数值分析(第五版) 第九章知识点总结

仅供大致参考，有许多定义存在不严谨的地方；不同学校的考察重点自然是不同的

第九章常微分方程初值问题数值解法

常微分方程的初值问题

{dydx=f(x,y)y(x0)=y0\left\{\begin{array}{l} \frac{\mathrm{d} y}{\mathrm{~d} x}=f(x, y) \\ y\left(x_{0}\right)=y_{0} \end{array}\right. { dxdy=f(x,y)y(x0)=y0 即，我们要求解函数yyy的数学表达式，然而我们目前只有yyy在某个点(初始点)的值(初值)，以及yyy的导数y′y^{\prime}y′的表达式，而y′y^{\prime}y′式中本身是包含yyy的。为此需要设计相应的数值计算方法进行近似。

显式Euler法

yn+1=yn+hf(xn,yn)y_{n+1}=y_{n}+h f\left(x_{n}, y_{n}\right) yn+1=yn+hf(xn,yn)

隐式Euler法

yn+1=yn+hf(xn+1,yn+1)y_{n+1}=y_{n}+h f\left(x_{n+1}, y_{n+1}\right) yn+1=yn+hf(xn+1,yn+1)

梯形公式

yn+1=yn+h2[f(xn,yn)+f(xn+1,yn+1)]y_{n+1}=y_{n}+\frac{h}{2}\left[f\left(x_{n}, y_{n}\right)+f\left(x_{n+1}, y_{n+1}\right)\right] yn+1=yn+2h[f(xn,yn)+f(xn+1,yn+1)]

改进Euler法

yn+1=yn+h2[f(xn,yn)+f(xn+1,yn+hf(xn,yn))]y_{n+1}=y_{n}+\frac{h}{2}\left[f\left(x_{n}, y_{n}\right)+f\left(x_{n+1}, y_{n}+h f\left(x_{n}, y_{n}\right)\right)\right] yn+1=yn+2h[f(xn,yn)+f(xn+1,yn+hf(xn,yn))]

龙格-库塔法

{yn+1=yn+h(c1K1+c2K2+⋯+crKr)(n=0,1,2,⋯)K1=f(xn,yn)Ki=f(xn+aih,yn+h∑j=1i−1bijKj)(i=2,3,⋯,r)\left\{\begin{array}{l} y_{n+1}=y_{n}+h\left(c_{1} K_{1}+c_{2} K_{2}+\cdots+c_{r} K_{r}\right) \quad(n=0,1,2, \cdots) \\ K_{1}=f\left(x_{n}, y_{n}\right) \\ K_{i}=f\left(x_{n}+a_{i} h, y_{n}+h \sum_{j=1}^{i-1} b_{i j} K_{j}\right) \quad(i=2,3, \cdots, r) \end{array}\right. ⎩⎪⎨⎪⎧yn+1=yn+h(c1K1+c2K2+⋯+crKr)(n=0,1,2,⋯)K1=f(xn,yn)Ki=f(xn+aih,yn+h∑j=1i−1bijKj)(i=2,3,⋯,r)

龙格-库塔法&二阶中点格式

{yn+1=yn+hK2(n=0,1,2,⋯)K1=f(xn,yn)K2=f(xn+12h,yn+12hK1)\left\{\begin{array}{l} y_{n+1}=y_{n}+h K_{2} \quad(n=0,1,2, \cdots) \\ K_{1}=f\left(x_{n}, y_{n}\right) \\ K_{2}=f\left(x_{n}+\frac{1}{2} h, y_{n}+\frac{1}{2} h K_{1}\right) \end{array}\right. ⎩⎨⎧yn+1=yn+hK2(n=0,1,2,⋯)K1=f(xn,yn)K2=f(xn+21h,yn+21hK1)

