内容体现经典与现代的紧密结合, 符合高校工科专业对数学的基本需求. 主要内容有距离与范数, 包括向量范数与矩阵范数; 矩阵的标准形与特征值计算, 包括矩阵的 Jordan标准形及特征值的幂迭代法; 矩阵分解与广义逆矩阵, 包括三角分解、满秩分解和奇异值分解; 线性方程组的数值解法, 包括直接解法与迭代解法; 最优化方法, 包括单纯形法、最优性条件、牛顿法、共轭梯度法、罚函数法、组合优化问题的模拟退火算法与遗传算法; 函数逼近与数据拟合, 包括多项式插值、最小二乘法、小波变换; 偏微分方程及其数值解法, 包括定解问题、解析方法、有限差分法、有限元方法; 统计分析, 包括一元及多元线性回归、贝叶斯统计、多元正态分布的参数估计与假设检验.
Advanced matrix algebra: mxn systems, linear independence, sparse matrices, introduction to second-order tensors. Further ordinary differential equations: systems of ODEs, variation of parameters; boundary-value problems. Fourier series: Euler formulae, convergence, half-range series, solution of ODEs, spectra. Further multivariable calculus: change of variables and chain rule, polar coordinates, line integrals; vector fields; del, divergence, curl and Laplacian; surface and volume integrals; Gauss and Stokes theorems. Partial differential equations: simple PDEs, Laplace, heat and wave equations, superposition, separation of variables, polar coordinates. Advanced numerical methods: solution of linear systems, numerical solution of ODEs and simple PDEs, accuracy, efficiency and stability; discrete Fourier transforms, introduction to PS and FE methods.
Content: The content reflects a close combination of classical and modern and meets the basic needs of engineering graduates from mathematics colleges and universities. Core content includes distance and paradigm, including vector paradigm and matrix paradigm; standard form and eigenvalue computation of matrices, including standard form of Jordan matrices and power iteration method of eigenvalues; matrix decomposition and generalised inverse matrix, including triangular decomposition, full rank decomposition and singular value decomposition; numerical solution of linear equations, including direct and iterative solution; optimisation methods, including the simplex method, the optimality condition method, Newton’s method, the conjugate gradient method, the penalty function method, the simulated annealing algorithm and the genetic algorithm for combinatorial optimisation problems.
1) The characteristic equation is $\lambda^2-2 \lambda+5=0$ with roots $\lambda=\frac{2 \pm \sqrt{4-20}}{2}=\frac{2 \pm 4 i}{2}=1 \pm 2 i$.
a basis for the set of solutions is $y_1=\mathfrak{e}^x \cos (2 x)$ and $y_2=\mathfrak{e}^x \sin (2 x)$ and the general solution is $y=e^x\left(c_1 \cos (2 x)+c_2 \sin (2 x)\right)$. To apply the initial conditions, we will need the derivative: $y^{\prime}=e^x\left(c_1 \cos (2 x)+c_2 \sin (2 x)\right)+e^x\left(-2 c_1 \sin (2 x)+2 c_2 \cos (2 x)\right)$ which we can write as $y^{\prime}=e^x\left(c_1+2 c_2\right) \cos (2 x)+e^x\left(-2 c_1+c_2\right) \sin (2 x)$. Now we can apply the initial conditions:
$$
\begin{aligned}
& y\left(\frac{\pi}{2}\right)=0 \Longrightarrow e^{\frac{\pi}{2}}\left(c_1 \cos (\pi)+c_2 \sin (\pi)\right)=0 \Longrightarrow e^{\frac{\pi}{2}}\left(-c_1\right)=0 \Longrightarrow c_1=0 . \
& y^{\prime}\left(\frac{\pi}{2}\right)=2 \Longrightarrow e^{\frac{\pi}{2}}\left(c_1+2 c_2\right)(-1)+e^{\frac{\pi}{2}}\left(-2 c_1+c_2\right)(0)=2 \Longrightarrow-2 c_2 e^{\frac{\pi}{2}}=2 \Longrightarrow c_2=-e^{-\frac{\pi}{2}} .
