第9讲常微分方程（Ordinary Differential Equations） · 练习 · 量化课程

金融数学微分方程练习题#

连续复利模型：解 $\frac{dA}{dt} = 0.05A + 200$ ，初始条件 $A(0) = 1000$
市场渗透模型： $\frac{dP}{dt} = 0.1P(500-P)$ ， $P(0)=50$ ，求：(a) 解析解 (b) 当 $t \to \infty$ 时的渗透率

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\begin{align*} \frac{dy}{dx} &= e^x y^{-2} \\ y^2 dy &= e^x dx \\ \int y^2 dy &= \int e^x dx \\ \frac{1}{3}y^3 &= e^x + C \\ y &= \sqrt[3]{3e^x + C} \end{align*}

\begin{align*} \frac{dy}{dx} &= (y-1)(y-3) \\ \frac{dy}{(y-1)(y-3)} &= dx \\ \int \left(\frac{1/2}{y-3} - \frac{1/2}{y-1}\right)dy &= \int dx \\ \frac{1}{2}\ln\left|\frac{y-3}{y-1}\right| &= x + C \\ \text{特解：} & \quad y=1,\ y=3 \end{align*}

\begin{align*} 令\ z = y/x &\Rightarrow y = xz \\ x\frac{dz}{dx} + z &= \frac{x^2z^2 + x^2}{x^2z} = \frac{z^2 + 1}{z} \\ \int \frac{z}{1} dz &= \int \frac{1}{x} dx \\ \frac{1}{2}z^2 &= \ln|x| + C \\ y &= x\sqrt{2\ln|x| + C} \end{align*}

\begin{align*} 令\ y = zx &\Rightarrow \frac{dy}{dx} = z + x\frac{dz}{dx} \\ z + x\frac{dz}{dx} &= \frac{2zx - x}{zx + 2x} = \frac{2z - 1}{z + 2} \\ x\frac{dz}{dx} &= -\frac{z^2+1}{z+2} \\ -\int \frac{z+2}{z^2+1}\,dz &= \int \frac{1}{x}\,dx \\ -\frac{1}{2}\ln(z^2+1)-2\arctan z &= \ln|x|+C,\qquad z=\frac{y}{x} \end{align*}

\begin{align*} 积分因子\ \mu(x) &= e^{\int 3 dx} = e^{3x} \\ \frac{d}{dx}[e^{3x}y] &= e^{5x} \\ e^{3x}y &= \frac{1}{5}e^{5x} + C \\ y &= \frac{1}{5}e^{2x} + Ce^{-3x} \end{align*}

\begin{align*} 积分因子\ \mu(t) &= e^{-0.05t} \\ \frac{d}{dt}[e^{-0.05t}A] &= 200e^{-0.05t} \\ A(t) &= e^{0.05t}\left(1000 + \frac{200}{0.05}(1 - e^{-0.05t})\right) \\ &= 5000e^{0.05t} - 4000 \end{align*}

\ln(1+x) \approx x - \frac{x^2}{2} + \frac{x^3}{3}

\begin{align*} x_0=0, y_0=2 \\ x_1=0.5:\ y_1 = 2 + 0.5(0 - 2) = 1 \\ x_2=1.0:\ y_2 = 1 + 0.5(0.5 - 1) = 0.75 \end{align*}

\begin{align*} x_0=0, y_0=1 \\ x_1=0.5:\ y_1 = 1 + 0.5(1 - 0) = 1.5 \\ x_2=1.0:\ y_2 = 1.5 + 0.5(1.5 - 0.5) = 2.0 \end{align*}