\documentclass{article}
\usepackage[utf8]{inputenc}

\title{hahaha}
\author{olekgryga }
\date{January 2019}

\usepackage{natbib}
\usepackage{graphicx}

\begin{document}

Aleksander Gryga
$${\bf Zadanie 50}$$
Rozwiazac rownanie lub uklad rownan przy uzyciu transformaty Laplacea.
$$p^2y-py(0)-y^\prime(0)-5(py-y(0))+14y=\frac{9}{p}+\frac{1}{p-3}+\frac{4}{(p-1)^2}.$$
$$p^2y-10-5py-14y=\frac{9}{p}+\frac{1}{p-3}+\frac{4}{(p-1)^2}.$$
$$y(p^2-5p-14)=\frac{9}{p}+\frac{1}{p-3}+\frac{4}{(p-1)^2}+10.$$
$$\Delta=25+56=81, \sqrt{\Delta}=9, p1=-2, p2=7.$$
$$y=\frac{9}{(p+2)(p-7)p}+\frac{1}{(p+2)(p-7)(p-3)}+\frac{4}{(p+2)(p-7)(p-1)^2}+\frac{10}{(p+2)(p-7)}.$$
Znajdujemy oryginaly lewych stron po kolei.
$$(1)\alpha^-1=(\frac{9}{(p+2)(p-7)p})=\alpha^-1(\frac{9}{p^3-5p^2-14p})$$
$$dla \hspace{2} p=0\hspace{2} mamy \hspace{2}V^\prime(p)=3p^2-10p-14, V^\prime(0)=-14$$
$$dla\hspace{2} p=-2 \hspace{2}mamy\hspace{2} V^\prime(-2)=18$$
$$dla \hspace{2}p=7\hspace{2} mamy \hspace{2}V^\prime(7)=63$$
$$f1(t)=\frac{-9}{14}e^(0t)+\frac{9}{18}e^(-2t)+\frac{9}{63}e^(7t)=\frac{-9}{14}+\frac{1}{2}e^(-2t)+\frac{1}{7}e^(7t)$$
$$(2)\alpha^-1=(\frac{1}{(p+2)(p-7)(p-3)}), V(p)=(p^2-5p+14)(p-3)=p^3-8p^2+p+42, V^\prime(p)=3p^2-16p+1$$
$$dla\hspace{2} p=-2\hspace{2} mamy\hspace{2} V^\prime(-2)=12+32+1=45$$
$$dla\hspace{2} p=7\hspace{2} mamy\hspace{2} V^\prime(7)=157-112+1=46$$
$$dla \hspace{2}p=3\hspace{2} mamy\hspace{2} V^\prime(3)=27-48+1=-20$$
$$f2(t)=\frac{1}{45}e^(-2t)+\frac{1}{46}e^(7t)-\frac{1}{20}e^(3t)$$
$$(3)\alpha^-1(\frac{10}{(p+2)(p-7)}$$
$$V(p)=p^2-5p-14, V^\prime(p)=2p-5$$
$$dla\hspace{2} p=-2\hspace{2} mamy\hspace{2} V^\prime(-2)=-9$$
$$dla \hspace{2}p=7\hspace{2} mamy\hspace{2} V^\prime(7)=9$$
$$f3(t)=-\frac{1}{9}e^(-2t)+\frac{1}{9}e^(7t)$$
$$(4)\frac{1}{(p+2)(p-7)(p-1)^2}=\frac{A}{p+2}+\frac{B}{p-7}+\frac{C}{p-1}+{D}{(p-1)^2}=$$
$$\frac{A(p^3-9p^2+15p-7)+B(p^3-3p+2)+C(p^3-7p^2-9p+14)+D(p^2-5p+14)}{(p+2)(p-7)(p-1)^2}$$\end{align}
$$\left\{ \begin{array}{ll}
A+B+C=0\\
-9A-7C+D=0\\
15A-3B-9C-5D=0\\
-7A+2B+14C+14D=4
\end{array}\right.$$
$$A=\frac{2}{3}, B=-\frac{1}{3}, C=-\frac{1}{3}, D=\frac{11}{3}$$
$$f4(t)=\frac{2}{3}e^(2t)-\frac{1}{3}e^(7t)+\frac{1}{3}e^t+\frac{11}{3}te^t$$
$$H=f1(t)+f2(t)+f3(t)+f4(t)$$


$${\bf Zadanie 18}$$
$$t^2(T+1)^\prime^\prime-2y=0$$
$$y_{1}(t)=1+\frac{1}{t}$$
$$y{2}(t)=y_{1}(t)\int{\frac{e^{-\int a_{n}(t)dt}}{y^\prime_{1}(t)}$$
$$y^\prime^\prime+a_{1}y^\prime+a_{0}y=0$$
$$a_{1}=0$$
Szukamy: $$y(t)=u(t)y_{1}(t)$$
Liczymy:
$$y^\prime_{1}(t)=-t^{-2}$$
$$y^\prime^\prime_{1}(t)=2t^{-3}$$
$$y_{2}(t)=(1+\frac{1}{t}\int{\frac{e^0}{(1+\frac{1}{t})^2})$$
$$y_{2}(t)=(1+\frac{1}{t})(t-\frac{1}{t+1}-2ln(t+1))$$


