{ "cells": [ { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "%matplotlib inline\n", "from pandas import DataFrame\n", "from IPython.display import display,Math,clear_output\n", "from numpy import array,linspace,zeros\n", "from sympy import latex\n", "import matplotlib.pyplot as plt\n", "from IPython.display import HTML" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ "\n", "
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''') " ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "### задача 1(118 а)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "$xy\\ d x- (1+x^2) d y=0\\\\\n", "x y\\ d x = (1+x^2) d y \\\\\n", "\\frac{x\\ dx}{1+x^2}=\\frac{dy}{y}\\\\\n", "\\int \\frac{x\\ dx}{1+x^2}=\\int \\frac{dy}{y}\\\\ \n", "производя\\ интегрирование \\\\\n", "\\ \\ \\int \\frac{x\\ dx}{1+x^2}=\\frac{1}{2} \\int \\frac{dx^2}{1+x^2}=\\frac{1}{2} ln(1+x^2)\\\\\n", "\\ \\ \\int \\frac{dy}{y} =ln(y)+C \\\\ \n", "получим \\\\\n", "ln(y)+C=\\frac{1}{2} ln(1+x^2) \\\\ \n", " или\\ собирая\\ логарифмы\\ (где\\ C=-ln(C_{1})): \\\\\n", " ln(y)=ln(C_{1} \\sqrt(1+x^2))\\\\\n", " y=C_{1} \\sqrt(1+x^2)\n", " $" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "### Задача 2 (118 б)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "$y'= \\frac{2 x}{y}+\\frac{y}{2},\\ y(2)=1,\\ h=0.2\\\\\n", "по\\ формуле:\\\\\n", "y_{i}=y_{i-1}+(x_{i}-x_{i-1})\\ f(x_{i-1},y_{i-1})\\ начиная\\ с\\ x_{0}=2, y_{0}=1\\ получим \\\\ \n", "$" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ "
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}, "execution_count": 3, "metadata": { }, "output_type": "execute_result" } ], "source": [ "x=linspace(2,3,6)\n", "y=zeros((6))\n", "f=lambda x,y: (2*x/y+y/2.)\n", "index=[\"i=0\"]\n", "columns=[]\n", "y[0]=1\n", "for i in range(len(y))[1:]: \n", " y[i]=y[i-1]+(x[i]-x[i-1])*(f(x[i-1],y[i-1]))\n", " index.append(\"i=%s\"%(i))\n", "#print(y)\n", "d=DataFrame(array([x,y]),columns=index,index=[r'$'+latex(\"x_{i}\")+'$',r'$'+latex(\"y_{i}\")+'$'])#\n", "display(d)\n", "plt.plot(x,y,'o')" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "### Задача 3( 118 в)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "$(x^2+y-y e^x)\\ d x +(x+2 y-e^x)\\ d y=0\\ (1)\\\\\n", "т.к.\\ (x^2+y-y e^x)_{y}=1-e^x=(x+2 y-e^x)_{x},то\\ выражение\\ (1)\\ является \\\\ полным\\ дифференциалом\\ некоторой\\ функции\\ F(x,y)\\ т.е:\\\\\n", "d F(x,y)=F_{x}(x,y) d x + F_{y}(x,y) d y \\\\\n", "где\\\\\n", "\\ \\ F_{x}(x,y)=(x^2+y-y e^x)\\\\\n", "\\ \\ F_{y}(x,y)=(x+2 y-e^x)\\\\\n", "(т.е.\\ выполняется\\ F_{x y}(x,y)=F_{y x}(x,y)\\ необходимое\\ усолвие\\ для\\\\ полного\\ дифференциала ) \\\\\n", "а\\ т.к.\\ d F(x,y)=0\\ то\\\\\n", "F(x,y)=const\\ (2)\\\\\n", "-неявное\\ решение\\ уравнения\\ (1)\\\\\n", "для\\ нахождения\\ F(x,y)\\ нужно\\ интегрировать\\ одну\\ из\\ частных\\ производных\\\\ по\\ соответсвующему\\ аргуметну.