龙格-库塔法&二阶休恩格式

{yn+1=yn+h4(K1+3K2)(n=0,1,2,⋯)K1=f(xn,yn)K2=f(xn+23h,yn+23hK1)\left\{\begin{array}{l} y_{n+1}=y_{n}+\frac{h}{4}\left(K_{1}+3 K_{2}\right) \quad(n=0,1,2, \cdots) \\ K_{1}=f\left(x_{n}, y_{n}\right) \\ K_{2}=f\left(x_{n}+\frac{2}{3} h, y_{n}+\frac{2}{3} h K_{1}\right) \end{array}\right. ⎩⎨⎧yn+1=yn+4h(K1+3K2)(n=0,1,2,⋯)K1=f(xn,yn)K2=f(xn+32h,yn+32hK1)

龙格-库塔法&四阶

{yn+1=yn+h6(K1+2K2+2K3+K4)(n=0,1,2,⋯)K1=f(xn,yn)K2=f(xn+h2,yn+h2K1)K3=f(xn+h2,yn+h2K2)K4=f(xn+h,yn+hK3)\left\{\begin{array}{l} y_{n+1}=y_{n}+\frac{h}{6}\left(K_{1}+2 K_{2}+2 K_{3}+K_{4}\right)(n=0,1,2, \cdots) \\ K_{1}=f\left(x_{n}, y_{n}\right) \\ K_{2}=f\left(x_{n}+\frac{h}{2}, y_{n}+\frac{h}{2} K_{1}\right) \\ K_{3}=f\left(x_{n}+\frac{h}{2}, y_{n}+\frac{h}{2} K_{2}\right) \\ K_{4}=f\left(x_{n}+h, y_{n}+h K_{3}\right) \end{array}\right. ⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧yn+1=yn+6h(K1+2K2+2K3+K4)(n=0,1,2,⋯)K1=f(xn,yn)K2=f(xn+2h,yn+2hK1)K3=f(xn+2h,yn+2hK2)K4=f(xn+h,yn+hK3)

单步法，收敛性

单步法的统一形式为：yn+1=yn+hφ(xn,yn,yn+1,h)y_{n+1}=y_{n}+h \varphi\left(x_{n}, y_{n}, y_{n+1}, h\right) yn+1=yn+hφ(xn,yn,yn+1,h) 如果不含yn+1y_{n+1}yn+1，则称为显示单步法，否则为隐式单步法。

如果对于xn=x0+nhx_{n}=x_{0}+n hxn=x0+nh，当h→0h \rightarrow 0h→0时有yn→y(xn)y_{n} \rightarrow y\left(x_{n}\right)yn→y(xn)，则称该数值方法是收敛的。

局部截断误差，阶，精度

设y(x)y(x)y(x)为微分方程的精确解，那么有精确解减去数值解：Tn+1=y(xn+1)−yn+1=y(xn+1)−y(xn)−hφ(xn,y(xn),y(xn+1),h)\begin{aligned} T_{n+1} &=y\left(x_{n+1}\right)-y_{n+1} \\ &=y\left(x_{n+1}\right)-y\left(x_{n}\right)-h \varphi\left(x_{n}, y\left(x_{n}\right), y\left(x_{n+1}\right), h\right) \end{aligned} Tn+1=y(xn+1)−yn+1=y(xn+1)−y(xn)−hφ(xn,y(xn),y(xn+1),h) 称Tn+1T_{n+1}Tn+1为单步法的局部截断误差。

