(t,\ y(t))\,\!}$ 的解，滿足 ${\displaystyle y(t_{0})=y_{0}\,\!}$
• 對於更高階的問題，可視 ${\displaystyle \mathbf {y} \,\!}$向量。每加高一個階，就増添一個分量給 ${\displaystyle \mathbf {y} \,\!}$

## 範例

${\displaystyle {\frac {dy}{dt}}=0.85y}$

${\displaystyle {\frac {dy}{y}}=0.85dt}$

${\displaystyle \ln |y|=0.85t+B}$

${\displaystyle |y|=e^{B}e^{0.85t}}$

${\displaystyle C}$為一個新的未知常數，${\displaystyle C=\pm e^{B}}$，因此

${\displaystyle y=Ce^{0.85t}}$

${\displaystyle 19=Ce^{0.85*0}}$
${\displaystyle C=19}$

${\displaystyle {\dot {y}}+3y=6t+5,\qquad y(0)=3}$

${\displaystyle sY(s)-y(0)+3Y(s)={\frac {6}{s^{2}}}+{\frac {5}{s}}}$
${\displaystyle \therefore Y(s)={\frac {y(0)s^{2}+5s+6}{s^{2}(s+3)}}}$

${\displaystyle Y(s)={\frac {\alpha }{s}}+{\frac {\beta }{s^{2}}}+{\frac {\gamma }{s+3}}}$
${\displaystyle \alpha =1,\beta =2,\gamma =y(0)-1}$
${\displaystyle Y(s)={\frac {1}{s}}+{\frac {2}{s^{2}}}+{\frac {y(0)-1}{s+3}}}$

${\displaystyle y(t)=2e^{-3t}+2t+1\,}$

## 參考資料

1. ^ Okamura, Hirosi. Condition nécessaire et suffisante remplie par les équations différentielles ordinaires sans points de Peano. Mem. Coll. Sci. Univ. Kyoto Ser. A. 1942, 24: 21–28 （法语）.
2. ^ Coddington, Earl A. and Levinson, Norman. Theory of ordinary differential equations. New York-Toronto-London: McGraw-Hill Book Company, Inc. 1955.Theorem 1.3
3. ^ Robinson, James C. Infinite-dimensional dynamical systems: An introduction to dissipative parabolic PDEs and the theory of global attractors. Cambridge: Cambridge University Press. 2001. ISBN 0-521-63204-8.Theorem 2.6