常微分方程

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

精确解总结

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

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

${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}=F(x)\,\!}$

${\displaystyle \mathrm {d} y=F(x)\,\mathrm {d} x\,\!}$

${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}=F(y)\,\!}$

${\displaystyle \mathrm {d} y=F(y)\,\mathrm {d} x\,\!}$

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

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

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

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

${\displaystyle yM(xy)+xN(xy)\,{\frac {\mathrm {d} y}{\mathrm {d} x}}=0\,\!}$

${\displaystyle yM(xy)\,\mathrm {d} x+xN(xy)\,\mathrm {d} y=0\,\!}$

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

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

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

, 一阶[2]

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

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

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

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

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

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

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

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

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

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

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

${\displaystyle n}$  阶线性，非齐次常系数[4]

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

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

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

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

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

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