泰勒展开式

研究局部截断误差的有关问题，基本上都是离不开泰勒展开。一般我们在处理局部截断误差精度时会去把精确解y(xn+1)y(x_{n+1})y(xn+1)展开，即：y(xn+1)=y(xn+h)=y(xn)+y′(xn)h+y′′(xn)2!h2+y′′′(xn)3!h3+⋯y\left(x_{n+1}\right)=y\left(x_{n}+h\right)=y\left(x_{n}\right)+y^{\prime}\left(x_{n}\right) h+\frac{y^{\prime \prime}\left(x_{n}\right)}{2 !} h^{2}+\frac{y^{\prime \prime \prime}\left(x_{n}\right)}{3 !} h^{3}+\cdots y(xn+1)=y(xn+h)=y(xn)+y′(xn)h+2!y′′(xn)h2+3!y′′′(xn)h3+⋯ 或者把f(xn+h,yn+k)f(x_{n}+h,y_{n}+k)f(xn+h,yn+k)给展开：f(xn+h,yn+k)=f(xn,yn)+∂f(xn,yn)∂xh+∂f(xn,yn)∂yk+12![∂2f(xn,yn)∂x2h2+2∂2f(xn,yn)∂x∂yhk+∂2f(xn,yn)∂y2k2]+⋯\begin{aligned} f\left(x_{n}+h, y_{n}+k\right) &=f\left(x_{n}, y_{n}\right)+\frac{\partial f\left(x_{n}, y_{n}\right)}{\partial x} h+\frac{\partial f\left(x_{n}, y_{n}\right)}{\partial y} k \\ +& \frac{1}{2 !}\left[\frac{\partial^{2} f\left(x_{n}, y_{n}\right)}{\partial x^{2}} h^{2}+2 \frac{\partial^{2} f\left(x_{n}, y_{n}\right)}{\partial x \partial y} h k+\frac{\partial^{2} f\left(x_{n}, y_{n}\right)}{\partial y^{2}} k^{2}\right]+\cdots \end{aligned} f(xn+h,yn+k)+=f(xn,yn)+∂x∂f(xn,yn)h+∂y∂f(xn,yn)k2!1[∂x2∂2f(xn,yn)h2+2∂x∂y∂2f(xn,yn)hk+∂y2∂2f(xn,yn)k2]+⋯

例1

对于初值问题：{y′=−1000(y−g(x))+g′(x)y(0)=g(0)\left\{\begin{array}{c} y^{\prime}=-1000(y-g(x))+g^{\prime}(x) \\ y(0)=g(0) \end{array}\right. {y′=−1000(y−g(x))+g′(x)y(0)=g(0) 其中g(x)g(x)g(x)为已知函数，其解y(x)=g(x)y(x)=g(x)y(x)=g(x)
1)若用显式Euler法求解，从稳定性考虑步长应在什么范围内选取
2)若用隐式Euler法求解，从稳定性考虑，步长有没有限制，为什么
3)若g(x)为不超过一次的多项式，用显式Euler法求解此问题时，从精确阶考虑，步长的选择有无限制，为什么

解 1) 根据绝对稳定的定义，有∣1+hλ∣≤1|1+h \lambda| \leq 1∣1+hλ∣≤1，而λ=−1000\lambda=-1000λ=−1000(y之前的系数)，解得0<h≤0.0020 < h \leq 0.0020<h≤0.002
2) 隐式欧拉法的绝对稳定域为(0,∞)(0,\infty)(0,∞)，因此无限制
3) 对于显式Euler法，可以写出迭代公式如下：yn+1=yn+h[−1000(yn−g(xn))+g′(xn)]y_{n+1}=y_n+h[-1000(y_n-g(x_n))+g^{\prime}(x_n)]yn+1=yn+h[−1000(yn−g(xn))+g′(xn)] 而既然g(x)g(x)g(x)不超过一次，那么令g(x)=ax+bg(x)=ax+bg(x)=ax+b，代入上式，可得yn+1=yn+h[−1000(yn−ax−b)+a]y_{n+1}=y_n+h[-1000(y_n-ax-b)+a]yn+1=yn+h[−1000(yn−ax−b)+a] 而y0=y(0)=g(0)=by_0=y(0)=g(0)=by0=y(0)=g(0)=b，可递推得y1=ah+by_1=ah+by1=ah+b，y2=a(2h)+by_2=a(2h)+by2=a(2h)+b，…，yn=a(nh)+by_n=a(nh)+byn=a(nh)+b，容易发现总是能得到精确解，因此无步长限制。

例题1(课后习题3)

