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- \documentclass[11pt,a4paper]{article}
- \usepackage{amsmath}
- \setlength{\parindent}{0mm}
- \begin{document}
- Lista 7 zadanie 39 \\
- Nastepujace uklady rownan rozwiazac metoda operatorowa lub sprowadzajac je do ukladow rownan rozniczkowych rzedu pierwszego w postaci normalnej. \\
- \begin{math}
- \\
- \begin{cases}
- x''+3x-y=t \\
- y''-9x-5y= \cos t
- \end{cases}
- \\ \\
- \begin{cases}
- D^2x+3x-y=t \\
- D^2y-9x-5y= \cos t
- \end{cases}
- \\ \\
- \begin{cases}
- \left(D^2+3 \right) x-y=t \\
- -9x+ \left(D^2-5 \right)y= \cos t
- \end{cases}
- \\ \\
- W= \left|
- \begin{array}{cc}
- D^2+3 & -1 \\
- -9 & D^2-5
- \end{array}
- \right| = \left(D^2+3 \right) \left(D^2-5 \right)-9=D^4-2D^2-24 \\
- \\ W_{x} = \left|
- \begin{array}{cc}
- t & -1 \\
- \cos t & D^2-5
- \end{array}
- \right| = \left(D^2-5 \right)t+ \cos t=D^2t-5t+ \cos t= \cos t -5t \\
- \\ W_{y} = \left|
- \begin{array}{cc}
- D^2+3 & t \\
- -9 & \cos t
- \end{array}
- \right| = \left(D^2+3 \right) \cos t+9t=D^2 \left( \cos t \right)+3 \cos t+9=
- \\ = - \cos t+3 \cos t +9t=2 \cos t+9t \\
- \\ (1) \hspace{10} \left(D^4-2D^2-24 \right)x= \cos t-5t -> x^{IV}-2x''-24x= \cost-5t \\
- \\ (2) \hspace{10} \left(D^4-2D^2-24 \right)y=2 \cos t+9t -> y^{IV}-2y''-24y=2 \cos t+9t \\
- \\ (1): \\
- \\ x^{IV}-2x''-24= \cos t-5t \\
- \\ l^4-2l^2-24=0 \\
- \\ l_1= -2i, \ l_2=2i, \ l_3= -\sqrt{6}, \ l_4= \sqrt{6} \\
- \\ x=C_1 \cos 2t+C_2 \sin 2t+C_3e^{- \sqrt{6}t}+C_4e^{\sqrt{6}t} \\
- \\ x^{IV}-2x''-24x= \cos t\\
- \\ A \cos t+2A \cos t-24A \cos t= \cos t\\
- \\ A= - \frac{1}{21} \\
- \\ x^{IV}-2x''-24x= -5t \\
- \\ 24 \left(Ct+D \right)= -5t \\
- \\ 24Ct+24D= -5t-> D=0, \ C= - \frac{5}{24} \\
- \\ x=C_1 \cos 2t+C_2 \sin 2t+C_3e^{- \sqrt{6}t}+C_4e^{\sqrt{6}t} + \frac{1}{21} \cos t- \frac{5}{24}t \\
- \\ (2): \\
- \\ y^{IV}-2y''-24y=2 \cos t+9t \\
- \\ A \cos t+ 2A \cos t-24A \cos t=2 \cos t \\
- \\ A= - \frac{2}{21} \\
- \\ y^{IV}-2y''-24y=9t \\
- \\ -24 \left(Ct+D \right)=9t-24Ct-24D=9t->D=0, \ C= - \frac{3}{8} \\
- \\ y=C_1 \cos 2t+C_2 \sin 2t+C_3e^{- \sqrt{6}t}+C_4e^{\sqrt{6}t}- \frac{2}{21} \cos t- \frac{3}{8}t
- \end{math}
- \newpage
- Lista 8 zadanie 54 \\
- Zbadac stabilnosc rozwiazania zerowego dla nastepujacych ukladow rownan. \\
- \begin{math}
- \\ X'= \left(
- \begin{array}{cc}
- 3 & 4 \\
- -4 & 3
- \end{array}
- \right)X \\
- \\ \left|
- \begin{array}{cc}
- 3-l & -4 \\
- 4 & 3-l
- \end{array}
- \right|=0 \\
- \\ \left(3-l \right)^2+16=0 \\
- \\ 9-6l+l^2+16=0 \\
- \\ l^2-6l+25=0 \\
- \\ l_1=3-4i, \ l_2=3+4i \\
- \\ rel_1=rel_2=3>0 \\
- \end{math}
- \\ Czesc rzeczywista pierwiastkow rownania charakterystycznego jest dodatnia, wiec punkt rownowagi $(0,0)$ jest niestabilny.