$${\bf Zadanie 60}$$
Zbadać stabilność rozwiązania zerowego dla następujących układów równań
$$X^\prime=
\left( \begin{array}{ccc}
{6} & {1}  \\
{-2} & {-3} \\
\end{array} \right)X
$$
$$det
\left( \begin{array}{ccc}
{-6-1} & {1}  \\
{-2} & {-3-1}  \\
\end{array} \right)=(6+1)(3+1)+2=l^2+9l+20
$$
$$l^2+9l+20=0 $$
$$\Delta=81-80=1, \sqrt{\Delta=1}$$
$$l1=\frec{-9-1}{2}=-5, l2=\frec{-9+1}{2}=-4$$
l1,l2 sa ujemne, wiec mamy wezel stabilny asymptotycznie

$${\bf Zadanie 37}$$
Wyznaczyć rozwiązanie ogólne równania niejednorodnego o stałych współczynnikach
$$y^\prime^\prime-8y^\prime+17y=e^(4t)(t^2-3tsint)$$
$$l^2-8l+17=0$$
$$\Delta=64-68=-4, \sqrt{\Delta}=2i$$
$$l1=\frac{8-2i}{2}=4-i, l2=4+i$$
$$l1=4-i, l2=4+i$$
$$v1=e^(4t)\cos{t},v2=e^(4t)(\cos{t}+4\sin{t})$$
$$y=C_{1}V_{1}+C_{2}V_{2}$$
Zastosujemy metode uzmienniania stalych C_{1}, C_{2}\\
$$
\left[
        \begin{array}{cc}
         v_{1} & v_{2}\\
         v_{1}^\prime & v_{2}^\prime
         \end{array}
      \right]
      \qquad
 \left[
       \begin{array}{cc}
          C_{1}^\prime\\
          C_{2}^\prime
       \end{array}
     \right]=
\left[
       \begin{array}{cc}
          0\\
         f(t)
       \end{array}
     \right]$$\\
$$\\$$
$$\left| \begin{array}{ccc}
v_{1} & v_{2} \\
v_{1}^\prime & v_{2}^\prime  \\
\end{array} \right|=
\left| \begin{array}{ccc}
e^{4t}\cos{t} & e^{4t}\sin{t}\\
e^{4t}(4\cos{t}-\sin{t}) & e^{4t}(\cos{t}+4\sin{t}) \\
\end{array} \right|$$
$$=e^{8t}(\cos^2{4t}+4\cos{t}\sin{t}-4\cos{t}\sin{t}+\sin^2{t})=e^{8t}>0$$\\
$$\left[
        \begin{array}{cc}
         v_{1} & 0\\
         v_{1}^\prime & f(t)
         \end{array}
      \right]
      \qquad=
      \left| \begin{array}{ccc}
e^{4t}\cos{t} & 0\\
e^{4t}(4\cos{t}-\sin{t}) & e^{4t}(t^2-3t\sin{t}) \\
\end{array} \right|=e^{8t}(t^2\cos{t}-3t\sin{t}\cos{t})$$
$$C_{1}^\prime=\frac{WC_{1}}{W}=-t^2\sin{t}+3t\sin^2{t}=-t^2\sin{t}+\fraq{3}{2}t(1-\cos{2t})$$\\
(1)
$$C_{1}(t)=\int C^\prime_{1}(t)=\int -t^2\sin{t}+\frac{3}{2}t(1-\cos{2t})$$
Obliczamy \hspace{2}najpierw
$$I_{1}=\int t^2\sin{t}dt=
\left| \begin{array}{ccc}
u=t^2 & dv=\sin{t}dt \\
du=2tdt & v=-\cos{t}  \\
\end{array} \right|
$$
$$=-t^2\cos{t}+2\int t\cos{t}dt=
\left| \begin{array}{ccc}
u=t & dv=\cos{t}dt \\
du=dt & v=-\sin{t}  \\
\end{array} \right|
=-t^2\cos{t}+2(t\sin{t}-\int \sin{t}dt)=$$
$$=-t^2\cos{t}+2t\sin{t}+2\cos{t}+D_{1}$$
Obliczamy\hspace{2}$$I_{2}=\int \frac{3}{2}t(1-\cos{2t})dt=\int (\frac{3}{2}t-\frac{3}{2}t\cos{2t})dt=$$
$$\frac{3}{4}t^2-\frac{3}{2}\int t\cos{2t}dt=
\left| \begin{array}{ccc}
u=t & dv=\cos{2t}dt \\
du=dt & v=\frac{1}{2}\sin{2t}  \\
\end{array} \right|
=$$
$$
=\frac{3}{4}t^2-\frac{3}{2}(\frac{1}{2}t\sin{2t}-\int \frac{1}{2}\sin{2t}dt)=$$
