# 常微分方程

（重定向自常微分方程式

${\displaystyle m{\frac {d^{2}s}{dt^{2}}}=f(s)}$

## 精确解总结

${\displaystyle P_{1}(x)Q_{1}(y)+P_{2}(x)Q_{2}(y)\,{\frac {dy}{dx}}=0\,\!}$

${\displaystyle P_{1}(x)Q_{1}(y)\,dx+P_{2}(x)Q_{2}(y)\,dy=0\,\!}$

${\displaystyle {\frac {dy}{dx}}=F(x)\,\!}$

${\displaystyle dy=F(x)\,dx\,\!}$

${\displaystyle {\frac {dy}{dx}}=F(y)\,\!}$

${\displaystyle dy=F(y)\,dx\,\!}$

${\displaystyle P(y){\frac {dy}{dx}}+Q(x)=0\,\!}$

${\displaystyle P(y)\,dy+Q(x)\,dx=0\,\!}$

${\displaystyle {\frac {dy}{dx}}=F\left({\frac {y}{x}}\right)\,\!}$

${\displaystyle y=ux}$，然后通过分离变量 ${\displaystyle u}$${\displaystyle x}$ 求解. ${\displaystyle \ln(Cx)=\int ^{\frac {y}{x}}{\frac {d\lambda }{F(\lambda )-\lambda }}\,\!}$

${\displaystyle yM(xy)+xN(xy)\,{\frac {dy}{dx}}=0\,\!}$

${\displaystyle yM(xy)\,dx+xN(xy)\,dy=0\,\!}$

${\displaystyle \ln(Cx)=\int ^{xy}{\frac {N(\lambda )\,d\lambda }{\lambda [N(\lambda )-M(\lambda )]}}\,\!}$

${\displaystyle M(x,y){\frac {dy}{dx}}+N(x,y)=0\,\!}$

${\displaystyle M(x,y)\,dy+N(x,y)\,dx=0\,\!}$

${\displaystyle M(x,y){\frac {dy}{dx}}+N(x,y)=0\,\!}$

${\displaystyle M(x,y)\,dy+N(x,y)\,dx=0\,\!}$

${\displaystyle {\frac {\partial (\mu M)}{\partial x}}={\frac {\partial (\mu N)}{\partial y}}\,\!}$

{\displaystyle {\begin{aligned}F(x,y)&=\int ^{y}\mu (x,\lambda )M(x,\lambda )\,d\lambda +\int ^{x}\mu (\lambda ,y)N(\lambda ,y)\,d\lambda \\&+Y(y)+X(x)=C\\\end{aligned}}\,\!}

${\displaystyle {\frac {d^{2}y}{dx^{2}}}=F(y)\,\!}$

${\displaystyle {\frac {dy}{dx}}+P(x)y=Q(x)\,\!}$

${\displaystyle {\frac {d^{2}y}{dx^{2}}}+b{\frac {dy}{dx}}+cy=r(x)\,\!}$

${\displaystyle y=y_{c}+y_{p}}$

${\displaystyle y_{c}=C_{1}e^{\left(-b+{\sqrt {b^{2}-4c}}\right){\frac {x}{2}}}+C_{2}e^{-\left(b+{\sqrt {b^{2}-4c}}\right){\frac {x}{2}}}\,\!}$

${\displaystyle y_{c}=(C_{1}x+C_{2})e^{-{\frac {bx}{2}}}\,\!}$

${\displaystyle y_{c}=e^{-{\frac {bx}{2}}}\left[C_{1}\sin {\left({\sqrt {\left|b^{2}-4c\right|}}{\frac {x}{2}}\right)}+C_{2}\cos {\left({\sqrt {\left|b^{2}-4c\right|}}{\frac {x}{2}}\right)}\right]\,\!}$

${\displaystyle n}$ 阶线性，非齐次常系数[4]

${\displaystyle \sum _{j=0}^{n}b_{j}{\frac {d^{j}y}{dx^{j}}}=r(x)\,\!}$

${\displaystyle y=y_{c}+y_{p}}$

${\displaystyle y_{c}=\sum _{j=1}^{n}C_{j}e^{\alpha _{j}x}\,\!}$

${\displaystyle y_{c}=\sum _{j=1}^{n}\left(\sum _{\ell =1}^{k_{j}}C_{\ell }x^{\ell -1}\right)e^{\alpha _{j}x}\,\!}$

${\displaystyle C_{j}e^{\alpha _{j}x}=C_{j}e^{\chi _{j}x}\cos(\gamma _{j}x+\phi _{j})\,\!}$

## 参考资料

1. Mathematical Handbook of Formulas and Tables (3rd edition), S. Lipschutz, M.R. Spiegel, J. Liu, Schuam's Outline Series, 2009, ISC_2N 978-0-07-154855-7
2. Elementary Differential Equations and Boundary Value Problems (4th Edition), W.E. Boyce, R.C. Diprima, Wiley International, John Wiley & Sons, 1986, ISBN 0-471-83824-1
3. ^ Further Elementary Analysis, R. Porter, G.Bell & Sons (London), 1978, ISBN 0-7135-1594-5
4. Mathematical methods for physics and engineering, K.F. Riley, M.P. Hobson, S.J. Bence, Cambridge University Press, 2010, ISC_2N 978-0-521-86153-3