Rozwiązać układ równań różniczkowych metodą współczyników nieoznaczonych i metodą przewidywań:
\(\displaystyle{ X' = \left[\begin{array}{ccc}1&-1&-2\\1&3&2\\1&-1&2\end{array}\right]X + \left[\begin{array}{c}t^{2}&t+1&2\end{array}\right]}\)
Rozwiązać metoda współczynników
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fluffiq
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Re: Rozwiązać metoda współczynników
\(\displaystyle{ X' = \left[\begin{array}{ccc}1&-1&-2\\1&3&2\\1&-1&2\end{array}\right]X + \left[\begin{array}{c}t^{2}&t+1&2\end{array}\right]}\)
\(\displaystyle{ \left[\begin{array}{c}t^{2}&t+1&2\end{array}\right] = X' - \left[\begin{array}{ccc}1&-1&-2\\1&3&2\\1&-1&2\end{array}\right]X}\)
\(\displaystyle{ X_{p}(t) = \left[\begin{array}{c}{a_{1}&a_{2}&a_{2}\end{array}\right]t^{2}+\left[\begin{array}{c}{b_{1}&b_{2}&b_{2}\end{array}\right]t + \left[\begin{array}{c}{c_{1}&c_{2}&c_{2}\end{array}\right]}\)
\(\displaystyle{ \left[\begin{array}{c}t^{2}&t+1&2\end{array}\right] = \left[\begin{array}{c}{2a_{1}&2a_{2}&2a_{2}\end{array}\right]t + \left[\begin{array}{c}{b_{1}&b_{2}&b_{2}\end{array}\right] - \left[\begin{array}{ccc}1&-1&-2\\1&3&2\\1&-1&2\end{array}\right] \cdot \Biggr( \left[\begin{array}{c}{a_{1}&a_{2}&a_{2}\end{array}\right]t^{2}+\left[\begin{array}{c}{b_{1}&b_{2}&b_{2}\end{array}\right]t + \left[\begin{array}{c}{c_{1}&c_{2}&c_{2}\end{array}\right] \Biggr)}\)
\(\displaystyle{ \left[\begin{array}{c}t^{2}&t+1&2\end{array}\right] = \left[\begin{array}{c}{2a_{1}&2a_{2}&2a_{2}\end{array}\right]t +\left[\begin{array}{c}{b_{1}&b_{2}&b_{2}\end{array}\right] - \Biggr( \left[\begin{array}{c}{a_{1}-a_{2}-2a_{3}&a_{1}+3a_{2}+2a_{3}&a_{1}-a_{2}+2a{3}\end{array}\right]t^{2} + \left[\begin{array}{c}{b_{1}-b_{2}-2b_{3}&b_{1}+3b_{2}+2b_{3}&b_{1}-b{2}+b{3}\end{array}\right]t^{2} + \left[\begin{array}{c}{c_{1}-c_{2}-2c_{3}&c_{1}+3c_{2}+2c_{3}&c_{1}-c{2}+c{3}\end{array}\right] \Biggr)}\)
\(\displaystyle{ \left[\begin{array}{c}1&0&0\end{array}\right]t^{2} + \left[\begin{array}{c}0&1&0\end{array}\right]t + \left[\begin{array}{c}0&1&2\end{array}\right]}\)
\(\displaystyle{ -\left[\begin{array}{c}{a_{1}-a_{2}-2a_{3}&a_{1}+3a_{2}+2a_{3}&a_{1}-a_{2}+2a_{3}\end{array}\right]t^{2} + \left[\begin{array}{c}{2a_{1}-b_{1}+b_{2}+2b_{3}&2a_{2}-b_{1}-3a_{2}-2_b_{3}&2a_{3}-b_{1}+b_{2}+2b_{3}\end{array}\right]t + \left[\begin{array}{c}{b_{1}-c_{1}+c_{2}+2c_{3}&b_{2}-c_{1}-3c_{2}-2c_{3}&b_{3}-c_{1}+c_{2}-2c_{3}\end{array}\right]}\)
\(\displaystyle{ \begin{cases} -a_{1}+a_{2}+2a_{3} = 1 \\ -a_{1}-3a_{2}-2a_{3}=0 \\ -a_{1}+a_{2}-2a_{3}=0 \end{cases}}\)
\(\displaystyle{ \begin{cases} a_{1} = -2a_{3} = -\frac{1}{2} \\ a_{2} = 0 \\ a_{3} = \frac{1}{4} \end{cases}}\)
czyli:
\(\displaystyle{ \left[\begin{array}{c}{a_{1}&a_{2}&a_{2}\end{array}\right] = \left[\begin{array}{c}-\frac{1}{2}&0&\frac{1}{4}\end{array}\right]}\)
\(\displaystyle{ \begin{cases}2a_{1}-b_{1}+b_{2}+2b_{3} = 0 \\ 2a_{2}-b_{1}-3a_{2}-2_b_{3} = 1\\ 2a_{3}-b_{1}+b_{2}+2b_{3} = 0\end{cases}}\)
