zestaw10

\documentclass{article}
 \usepackage{longtable}
\usepackage[intlimits]{amsmath}
\author{Piotr Wegrzyn}
\title{Zestaw nr.10.\\
Zadanie nr.39}
\date{}
\begin{document}
\maketitle
\begin{center}

\[F(u)= \int_{0}^{1}((1+x)e^x u + \frac{1}{2}e^x u'^2)dx \quad u(0)=1, \,u(1)=\frac{3}{2}\]
\[F(u,u',x)=((1+x)e^x u + \frac{1}{2}e^x u'^2)=e^xu +xe^xu+\frac{1}{2}e^xu'^2\]

\[\frac{\partial F}{\partial u} = e^x + xe^x\]
\[\frac{\partial F}{\partial u'} = e^xu'\]
\[\frac{\partial F}{\partial x} = e^xu+e^xu+xe^xu+\frac{1}{2}e^xu'^2\]

\[\frac{\partial' F}{\partial u' \partial x} = e^x \]
\[\frac{\partial' F}{\partial u' \partial u} = 0\]
\[\frac{\partial' F}{\partial u' \partial u'} = e^x\]

\[e^x+xe^x -e^x -e^xu''=0\]
\[xe^x -e^xu''=0\]
\[e^xu''=xe^x\]
\[u''=x\]
\[u'=\frac{1}{2}x^2 +c_1\]
\[u=\frac{1}{6}x^3+c_1x+c_2\]

Wstawiamy warunki poczatkowe:

\[U(x)= \frac{1}{6}x^3 + c_1x+c_2  \quad u(0)=1,u(1)=\frac{3}{2}\]

\[1=0+0+c_2 => c_2 =1\]
\[\frac{3}{2}=\frac{1}{6}+c_1 +1\]
\[\frac{9}{6}=\frac{7}{6}+c_1 => c_1 =\frac{1}{3}\]
\[u(x)=\frac{1}{6}x^3 + \frac{1}{3}x +1\]


\end{center}
\end{document}