# 常微分方程

$\int_M \mathrm{d}\omega = \oint_{\partial M} \omega$

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

## 精确解总结

$P_1(x)Q_1(y) + P_2(x)Q_2(y)\,\frac{dy}{dx} = 0 \,\!$

$P_1(x)Q_1(y)\,dx + P_2(x)Q_2(y)\,dy = 0 \,\!$

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

$dy= F(x) \, dx\,\!$

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

$dy= F(y) \, dx\,\!$

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

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

Integrate throughout. $\int^y P(\lambda)\,{d\lambda} + \int^x Q(\lambda)\,d\lambda = C\,\!$

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

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

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

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

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

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

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

Integrate throughout. \begin{align} F(x,y) & = \int^y M(x,\lambda)\,d\lambda + \int^x N(\lambda,y)\,d\lambda \\ & + Y(y) + X(x) = C \end{align} \,\!

where Y(y) and X(x) are functions from the integrals rather than constant values, which are set to make the final function F(x, y) satisfy the initial equation.

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

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

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

\begin{align} 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{align} \, \!

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

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

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

Particular integral yp: in general the method of variation of parameters, though for very simple r(x) inspection may work.[2]

$y=y_c+y_p$

$y_c=C_1e^{ \left ( -b+\sqrt{b^2 - 4c} \right )\frac{x}{2}} + C_2e^{-\left ( b+\sqrt{b^2 - 4c} \right )\frac{x}{2}}\,\!$

$y_c = (C_1x + C_2)e^{-bx/2}\,\!$

$y_c = e^{ -b\frac{x}{2}} \left [ C_1 \sin{\left ( \sqrt{\left | b^2-4c \right |}\frac{x}{2} \right )} + C_2\cos{\left ( \sqrt{\left | b^2-4c \right |}\frac{x}{2} \right )} \right ] \,\!$

n阶, 线性, 非齐次, 常系数[4]

$\sum_{j=0}^n b_j \frac{d^j y}{dx^j} = r(x)\,\!$

Complementary function yc: assume yc = eαx, substitute and solve polynomial in α, to find the linearly independent functions $e^{\alpha_j x}$.

Particular integral yp: in general the method of variation of parameters, though for very simple r(x) inspection may work.[2]

$y=y_c+y_p$

Since αj are the solutions of the polynomial of degree n: $\prod_{j=1}^n \left ( \alpha - \alpha_j \right ) = 0 \,\!$, then:

for αj all different,

$y_c = \sum_{j=1}^n C_j e^{\alpha_j x} \,\!$

for each root αj repeated kj times,

$y_c = \sum_{j=1}^n \left( \sum_{\ell=1}^{k_j} C_\ell x^{\ell-1}\right )e^{\alpha_j x} \,\!$

for some αj complex, then setting α = χj + iγj, and using Euler's formula, allows some terms in the previous results to be written in the form

$C_je^{\alpha_j x} = C_j e^{\chi_j x}\cos(\gamma_j x + \phi_j)\,\!$

where ϕj is an arbitrary constant (phase shift).

## 参考资料

1. ^ 1.0 1.1 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. ^ 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 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. ^ 4.0 4.1 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