\end{aligned}
$$
So the (unique) solution to the initial value problem is $y=e^x\left(-e^{-\frac{\pi}{2}} \sin (2 x)\right)=-e^{x-\frac{\pi}{2}} \sin (2 x)$.
OR
Substitute $y\left(\frac{\pi}{2}\right)=0$ in $y=e^x\left(c_1 \cos (2 x)+c_2 \sin (2 x)\right)$ to get $0=e^{\frac{\pi}{2}}\left(-c_1\right) \Longrightarrow c_1=0$.
So at this point, we know that $y=c_2 e^x \sin (2 x)$. Now find the derivative: $y^{\prime}=c_2 e^x \sin (2 x)+2 c_2 e^x \cos (2 x)$ and apply the condition $y^{\prime}\left(\frac{\pi}{2}\right)=2$ to get $2=2 c_2 e^{\frac{\pi}{2}}(-1) \Longrightarrow c_2=-e^{-\frac{\pi}{2}}$ and so the solution to the IVP is $y=-e^{-\frac{\pi}{2}} e^x \sin (2 x)$, as before.
Outer solution: To zeroth order we must have $-y^{\prime}+x y=0$, or
$$
\frac{d y}{y}=x d x \quad \Rightarrow \quad \ln y=\frac{1}{2} x^2+c \quad \Rightarrow \quad y=C e^{x^2 / 2} .
$$
mpose the left-end boundary condition: $1=y(0)=C$. Thus
$$
y_0=e^{x^2 / 2} .
$$
Inner solution: Let $z=\frac{x-1}{\epsilon}$, so that $x=1+\epsilon z$ and $z$ is negative inside the interval. Then $\frac{d}{d x}=\frac{1}{\epsilon} \frac{d}{d z}$, so the ODE transforms (after multiplication by $\epsilon$ ) to
$$
\frac{d^2 y}{d z^2}-\frac{d y}{d z}+\left(\epsilon+\epsilon^2 z\right) y=0 .
$$
o to zeroth order we must have $\left(y^{\prime}\right)^{\prime}-y^{\prime}=0$, so $y^{\prime}=A e^z$, hence
$$
y=A e^z+B=A e^{(x-1) / \epsilon}+B .
$$
mpose the right-end boundary condition: $0=y(1)=A+B$, so $B=-A$. Thus
$$
y_{\mathrm{i}}=A\left[e^{(x-1) / \epsilon}-1\right] .
$$
Composite solution: Let $\eta=\sqrt{\epsilon} z=\frac{x-1}{\sqrt{\epsilon}}$. Study the limit $\epsilon \rightarrow 0^{+}$with $\eta$ fixed (and negative).
$$
y_{\circ}=e^{(1+\sqrt{\epsilon} \eta)^2 / 2} \rightarrow e^{1 / 2} .
$$
$$
y_{\mathrm{i}}=A\left[e^{\eta / \sqrt{\epsilon}}-1\right] \rightarrow-A .
$$
Therefore, $A=-\sqrt{\epsilon}$, Construct the uniform, composite solution by adding the two nonuniform solutions and subtracting their common limit, $\sqrt{\epsilon}$ :
$$
\begin{aligned}
y & \sim y_{\circ}+y_{\mathrm{i}}-\sqrt{e} \
& =e^{x^2 / 2}+\sqrt{e}\left[1-e^{\frac{(x-1)}{\epsilon}}\right]-\sqrt{e} \
& =e^{x^2 / 2}-\sqrt{e} e^{\frac{(x-1)}{\epsilon}} .
\end{aligned}
$$
学习资料推荐
针对 Advanced engineering mathematics 推荐三本教材：
✅Advanced mathematics for engineering and science
✅Solutions manual for Advanced engineering mathematics 8ed
✅ Instructor’s Manual For Advanced Engineering Mathematics 9th Edition
高级工程数学预测介绍：
- “工程数学 “肯定包括微积分，但也可能涉及许多其他数学领域：算术、代数、三角法、几何、会计、经济学。了解所有这些都是必要的，但哪个最重要取决于工程类型和工作描述。