$$=\frac{3}{4}t^2-\frac{3}{4}t\sin{2t}-\frac{3}{8}\cos{2t}+D_{2}
$$
$$C_{1}(t)=I_{1}+I_{2}=-t^2\cos{t}+2t\sin{t}+2\cos{t}+\frac{3}{4}t^2-\frac{3}{4}t\sin{2t}-\frac{3}{8}\cos{2t}+D$$
$$D=D_{1}+D_{2}
$$(2)C_{2}(t)=\int C_{2}^\prime(t)=\int(t^2\cos{t}-\frac{3}{2}t\sin{2t})dt$$\\
Obliczamy J_{1}:
$$J_{1}=\int t^2\cos{t}dt=
\left| \begin{array}{ccc}
u=t^2 & dv=\cos{t}dt &\\
du=2tdt & v=-\sin{t} & \\
\end{array} \right|=-t^2\sin{t}+2\int t\sin{t}dt=
\left| \begin{array}{ccc}
u=t & dv=\sin{t}dt  \\
du=dt & v=-\cos{t} \\
\end{array} \right|=
-t^2\sin{t}=2(-t\cos{t}+\int \cos{t}dt)=
-t^2\sin{t}-2t\cos{t}-\sin{t}+E_{1}$$\\
Obliczamy J_{2}:
$$J_{2}=-\frac{3}{2}\int t\sin{2t}dt=
\left| \begin{array}{ccc}
u=t & dv=\sin{2t}dt \\
du=dt & v=-\frac{1}{2}\cos{2t} \\
\end{array} \right|= -\frac{3}{2}(-\frac{1}{2}t\cos{2t}+\frac{1}{2}\int \cos{2t}dt)= \frac{3}{4}t\cos{2t}+\frac{1}{4}\sin{2t}+E2$$
$$C_{2}(t)=J_{1}+J_{2}=-t^{2}\sin{t}-2t{t}\cos{t}-\sin{t}+\frac{3}{4}t\cos{2t}+\frac{1}{4}\sin{2t}+E$$\\
Wstawiamy\hspace{2} znalezione\hspace{2} wartosci \hspace{2}C_{1} i C_{2}\hspace{2} do\hspace{2} wzoru
$$y=C_{1}V_{1}+C_{2}V_{2}$$\\
i \hspace{2}otrzymujemy\hspace{2} rozwiazanie\hspace{2} ogolne \hspace{2}rownania \hspace{2}niejednorodnego.\\
$$\\
 $${\bf Zadanie 45}$$\end{align}
Następujące układy równań rozwiązać metodą operatorową lub sprowadzając je do układów równań różniczkowych rzędu pierwszego w postaci normalnej.
$$ \left\{ \begin{array}{ll}
X^\prime^\prime-2y^\prime^\prime+y^\prime+x-3y=0\\
-2x^\prime^\prime+4y^\prime^\prime-x^\prime-2x+5y=0}
\end{array} \right.$
$$\\
\left\{ \begin{array}{ll}
D^2x-2d^2y+Dy+x-3y=0\\
-2D^2x+4D^2y-Dx-2x+5y=0
\end{array} \right.
$$\\
$$
\left\{ \begin{array}{ll}
(D^2+1)x-(2D^2-D+3)y=0\\
-(2D^2+D+2)x+(4D^2+5)y=0
\end{array} \right.
$$\\
$$
\mathbf{W} =
\left| \begin{array}{ccc}
D^2+1 & -(2D^2-D+3)  \\
-(2D^2+D+2) & 4D^2+5
\end{array} \right|
=
(D^2+1)(4D^2+5)-(2D^2-D+3)(2D^2+D+2)=$$
$$=4D^4+(D^2+5-(4D^4+4D^2-D^2-2D+6D^2+3D+6)=4D^4+9D^2+5-(4D^4+9D^2+D+6)=$$
$$=5-D-6=-D-1=0$$
$$-(D+1)x=0 /*(-1)$$
$$(D+1)x=0$$
$$x^\prime+x=0$$
$$i+1=0$$
$$i=-1$$
$$x=C_{1}e^{-t}$$//
analogicznie \hspace{2}mamy
$$y=c_{2}e^{-t}$$
Podstawiamy \hspace{2}do\hspace{2} układu\hspace{2} równań\hspace{2} i skracamy\hspace{2} przez\hspace{2} e^{-t}\hspace{2} otrzymujemy:
\end{align}
$$\left\{ \begin{array}{ll}
2C_{1}-6C_{2}=0 /:2\\
-3C_{1}+9C_{2}=0 /:3\
\end{array} \right.$$\end{align}
$$\left\{ \begin{array}{ll}
C_{1}=3C_{3}\\
C_{2}=C\\
C_{1}=3C
\end{array} \right.
\end{array} \right$$\end{align}
$$\left\{ \begin{array}{ll}
x=3Ce^{-t}\\
y=Ce^{-t}
\end{array} \right.
\end{array} \right$$
\end{document}