用梯形法解初值问题{y′+y=0y(0)=1\left\{\begin{array}{l} y^{\prime}+y=0 \\ y(0)=1 \end{array}\right. {y′+y=0y(0)=1 证明其近似解为yn=[2−h2+h]ny_{n}=\left[\frac{2-h}{2+h}\right]^{n}yn=[2+h2−h]n，并证明当h→0h \rightarrow 0h→0时，它收敛于原初值问题的准确解y=e−xy=e^{-x}y=e−x
解：这里我们给出梯形法的公式：yn+1=yn+h2[f(xn,yn)+f(xn+1,yn+1)]y_{n+1}=y_{n}+\frac{h}{2}\left[f\left(x_{n}, y_{n}\right)+f\left(x_{n+1}, y_{n+1}\right)\right] yn+1=yn+2h[f(xn,yn)+f(xn+1,yn+1)] 那么由于f(x,y)=y′=−yf(x,y)=y^{\prime}=-yf(x,y)=y′=−y，代进去，有：yn+1=yn+h2(−yn−yn+1)y_{n+1}=y_{n}+\frac{h}{2}\left(-y_{n}-y_{n+1}\right)yn+1=yn+2h(−yn−yn+1) 移项，可以得到：yn+1=[2−h2+h]yn=[2−h2+h]2yn−1=[2−h2+h]n+1y0y_{n+1}=\left[\frac{2-h}{2+h}\right] y_{n}=\left[\frac{2-h}{2+h}\right]^{2} y_{n-1}=\left[\frac{2-h}{2+h}\right]^{n+1} y_{0} yn+1=[2+h2−h]yn=[2+h2−h]2yn−1=[2+h2−h]n+1y0 代入y0=1y_{0}=1y0=1，可以证得近似解。而由于x=nhx=nhx=nh，n为运算次数，h为步长，因此该式可化为：yn=[2−h2+h]x/hy_{n}=\left[\frac{2-h}{2+h}\right]^{x / h}yn=[2+h2−h]x/h lim⁡h→0yn=lim⁡h→0[(1−2h2+h)2+h2h]2h2+hxh=e−x\lim _{h \rightarrow 0}{y_n} =\lim _{h \rightarrow 0}\left[\left(1-\frac{2 h}{2+h}\right)^{\frac{2+h}{2 h}}\right]^{\frac{2 h}{2+h} \frac{x}{h}}=\mathbf{e}^{-x} h→0limyn=h→0lim[(1−2+h2h)2h2+h]2+h2hhx=e−x

例题2(课后习题6)

证明对任意参数ttt，下列龙格-库塔方法是二阶的：
{yn+1=yn+12h(K2+K3)K1=f(xn,yn)K2=f(xn+th,yn+thK1)K3=f(xn+(1−t)h,yn+(1−2)hK1\left\{\begin{array}{l} y_{n+1}=y_{n}+\frac{1}{2} h\left(K_{2}+K_{3}\right) \\ K_{1}=f\left(x_{n}, y_{n}\right) \\ K_{2}=f\left(x_{n}+t h, y_{n}+t h K_{1}\right) \\ K_{3}=f\left(x_{n}+(1-t) h, y_{n}+(1-2) h K_{1}\right. \end{array}\right. ⎩⎪⎪⎨⎪⎪⎧yn+1=yn+21h(K2+K3)K1=f(xn,yn)K2=f(xn+th,yn+thK1)K3=f(xn+(1−t)h,yn+(1−2)hK1 解：Tn+1=y(xn+1)−yn+1T_{n+1}=y(x_{n+1})-y_{n+1}Tn+1=y(xn+1)−yn+1 首先利用泰勒展开处理y(xn+1)=y(xn+h)y(x_{n+1})=y(x_n + h)y(xn+1)=y(xn+h)，有：y(xn+h)=y(xn)+hy′(xn)+12h2y′′(xn)+13!h3y′′′(ξ)y(x_n + h)= y(x_n)+h y^{\prime}(x_n)+\frac{1}{2} h^{2} y^{\prime \prime}(x_n)+\frac{1}{3 !} h^{3} y^{\prime \prime \prime}(\xi) y(xn+h)=y(xn)+hy′(xn)+21h2y′′(xn)+3!1h3y′′′(ξ) 而yn+1=y(xn)+h2(K2+K3)=y(xn)+12h[f(xn+th,yn+thy′(xn))+f(xn+(1−t)h,yn+(1−t)hy′(xn))])y_{n+1}=y(x_n)+\frac{h}{2}(K_2+K_3)=y(x_n)+\frac{1}{2}h\left[f\left(x_n+t h, y_n+t h y^{\prime}(x_n)\right)+f\left(x_n+(1-t) h, y_n+(1-t) h y^{\prime}(x_n)\right)\right])yn+1=y(xn)+2h(K2+K3)=y(xn)+21h[f(xn+th,yn+thy′(xn))+f(xn+(1−t)h,yn+(1−t)hy′(xn))]) 对f(xn+th,yn+thy′(xn))f\left(x_n+t h, y_n+t h y^{\prime}(x_n)\right)f(xn+th,yn+thy′(xn))与f(xn+(1−t)h,yn+(1−t)hy′(xn))f\left(x_n+(1-t) h, y_n+(1-t) h y^{\prime}(x_n)\right)f(xn+(1−t)h,yn+(1−t)hy′(xn))各自做二元泰勒展开，带回原式，可解得最后Tn+1=O(h3)T_{n+1}=O(h^3)Tn+1=O(h3)，即方法是二阶的。

