A Basic Course in Partial Differential Equations, Qing Han, 2011 [下载说明:点击链接,等待5秒, 点击右上角的跳过广告后调至下载页面, 点击电信下载即可] http://adf.ly/dNpi7           http://adf.ly/dNplx 习题解答请见:http://bbs.sciencenet.cn/thread-1337491-1-1.html 或者在线观看: http://www.cnblogs.com/zh…
1.Introduction 2.First-order Differential Equations Exercise2.1. Find solutons of the following intial-value problems in $\bbR^2$: (1)$2u_y-u_x+xu=0$ with $u(x,0)=2xe^{x^2/2}$; (2)$u_y+(1+x^2)u_x-u=0$ with $u(x,0)=\arctan x$. Solution: (1)Since $(-1,…
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title: [线性代数]6-3:微分方程的应用(Applications to Differential Equations) categories: Mathematic Linear Algebra keywords: Eigenvalues Eigenvectors Differential Equations toc: true date: 2017-11-22 15:09:04 Abstract: 本文主要介绍线性代数在微分方程中的应用 Keywords: Eigenvalues,E…
NIPS2018最佳论文解读:Neural Ordinary Differential Equations 雷锋网2019-01-10 23:32     雷锋网 AI 科技评论按,不久前,NeurIPS 2018 在加拿大蒙特利尔召开,在这次著名会议上获得最佳论文奖之一的论文是<Neural Ordinary Differential Equations>,论文地址:https://arxiv.org/abs/1806.07366.Branislav Holländer 在 towards…
此文是对 [Introduction to Differential Equations,Michael E.Taylor] 第3页的一个注记.在该页中,作者给了微分方程$$\frac{dx}{dt}=x,x(0)=1.$$一个幂级数的解法.设$$x(t)=a_0t^0+a_1t^1+a_2t^2+a_3t^3+a_4t^4+\cdots$$注意,作者这样设之后,其实已经假定存在一个实解析函数满足该微分方程,剩下的就是解出该实解析函数.为此,作者进行逐项微分.$$x'(t)=a_1+2a_2t+…
Instead of specifying a discrete sequence of hidden layers, we parameterize the derivative of the hidden state using a neural network. Before: a discrete sequence of hidden layers. After: the derivative of the hidden state. Traditional methods: resid…
(Newton 1671, “Problema II, Solutio particulare”). Solve the total differential equation $$3x^2-2ax+ay-3y^2y'+axy'=0.$$Solve:We have $$y'(3y^2-ax)=3x^2-2ax+ay.$$So$$dy(3y^2-ax)=(3x^2-2ax+ay)dx.$$So $$y^{3}-axy=x^3-ax^2+axy+C$$where $c$ is a constant.…
学习微分方程中,一个很常见的疑惑就是,我们所熟悉的非齐次微分方程的通解是对应齐次方程的通解加特解,但是更为重要的是,我们需要知道这句话是怎么得来的. 我们探讨一个未知问题的一般思路是将其不断的与已知已解决的问题进行靠拢,关于微分方程,最简单的不过是可分离变量的微分方程,那么我们就尝试将(1)方程与之靠拢. 这样我们就很容易理解文章已开始提出的疑问:一阶非齐次线性方程的通解是对应齐次方程的通解和非齐次方程的特解这句话了.同时也能够很彻底理解一阶非齐次线性方程的通解公式的形式了. 而上面这化腐朽为神…
Solve equation $y'=1-3x+y+x^2+xy$ with another initial value $y(0)=1$. Solve: We solve this by using Newton's extraordinary method.We assume that the solution is analytic,which means it can be expanded in Taylor series.$y(0)=1$ means that $$ y'(0)=2…