\(\displaystyle{ 2a_{1}-b_{1}+b_{2}+2b_{3} - 2a_{2}+b_{1}+3a_{2}+2_b_{3} = 0}\)
\(\displaystyle{ 4b_{2} + 4b_{3} = 0}\)
\(\displaystyle{ b_{2} = -b_{3}}\)
\(\displaystyle{ 1 - 4b_{3} = -\frac{1}{2}}\)
\(\displaystyle{ -4b_{3} = -\frac{3}{2}}\)
\(\displaystyle{ b_{3} = \frac{3}{8}}\)
\(\displaystyle{ b_{1} = -\frac{5}{8}}\)
\(\displaystyle{ b_{2} = -\frac{3}{8}}\)
czyli:
\(\displaystyle{ \left[\begin{array}{c}{b_{1}&b_{2}&b_{2}\end{array}\right] = \left[\begin{array}{c}-\frac{5}{8}&-\frac{3}{8}&\frac{3}{8}\end{array}\right] \\}\)
czyli:
\(\displaystyle{ \left[\begin{array}{c}{c_{1}&c_{2}&c_{2}\end{array}\right] = \left[\begin{array}{c}-\frac{27}{16}&-\frac{1}{16}&\frac{-9}{16}\end{array}\right] \\}\)
Stad :
\(\displaystyle{ X_{p}(t) = \left[\begin{array}{c}-\frac{1}{2}&0&\frac{1}{4}\end{array}\right]t^{2}+\left[\begin{array}{c}-\frac{5}{8}&-\frac{3}{8}&\frac{3}{8}\end{array}\right]t + \left[\begin{array}{c}-\frac{27}{16}&-\frac{1}{16}&\frac{-9}{16}\end{array}\right]}\)
Obliczmy:
\(\displaystyle{ X_{h}(t) = ?}\)
\(\displaystyle{ X' = Ax}\)
\(\displaystyle{ \frac{1}{x} \mbox{d}x = A \mbox{d}t}\)
\(\displaystyle{ \ln{|x|} = At + C}\)
\(\displaystyle{ |x| = e^{c} \cdot e^{At}}\)
\(\displaystyle{ x = \pm e^{c} \cdot e^{At}}\)
czyli:
\(\displaystyle{ X_{h}(t) = Ce^{At} \\}\)
Ostatecznie: \(\displaystyle{ X = X_{h}(t) + X_{p}(t) \\}\)
czyli:
\(\displaystyle{ X = Ce^{At} + \Biggr(\left[\begin{array}{c}-\frac{1}{2}&0&\frac{1}{4}\end{array}\right]t^{2}+\left[\begin{array}{c}-\frac{5}{8}&-\frac{3}{8}&\frac{3}{8}\end{array}\right]t + \left[\begin{array}{c}-\frac{27}{16}&-\frac{1}{16}&\frac{-9}{16}\end{array}\right] \Biggr)}\)
Rozwiązałem to taka metoda, ktos jest w stanie powiedzieć jak to rozwiązać tą drugą?
\(\displaystyle{ \left[\begin{array}{c}t^{2}&t+1&2\end{array}\right] = X' - \left[\begin{array}{ccc}1&-1&-2\\1&3&2\\1&-1&2\end{array}\right]X}\)
\(\displaystyle{ X_{p}(t) = \left[\begin{array}{c}{a_{1}&a_{2}&a_{2}\end{array}\right]t^{2}+\left[\begin{array}{c}{b_{1}&b_{2}&b_{2}\end{array}\right]t + \left[\begin{array}{c}{c_{1}&c_{2}&c_{2}\end{array}\right]}\)
\(\displaystyle{ \left[\begin{array}{c}t^{2}&t+1&2\end{array}\right] = \left[\begin{array}{c}{2a_{1}&2a_{2}&2a_{2}\end{array}\right]t + \left[\begin{array}{c}{b_{1}&b_{2}&b_{2}\end{array}\right] - \left[\begin{array}{ccc}1&-1&-2\\1&3&2\\1&-1&2\end{array}\right] \cdot \Biggr( \left[\begin{array}{c}{a_{1}&a_{2}&a_{2}\end{array}\right]t^{2}+\left[\begin{array}{c}{b_{1}&b_{2}&b_{2}\end{array}\right]t + \left[\begin{array}{c}{c_{1}&c_{2}&c_{2}\end{array}\right] \Biggr)}\)
\(\displaystyle{ \left[\begin{array}{c}t^{2}&t+1&2\end{array}\right] = \left[\begin{array}{c}{2a_{1}&2a_{2}&2a_{2}\end{array}\right]t +\left[\begin{array}{c}{b_{1}&b_{2}&b_{2}\end{array}\right] - \Biggr( \left[\begin{array}{c}{a_{1}-a_{2}-2a_{3}&a_{1}+3a_{2}+2a_{3}&a_{1}-a_{2}+2a{3}\end{array}\right]t^{2} + \left[\begin{array}{c}{b_{1}-b_{2}-2b_{3}&b_{1}+3b_{2}+2b_{3}&b_{1}-b{2}+b{3}\end{array}\right]t^{2} + \left[\begin{array}{c}{c_{1}-c_{2}-2c_{3}&c_{1}+3c_{2}+2c_{3}&c_{1}-c{2}+c{3}\end{array}\right] \Biggr)}\)
