二阶微分方程求解

求解过程的书写

For f(x)=0, the characteristic equation is (特征方程$\Rightarrow$解)$\Rightarrow$general solution is $x=$... for arbitrary constants A and B.

For f(x)=..., a good guess for the particular integral is $x=$ (x猜的特解). Substituting this into the equation gives (代入后的式子). So the general solution is $x=$... Since we know the initial conditions, we can fix A and B , using $x(0)=...=$(题给常数). This gives A=... and B=... . Therefore, $x=$...

通解

特征值重合时, 另一通解为 $te^{\lambda t}$. 为共轭复数对$\lambda=\mu\pm i\omega$时, $x=e^{\mu t}(\alpha\cos(\omega t)+\beta\sin(\omega t))$.

猜特解

对于$\frac{d^2x}{dt^2}+A\frac{dx}{dt}+Bx=f(t)$,

$f(t)$	$x_P(t)$
$Ce^{kt}$	$ce^{kt}$
$C\cos kt+D\sin kt$	$c\cos kt+d\sin kt$
$C_nt^n+\ldots+C_1t+C_0$	$c_nt^n+\ldots+c_1t+c_0$

而对于更复杂的形式, 如$f(t)=t^2e^{2t}$, 就可以用分解法

.$(\frac{d}{dt}-\lambda_+)(\frac{d}{dt}-\lambda_-)x=f(t)$

If we define $z=\frac{dx}{dt}-\lambda_-x$, then $(\frac{d}{dt}-\lambda_+)z=f(t)\Rightarrow\frac{dz}{dt}-\lambda_+z=f(t)$. 然后用积分因子法得出答案.

具体过程如下.