例题3(课后习题7)

证明中点公式yn+1=yn+hf(xn+h2,yn+12hf(xn,yn))y_{n+1}=y_{n}+h f\left(x_{n}+\frac{h}{2}, y_{n}+\frac{1}{2} h f\left(x_{n}, y_{n}\right)\right) yn+1=yn+hf(xn+2h,yn+21hf(xn,yn))是二阶的。
解： Tn+1=y(xn+1)−y(xn)−hf(xn+h2,y(xn)+h2y′(xn))T_{n+1}=y\left(x_{n+1}\right)-y\left(x_{n}\right)-h f\left(x_{n}+\frac{h}{2}, y\left(x_{n}\right)+\frac{h}{2} y^{\prime}\left(x_{n}\right)\right) Tn+1=y(xn+1)−y(xn)−hf(xn+2h,y(xn)+2hy′(xn)) 根据泰勒展开式将y(xn+1)y(x_{n+1})y(xn+1)展开，有：y(xn+1)=y(xn)+hy′(xn)+h22y′′(xn)+h36y′′′(xn)+O(h4)y(x_{n+1})=y(x_{n})+hy^{\prime}(x_{n})+\frac{h^{2}}{2}y^{\prime\prime}(x_{n})+\frac{h^{3}}{6}y^{\prime\prime\prime}(x_{n})+O(h^{4})y(xn+1)=y(xn)+hy′(xn)+2h2y′′(xn)+6h3y′′′(xn)+O(h4) 同理，利用二元泰勒公式将f(xn+h2,y(xn)+h2y′(xn))f\left(x_{n}+\frac{h}{2}, y\left(x_{n}\right)+\frac{h}{2} y^{\prime}\left(x_{n}\right)\right)f(xn+2h,y(xn)+2hy′(xn))展开，有：f(xn+h2,y(xn)+h2y′(xn))=f(xn,y(xn))+h2∂f(xn,y(xn))∂x+h2y′(xn)∂f(xn,y(xn))∂y+12![(h2)2∂2f(xn,y(xn))∂x2+h2h2y′(xn)∂2f(xn,y(xn))∂x∂y+[h2y′(xn)]2∂2f(xn,y(xn))∂y2]+O(h3)f\left(x_{n}+\frac{h}{2}, y\left(x_{n}\right)+\frac{h}{2} y^{\prime}\left(x_{n}\right)\right) = f\left( {{x_n},y\left( {{x_n}} \right)} \right) + {h \over 2}{{\partial f\left( {{x_n},y\left( {{x_n}} \right)} \right)} \over {\partial x}} + {h \over 2}{y^\prime }\left( {{x_n}} \right){{\partial f\left( {{x_n},y\left( {{x_n}} \right)} \right)} \over {\partial y}} + {1 \over {2!}}\left[ {{{\left( {{h \over 2}} \right)}^2}{{{\partial ^2}f\left( {{x_n},y\left( {{x_n}} \right)} \right)} \over {\partial {x^2}}}} \right. + {h \over 2}{h \over 2}{y^\prime }\left( {{x_n}} \right){{{\partial ^2}f\left( {{x_n},y\left( {{x_n}} \right)} \right)} \over {\partial x\partial y}}\left. { + {{\left[ {{h \over 2}{y^\prime }\left( {{x_n}} \right)} \right]}^2}{{{\partial ^2}f\left( {{x_n},y\left( {{x_n}} \right)} \right)} \over {\partial {y^2}}}} \right] + O\left( {{h^3}} \right)f(xn+2h,y(xn)+2hy′(xn))=f(xn,y(xn))+2h∂x∂f(xn,y(xn))+2hy′(xn)∂y∂f(xn,y(xn))+2!1[(2h)2∂x2∂2f(xn,y(xn))+2h2hy′(xn)∂x∂y∂2f(xn,y(xn))+[2hy′(xn)]2∂y2∂2f(xn,y(xn))]+O(h3) 把这两个泰勒展开后的式子代回去，由于y′(xn)y^{\prime}(x_{n})y′(xn)其实就是f(xn,yn)f(x_{n},y_{n})f(xn,yn)，因此消掉一堆东西后可以得到：=h33!y′′′(xn)−h38[∂2f∂x2+y′(x)∂2f∂x∂y+(y′(x))2∂2∂y2](xn,y(xn))+O(h4)\begin{aligned} &=\frac{h^{3}}{3 !} y^{\prime \prime \prime}\left(x_{n}\right)-\frac{h^{3}}{8}\left[\frac{\partial^{2} f}{\partial x^{2}}+y^{\prime}(x) \frac{\partial^{2} f}{\partial x \partial y}+\left(y^{\prime}(x)\right)^{2} \frac{\partial^{2}}{\partial y^{2}}\right]_{\left(x_{n}, y\left(x_{n}\right)\right)}+O\left(h^{4}\right) \end{aligned} =3!h3y′′′(xn)−8h3[∂x2∂2f+y′(x)∂x∂y∂2f+(y′(x))2∂y2∂2](xn,y(xn))+O(h4) 因此是二阶的。