- \newpage
- Lista 9 zadanie 54 \\
- Wyznaczyc rozwiazanie ogolne rownania \\
- $a_2(t)y''+a_1(t)y'+a_0(t)y=0$ \\
- jezeli $y_1$ jest rozwiazaniem tego rownania. \\
- \begin{math}
- \\ \left(1+t \right)y''+ty'-y=0, \ y_1(t)=t \\
- \\ a_2(t)y''+a_1(t)y'+a_0(t)y=0 \ / \ :a_2(t), \ a_2(t) \neq 0 \\
- \\ y''+ \frac{a_1(t)}{a_2(t)}y'+ \frac{a_0(t)}{a_2(t)}y=0 \\
- \\ p(t)= \frac{a_1(t)}{a_2(t)}, \ q(t)= \frac{a_0(t)}{a_2(t)} \\
- \\ y''+p(t)y'+q(t)=0 \\
- y=y_1 \int udt \\
- \\ y=y_1 \int udt, \ y'=y_1' \int udt+y_1u, \ y''=y_1'' \int udt+2y_1'+y_1u' \\
- \\ y_1u'+[2y_1'+p(x)y_1]u=0 \\
- \\ u'+[\frac{2y_1'}{y_1}+p(t)]u=0 \\
- \\ u= \frac{C}{y_1^2}e^{- \int p(t)dt}, \ C=1 \\
- \\ y_2=y_1 \int \frac{e^{- \int p(t)dt}}{y_1^2}dt \\
- \\ y=y_1 \left[C_1+C_2 \int \frac{e^{- \int p(t)dt}}{y_1^2}dt \right] \\
- \\ \left(1+t \right)y''+ty'-y=0, \ y_1(t)=t \\
- y''+ \frac{t}{1+t}y'- \frac{1}{1+t}y=0, \ t \neq -1 \\
- \\ p(t)= \frac{t}{1+t}, \ q(t)= - \frac{1}{1+t} \\
- \\ e^{- \int p(t)dt}=e^{- \int \frac{t}{1+t}dt}=e^{-t}(t+1) \\
- \\ \int \frac{e^{-t}(t+1)}{t^2}dt= \int e^{-t} \left( \frac{1}{t}+ \frac{1}{t^2} \right) \\
- \\ y=t \left[C_1+C_2 \int e^{-t} \left( \frac{1}{t}+ \frac{1}{t^2} \right)dt \right] \\
- \end{math}
- \newpage
- Lista 10 zadanie 31 \\
- Wyznaczyc rozwiazanie ogolne rownania niejednorodnego o stalych wspolczynnikach. \\
- \begin{math}
- \\ y''-2y'+2y=e^t \operatorname{tg} t \\
- \\ l^2-2l+2=0 \\
- \\ l_1= \frac{2-2i}{2}=1-i, \ l_2=1+i \\
- \\u_1=e^t cos t, \ u_2=e^t \sin t \\
- \\ u_1'=e^t( \cos t- \sin t), \ u_2'=e^t( \cos t + \sin t) \\
- \\ \left|
- \begin{array}{cc}
- u_1 & u_2 \\
- u_1' & u_2'
- \end{array}
- \right| \left|
- \begin{array}{cc}
- C_1(t) \\
- C_2'(t)
- \end{array}
- \right| = \left|
- \begin{array}{cc}
- 0 \\
- e^t \operatorname{tg} \end{array}
- \right| \\ \\ \left|
- \begin{array}{cc}
- e^t \cos t & e^t \sin t \\
- e^t( \cos t- \sin t) & e^t( \cos t+ \sin t)
- \end{array}
- \right| \left| \begin{array}{cc}
- C_1(t) \\
- C_2'(t) \operatorname{tg} \end{array}
- \right| \\
- \\ W=e^{2t} \cos t( \cos t+\sin t)-e^{2t} \sin t( \cos t- \sin t)= \\
- \\ =e^{2t}[\cos^2t+ \cos t \sin t- \sin t \cos t + \sin^2t]= e^{2t}(\cos^2t+ \sin^2t)=e^{2t}>0 \\
- \\ W_C_1'= \left|
- \begin{array}{cc}
