\documentclass[12pt]{article}
\usepackage[a4paper,left=2.5cm,right=2.5cm,top=1.5cm,bottom=1.5cm]{geometry}
\usepackage[T1]{fontenc}
\usepackage[latin2]{inputenc}
\usepackage[magyar]{babel}
\usepackage{times}

\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{array, multicol}
\usepackage{subfigure}

\begin{document}

\begin{center}
  \Large{\textbf{Bsc/A3/Gé/2.HF/2011 őszi félév (V. Nagy)}}\\
  \Large{Leadási határidő: a 2. ZH hetének gyakorlatán (12. hét) a gyak.~vezér részére (olvashatóan, kézzel írva!)}
\end{center}

\begin{enumerate}
 
 \item Milyen $\beta$-ra lesz egzakt a diff. egyenlet? Ekkor oldja meg!
 \[
   \left(\beta x y^2 e^{x^2}+3x^2\right) \operatorname{d}x+\left(\beta y\cdot  e^{x^2}+\beta \sin y \cos y\right)\operatorname{d}y=0, \ y(2)=0
 \]
 
 \item Oldja meg!
	  \begin{enumerate}
	   \item $e^x y^2-2x y+(2y e^x-x^2) y'=0, \ y(0)=2$
	   \item $(y \cos^2 x-\sin x)\operatorname{d}y-y\cos x(y \sin x+1)\operatorname{d}x=0$
	   \item $y' e^{-x} \cos y=e^{-x}\sin y-\cos x$
	  \end{enumerate}
 
 \item Oldja meg!
	  \begin{enumerate}
	   \item $y'=\frac{x^2}{y^3}, \ y'=2^{x-3y}, \ y'=x^2 \cdot y^2$
	   \item $y'= \frac{(x^2-2x)e^{-x}}{(2y+1)\sin 2y}, \ y'= \frac{(2x+3)\cos 3x}{(2y^2+3y)\ln^2 y}, \ y'= \frac{x^5+3x^2+1}{(e^y\cdot \sin 2y)(x^2+1)}$
	  \end{enumerate}
 
 \item Oldja meg!
	  \begin{enumerate}
	   \item $(5x-7y)-(x+y)y'=0, \ y(1)=1$
	   \item $y'=\frac{y}{x}\sin \frac{y}{x}, \ y(\frac12)=\frac{\pi}{4}$
	   \item $y'=\frac{y}{x}\left(1+\frac{y-6x}{2y+6x}\right), \ y(1)=2$
	   \item $(x+y)^2y'=x^2-2xy +5y^2, \ y(1)=1$
	  \end{enumerate}
	  
 \item Oldja meg!
	  \begin{enumerate}
	   \item $y'=\sqrt{2x+3y-1}+4x+6y+3$
	   \item $y'=4+\cos(-5x+2y+3), \ y(1)=1$
	   \item $y'=e^{2y-6x+1}+3, \ y(0)=0$
	  \end{enumerate}

 \item Oldja meg!
      \begin{enumerate}
	    \item $y'+\frac{y}{x}=\sqrt{x}, \ y(1)=1$
	    \item $y' x-\frac{y}{x+1}=x$
		\item $y' \sin x+\cos x=1+y$
		\item $y' \cos x+y \sin x=1, \ y(0)=1$
		\item $y'+\frac{y}{x}+e^x =0, \ y(1)=0$
      \end{enumerate}
	  
 \item 
      \begin{enumerate}
        \item $y(y-1)y''+(y')^2=0$
		\item $(1+x^2)y''+(y')^2+1=0$
		\item $y\cdot y'' +(y')^2=1, \ y(0)=1, y'(0)=0$
	  \end{enumerate}

 \item 
      \begin{enumerate}
        \item $ 2y'' -y' -3y= 2\sin x+\frac{5}{e^x}$
        \item $ y'' -4y' +4y= e^{3x} + 2 e^{2x}, \ y(0)=1, y'(0)=1$
		\item $ y'' -y' -6y= \sin x+2 \cos 3x$
		\item $ y'' -2y' +2y= 4 e^{2x}+2$
	  \end{enumerate}
	  
 \item 
      \begin{enumerate}
	    \item $\dot{\underline{x}} =\left(\begin{array}{cc}
		         6 & -1 \\ 4 & 2
		      \end{array}\right) \underline{x}+ e^{4t}
			   \left(\begin{array}{c}
			   2 \\ 1
			   \end{array}\right)$
		\item $\dot{x}=x+2y+1$ \\ $\dot{y}=x+y+t$
		\item $\dot{x}=-7x+y+t, \ x(0)=3$ \\ $\dot{y}=-2x-5y+2, \ y(0)=-5$
	  \end{enumerate}
 
 \item $\mathcal{L}(f(t))(s)=?$
      \begin{enumerate}
	    \item $t \sinh 3t$
		\item $ t e^t +\cosh t$
		\item $ t^2 \cos t $
		\item $ \int_0^t \tau^2 e^{-\tau} \operatorname{d}\tau$
      \end{enumerate}
	   \begin{figure}[h!]
         \centering
		 \setcounter{subfigure}{4}
         \subfigure[ ]{\includegraphics[height=0.25\textwidth]{kép3} \label{kep3}} \quad
         \subfigure[ ]{\includegraphics[height=0.17\textwidth]{kép4} \label{kep4}}
		 \setcounter{figure}{9}
		 \caption{ }
       \end{figure}

 \item $\mathcal{L}(f)(s)=\ln \frac{s^2+1}{(s-1)^2}, \ f(t)=?$
 
 \item $f(t)=?, \ \mathcal{L}(f(t))(s)=?$
      \begin{enumerate}
	    \item $1 \ast \cos t$
		\item $ e^t \ast t$
		\item $ \sin^2 t \ast t $
      \end{enumerate}
	  
 \item 
      \begin{enumerate}
	    \item $\ddot{x}(t)-3 \dot{x}(t)+2x(t)=f(t), \ x(0)=\dot{x}(0)=1$
	     \begin{figure}[h!]
           \centering
		   \includegraphics[width=0.4\textwidth]{kép5}
		   \setcounter{figure}{12}
		   \caption{ }
         \end{figure}
		\item $ x^{(4)}(t)+2\ddot{x}(t)+x(t)=\cos t, \ x(0)=\dot{x}(0)=\ddot{x}(0)=\dddot{x}(0)=0$
		\item $  t \ddot{x}(t)+(1-t)\dot{x}(t)+x(t)=0, \ x(0)=1, \dot{x}(0)=-1$
      \end{enumerate}
	  
 \item
      \begin{enumerate}
	    \item $1-\cos t=\int_0^t y(\tau) \sinh(t-\tau) \operatorname{d}\tau, \ y(t)=?$
		\item $ \dot{y}(t)-2 \int_0^t y(\tau) e^{t- \tau} \operatorname{d}\tau, \ y(0)=1, y(t)=?$
		\item $ y(t)=\sin t +\int_0^t y(\tau) e^{t- \tau} \operatorname{d}\tau, \ y(t)=?$
      \end{enumerate}
\end{enumerate}
\end{document}