例题4(课后习题11)

证明解y′=f(x,y)y^{\prime}=f(x,y)y′=f(x,y)的下列差分公式yn+1=12(yn+yn−1)+h4(4yn+1′−yn′+3yn−1′)y_{n+1}=\frac{1}{2}\left(y_{n}+y_{n-1}\right)+\frac{h}{4}\left(4 y^{\prime}_{n+1}-y^{\prime}_{n}+3 y^{\prime}_{n-1}\right) yn+1=21(yn+yn−1)+4h(4yn+1′−yn′+3yn−1′)是二阶的，并求出截断误差的主项。
解：与上一问类似，首先Tn+1=y(xn+1)−yn+1T_{n+1}=y(x_{n+1})-y_{n+1}Tn+1=y(xn+1)−yn+1 对于y(xn+1)=y(xn+h)y(x_{n+1})=y(x_{n}+h)y(xn+1)=y(xn+h)，利用泰勒展开，有：y(xn+h)=y(xn)+hy′(xn)+12h2y′′(xn)+13!h3y′′′(xn)+O(h4)y(x_{n}+h)=y\left(x_{n}\right)+h y^{\prime}\left(x_{n}\right)+\frac{1}{2} h^{2} y^{\prime \prime}\left(x_{n}\right)+\frac{1}{3 !} h^{3} y^{\prime \prime \prime}\left(x_{n}\right)+O\left(h^{4}\right)y(xn+h)=y(xn)+hy′(xn)+21h2y′′(xn)+3!1h3y′′′(xn)+O(h4) 现在的问题就是怎么处理12(y(xn)+y(xn−h))+14h[4y′(xn+h)−y′(xn)+3y′(xn−h)]{1 \over 2}\left( {y\left( {{x_n}} \right) + y\left( {{x_n} - h} \right)} \right){\rm{ + }}{1 \over 4}h\left[ {4{y^\prime }\left( {{x_n} + h} \right) - {y^\prime }\left( {{x_n}} \right) + 3{y^\prime }\left( {{x_n} - h} \right)} \right]21(y(xn)+y(xn−h))+41h[4y′(xn+h)−y′(xn)+3y′(xn−h)]。其实依然完全是泰勒展开，这里给出一个例子，展开y′(xn+h)y^{\prime}({x_{n}+h})y′(xn+h)：y′(xn+h)=y(xn′)+hy′′(xn)+12h2y′′′(xn)+O(h3)y^{\prime}({x_{n}+h})=y\left(x^{\prime}_{n}\right)+h y^{\prime\prime}\left(x_{n}\right)+\frac{1}{2} h^{2} y^{\prime \prime\prime}\left(x_{n}\right)+O\left(h^{3}\right)y′(xn+h)=y(xn′)+hy′′(xn)+21h2y′′′(xn)+O(h3) 都展来开，代进去的话，最终消去得：−58h3y′′′(xn)+O(h4)-\frac{5}{8} h^{3} y^{\prime \prime \prime}\left(x_{n}\right)+O\left(h^{4}\right) −85h3y′′′(xn)+O(h4) 因此差分公式是二阶的。