- 0 \cos t & e^t \sin t \\
- e^t \operatornane{tg} t & e^t \operatorname{tg} t
- \end{array}
- \right| = -e^{2t} \sin t \operatorname{tg} t= -e^{2t} \frac{ \sin^2t}{ \cos t} \\
- W_C_2'= \left|
- \begin{array}{cc}
- e^t \cos t & 0 \\
- e^t(\cos t- \sin t) & e^t \operatorname{tg} t
- \end{array}
- \right| =e^{2t} \cos t \operatorname{tg} t= e^{2t} \sin t \\
- \\ C_1'(t)= \frac{W_C_1'}{W}= \frac{-e^{2t} \sin^2t}{e^{2t} \cos t}= - \frac {\sin^2t}{\cos t} \\
- \\ C_2'(t)= \frac{e^{2t} \sint}{e^{2t}}= \sin t \\
- \\ C_2(t)= \int \sin tdt= - \cos t+D_2 \\
- \\ C_1(t)=- \int \frac{\sin^2t}{\cos t}dt= \int \frac{ \cos^2t-1}{\cos t}dt= \sin t - \ln \left| \operatorname{tg} \frac{x}{2}+ \frac{\pi}{4} \right| + D_1 \\
- \\y(t)=C_1(t)U_1(t)+C_2(t)U_2(t)=\left(\sin t- \ln \left| \operatorname{tg} \frac{x}{2}+ \frac{\pi}{4} \right| +D_1 \right)e^t \cos t+ \left(- \cos t +D_2 \right)e^t \sin t \\
- \end{math}
- \newpage
- Lista 11 zadanie 44 \\
- Rozwiazac rownanie lub uklad rownan przy uzyciu transformaty Laplace'a \\
- \begin{math}
- y''+y=te^{-t}+t, \ y(0)=1, \ y'(0)=0 \\
- \\ \mathcal{L}[y'']+\mathcal{L}[y]=\mathcal{L}[te^{-t}]+\mathcal{L}[t] \\
- \\ z^2\mathcal{L}[y]-z\mathcal{L}[y(0)]-y'(0)+ \mathcal{L}[y]= \frac{1}{(z+1)^2}+ \frac{1}{z^2} \\
- \\ \left( z^2+1 \right) \mathcal{L}[y]=zy(0)+y'(0)+ \frac{1}{(z+1)^2}+ \frac{1}{z^2}\\
- \\ \mathcal{L}[y]= \frac{z}{z^2+1}+ \frac{1}{(z^2+1)(z+1)^2}+ \frac{1}{(z^2+1)z^2} \\
- \\ \frac{1}{(z^2+1)(z+1)^2} = \frac{Az+B}{z^2+1}+ \frac{C}{z+1}+ \frac{D}{z+1} \\
- \\ \begin{cases}
- A+C=0 \\
- 2A+B+C+D=0 \\
- A+2B+C=0 \\
- B+C+D=1
- \end{cases} \ \begin{cases}
- A= -\frac{1}{2} \\
- B=0 \\
- C= \frac{1}{2} \\
- D= \frac{1}{2}
- \end{cases} \\
- \\ \frac{1}{(z^2+1)(z+1)^2} = \frac{-\frac{1}{2}z}{z^2+1}+\frac{\frac{1}{2}}{z+1}+\frac{\frac{1}{2}}{(z+1)^2} \\
- \\ \frac{1}{(z^2+1)z^2}= \frac{Az+B}{z^2+1}+ \frac{C}{z}+ \frac{D}{z^2} \\
- \\ \begin{cases}
- A= -C \\
- B=-D \\
- C=0 \\
- D=1
- \end{cases} \ \begin{cases}
- A=0 \\
- B=-1 \\
- C=0 \\
- D=1
- \end{cases} \\
- \\ \frac{1}{(z^2+1)(z+1)^2}= - \frac{1}{z^2+1}+ \frac{1}{z^2} \\
- \\ y(t)= \cos t - \frac{1}{2} \cos t + \frac{1}{2}e^{-t}+ \frac{1}{2}te^{-t}- \sin t +t
- \end{math}
- \end{document}
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