\\ Например\\ по\\ x\\ получим:\\\\\n", "F(x,y)=\\int F_{x}(x,y) d x = \\int (x^2+y-y e^x) d x=\n", "\\frac{x^3}{3}+x y-y e^x+C(y) \\ \\ \\ (3)\\\\\n", "здесь\\ С(у)-постоянная\\ интегрирования\\ (не\\ зависит\\ от\\ x\\ но\\ зависит\\ от\\ y).\\\\\n", "Для\\ нахождения\\ C(y)\\ надо\\ продифференцировать\\ (3)\\ по\\ y\\ и\\ прировнять\\ к\\\\\n", "F_{y}(x,y)=(x+2 y-e^x)\\\\\n", "получим\\ дифф\\ уравнение:\\\\\n", "C'(y)+x-e^x=(x+2 y-e^x)\\\\\n", "C'(y)=2 y\\\\\n", "\\frac{d C(y)}{d y}=2 y\\\\\n", "d C(y)=2 y d y\\\\\n", "\\int d C(y)=\\int 2 y d y\\\\\n", "C(y)=y^2+C\\\\\n", "Подставляя\\ C(y)\\ в\\ (3)\\ найдем\\ F(x,y)\\ с\\ точностью\\ до\\ константы:\\\\\n", "F(x,y)=\\frac{x^3}{3}+x y-y e^x+y^2+C\\\\\n", "а\\ затем,\\ воспользовавшись\\ (2)\\ получим\\ решение\\ в\\ неявной\\ форме,\\ с\\ точностью\\ до\\ константы\\ const:\\\\\n", "\\frac{x^3}{3}+x y-y e^x+y^2=const\n", "$" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "### Задача 4 (127)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "В партии из 10 деталей 8 стандартных. Выбрали 2-е детали. Найти мат \n", "одидание ($E(X)$) и дисперсию ($D(X)$) числа стандартных деталий\n", "$Решение:\\\\\n", "E(X)=\\sum(x_{i}p(X=x_{i}))=0\\ p(0)+1\\ p(1)+2\\ p(2)\\\\\n", "p(1)\\ можно\\ найти\\ как\\ число\\ способов\\ выбрать\\ 1\\ елемент\\ из\\ 8\\\\\n", "умножить\\ на\\ число\\ способов\\ выбрать\\ 1\\ елемент\\ из 2\\\\\n", "деленное\\ на\\ число\\ способов\\ выбрать\\ 2\\ елемента\\ из\\ 10:\\\\\n", "p(1)=\\frac{\n", "\\begin{pmatrix}\n", "8\\\\\n", "2\n", "\\end{pmatrix}\n", "\\begin{pmatrix}\n", "2\\\\\n", "1\n", "\\end{pmatrix}\n", "}\n", "{\n", "\\begin{pmatrix}\n", "10\\\\\n", "2\n", "\\end{pmatrix}\n", "}=2 \\frac{8}{10}\\frac{2}{9}\\\\\n", "p(2)\\ можно\\ найти\\ как\\ число\\ способов\\ выбрать\\ 2\\ елемента\\ из\\ 8\\\\\n", "деленное\\ на\\ число\\ способов\\ выбрать\\ 2\\ елемента\\ из\\ 10:\\\\\n", "p(2)=\\frac{\n", "\\begin{pmatrix}\n", "8\\\\\n", "2\n", "\\end{pmatrix}}\n", "{\n", "\\begin{pmatrix}\n", "10\\\\\n", "2\n", "\\end{pmatrix}\n", "}=\\frac{\\frac{8!}{(8-2)! 2!}}{\\frac{10!}{(10-2)! 2!}}=\\frac{8}{10}*\\frac{7}{9}\\\\\n", "т.о.\\\\\n", "E(X)=1\\ p(1)+2\\ p(2)=1.6\\\\\n", "D(X)=E(X^2)-E^2(x)=(1^2 p(1)+2^2 p(2))-1,6\n", "^2=0.284$\n" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ ] } ], "metadata": { "kernelspec": { "display_name": "Python 3 (Ubuntu Linux)", "language": "python", "metadata": { "cocalc": { "description": "Python 3 programming language", "priority": 100, "url": "https://www.python.org/" } }, "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.6.8" }, "widgets": { "state": { }, "version": "1.1.1" } }, "nbformat": 4, "nbformat_minor": 0 }