\(\displaystyle{ \left[\begin{array}{c}1&0&0\end{array}\right]t^{2} + \left[\begin{array}{c}0&1&0\end{array}\right]t + \left[\begin{array}{c}0&1&2\end{array}\right]}\)
\(\displaystyle{ -\left[\begin{array}{c}{a_{1}-a_{2}-2a_{3}&a_{1}+3a_{2}+2a_{3}&a_{1}-a_{2}+2a_{3}\end{array}\right]t^{2} + \left[\begin{array}{c}{2a_{1}-b_{1}+b_{2}+2b_{3}&2a_{2}-b_{1}-3a_{2}-2_b_{3}&2a_{3}-b_{1}+b_{2}+2b_{3}\end{array}\right]t + \left[\begin{array}{c}{b_{1}-c_{1}+c_{2}+2c_{3}&b_{2}-c_{1}-3c_{2}-2c_{3}&b_{3}-c_{1}+c_{2}-2c_{3}\end{array}\right]}\)
\(\displaystyle{ \begin{cases} -a_{1}+a_{2}+2a_{3} = 1 \\ -a_{1}-3a_{2}-2a_{3}=0 \\ -a_{1}+a_{2}-2a_{3}=0 \end{cases}}\)
\(\displaystyle{ \begin{cases} a_{1} = -2a_{3} = -\frac{1}{2} \\ a_{2} = 0 \\ a_{3} = \frac{1}{4} \end{cases}}\)
czyli:
\(\displaystyle{ \left[\begin{array}{c}{a_{1}&a_{2}&a_{2}\end{array}\right] = \left[\begin{array}{c}-\frac{1}{2}&0&\frac{1}{4}\end{array}\right]}\)
\(\displaystyle{ \begin{cases}2a_{1}-b_{1}+b_{2}+2b_{3} = 0 \\ 2a_{2}-b_{1}-3a_{2}-2_b_{3} = 1\\ 2a_{3}-b_{1}+b_{2}+2b_{3} = 0\end{cases}}\)
\(\displaystyle{ 2a_{1}-b_{1}+b_{2}+2b_{3} - 2a_{2}+b_{1}+3a_{2}+2_b_{3} = 0}\)
\(\displaystyle{ 4b_{2} + 4b_{3} = 0}\)
\(\displaystyle{ b_{2} = -b_{3}}\)
\(\displaystyle{ 1 - 4b_{3} = -\frac{1}{2}}\)
\(\displaystyle{ -4b_{3} = -\frac{3}{2}}\)
\(\displaystyle{ b_{3} = \frac{3}{8}}\)
\(\displaystyle{ b_{1} = -\frac{5}{8}}\)
\(\displaystyle{ b_{2} = -\frac{3}{8}}\)
czyli:
\(\displaystyle{ \left[\begin{array}{c}{b_{1}&b_{2}&b_{2}\end{array}\right] = \left[\begin{array}{c}-\frac{5}{8}&-\frac{3}{8}&\frac{3}{8}\end{array}\right] \\}\)
czyli:
\(\displaystyle{ \left[\begin{array}{c}{c_{1}&c_{2}&c_{2}\end{array}\right] = \left[\begin{array}{c}-\frac{27}{16}&-\frac{1}{16}&\frac{-9}{16}\end{array}\right] \\}\)
Stad :
\(\displaystyle{ X_{p}(t) = \left[\begin{array}{c}-\frac{1}{2}&0&\frac{1}{4}\end{array}\right]t^{2}+\left[\begin{array}{c}-\frac{5}{8}&-\frac{3}{8}&\frac{3}{8}\end{array}\right]t + \left[\begin{array}{c}-\frac{27}{16}&-\frac{1}{16}&\frac{-9}{16}\end{array}\right]}\)
Obliczmy:
\(\displaystyle{ X_{h}(t) = ?}\)
\(\displaystyle{ X' = Ax}\)
\(\displaystyle{ \frac{1}{x} \mbox{d}x = A \mbox{d}t}\)
\(\displaystyle{ \ln{|x|} = At + C}\)
\(\displaystyle{ |x| = e^{c} \cdot e^{At}}\)
\(\displaystyle{ x = \pm e^{c} \cdot e^{At}}\)
czyli:
\(\displaystyle{ X_{h}(t) = Ce^{At} \\}\)
Ostatecznie: \(\displaystyle{ X = X_{h}(t) + X_{p}(t) \\}\)
czyli:
\(\displaystyle{ X = Ce^{At} + \Biggr(\left[\begin{array}{c}-\frac{1}{2}&0&\frac{1}{4}\end{array}\right]t^{2}+\left[\begin{array}{c}-\frac{5}{8}&-\frac{3}{8}&\frac{3}{8}\end{array}\right]t + \left[\begin{array}{c}-\frac{27}{16}&-\frac{1}{16}&\frac{-9}{16}\end{array}\right] \Biggr)}\)
Rozwiązałem to taka metoda, ktos jest w stanie powiedzieć jak to rozwiązać tą drugą?