例题5(课后习题12)

试证明线性二步法yn+2+(b−1)yn+1−byn=h4[(b+3)fn+2+(3b+1)fn]y_{n+2}+(b-1) y_{n+1}-b y_{n}=\frac{h}{4}\left[(b+3) f_{n+2}+(3 b+1) f_{n}\right] yn+2+(b−1)yn+1−byn=4h[(b+3)fn+2+(3b+1)fn]当b≠−1b \neq-1b=−1时方法为二阶，当b=−1b = -1b=−1时方法为三阶。
解：注意这里是求Tn+2=y(xn+2)−yn+2T_{n+2}=y(x_{n+2})-y_{n+2}Tn+2=y(xn+2)−yn+2。有：Tn+2=y(xn+2h)+(b−1)y(xn+h)−by(xn)−h4[(b+3)y′(xn+2h)+(3b+1)y′(xn)]\begin{aligned} T_{n+2}=& y\left(x_{n}+2h\right)+(b-1) y\left(x_{n}+h\right)-b y\left(x_{n}\right) \\ &-\frac{h}{4}\left[(b+3) y^{\prime}\left(x_{n}+2h\right)+(3 b+1) y^{\prime}\left(x_{n}\right)\right] \end{aligned} Tn+2=y(xn+2h)+(b−1)y(xn+h)−by(xn)−4h[(b+3)y′(xn+2h)+(3b+1)y′(xn)] 与前几问类似，各自进行泰勒展开，最终得到：−13(b+1)h3y′′′(xn)−[38+724b]h4y(4)(xn)+O(h5)-\frac{1}{3}(b+1) h^{3} y^{\prime \prime \prime}\left(x_{n}\right)-\left[\frac{3}{8}+\frac{7}{24} b\right] h^{4} y^{(4)}\left(x_{n}\right)+O\left(h^{5}\right) −31(b+1)h3y′′′(xn)−[83+247b]h4y(4)(xn)+O(h5) 当b≠−1b \neq -1b=−1时，有Tn+2=−13(b+1)h3y′′′(xn)+O(h4)T_{n+2}=-\frac{1}{3}(b+1) h^{3} y^{\prime \prime \prime}\left(x_{n}\right)+O\left(h^{4}\right) Tn+2=−31(b+1)h3y′′′(xn)+O(h4)为二阶；反正，为：Tn+2=−[38+724b]h4y(4)(xn)+O(h5)T_{n+2}=-\left[\frac{3}{8}+\frac{7}{24} b\right] h^{4} y^{(4)}\left(x_{n}\right)+O\left(h^{5}\right) Tn+2=−[83+247b]h4y(4)(xn)+O(h5)是四阶。