Akademisyenler öncülüğünde matematik/fizik/bilgisayar bilimleri soru cevap platformu
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Ordinary, Linear, variable coefficient, 2nd order dif. eqn. Hoca ödev verdi. eigen value yi bulup çözebilirsiniz dedi. Fakat ne kadar uğraştıysam bir türlü sonuca ulaşamadım. yardımcı olursanız sevinirim. Teşekkür ederim. Kolay gelsin.

 

$\left[\begin{array}{c} x'_1(t)\\ x'_2(t)\\ \end{array} \right]=\left[ \begin{array}{cc}  \cos (2 t) & 1-\sin (2 t) \\  -1-\sin (2 t) & -\cos (2 t) \\ \end{array} \right]\left[ \begin{array}{c} x_1 \\ x_2\\ \end{array} \right]$

 

$X'(t)=A(t)X(t)$
Akademik Matematik kategorisinde (27 puan) tarafından 
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$\dot{X}(t)$ olarak ifade ettiğiniz şey $X'(t)$ mi?

evet hocam fiziksel gösterimi öyle

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2 beğenilme 0 beğenilmeme

Eger soru soyle olsaydi (soruda typo olabilecegini dusunuyorum) biraz daha kolay olabilirdi. Not: Butun adimlar Mathematica ile hesaplandi. Malesef Mathematica analitik olarak cozemedi ama bazi baslangic degerleri icin numerik cozumler karsilastirildi.

 

$X'(t)=A(t)X(t)$

 

$A(t)=\left[

\begin{array}{cc}

\cos (2 t) & 1-\sin (2 t) \\

1-\sin (2 t) & \cos (2 t) \\

\end{array}

\right]$

 

 

Bu durumda $\displaystyle\int_0^tA(s)ds=\displaystyle\int_0^t\left[

\begin{array}{cc}

\cos (2 s) & 1-\sin (2 s) \\

1-\sin (2 s) & \cos (2 s) \\

\end{array}

\right]ds=\left[

\begin{array}{cc}

\sin (t) \cos (t) & t-\sin ^2(t) \\

t-\sin ^2(t) & \sin (t) \cos (t) \\

\end{array}

\right]$ olmak uzere

 

 

$A(t)\cdot\displaystyle\int_0^tA(s)ds=\left[

\begin{array}{cc}

\cos (2 t) & 1-\sin (2 t) \\

1-\sin (2 t) & \cos (2 t) \\

\end{array}

\right]\left[

\begin{array}{cc}

\sin (t) \cos (t) & t-\sin ^2(t) \\

t-\sin ^2(t) & \sin (t) \cos (t) \\

\end{array}

\right]$

 

 

$=\left[

\begin{array}{cc}

(\cos (t)-\sin (t)) (-t \sin (t)+\sin (t)+t \cos (t)) & \left[t-\sin ^2(t)\right] \cos (2 t)-\sin (t) (\sin

(2 t)-1) \cos (t) \\

\left[t-\sin ^2(t)\right] \cos (2 t)-\sin (t) (\sin (2 t)-1) \cos (t) & (\cos (t)-\sin (t)) (-t \sin (t)+\sin

(t)+t \cos (t)) \\

\end{array}

\right]$

 

 

$\displaystyle\int_0^tA(s)ds\cdot A(t)=\left[

\begin{array}{cc}

\sin (t) \cos (t) & t-\sin ^2(t) \\

t-\sin ^2(t) & \sin (t) \cos (t) \\

\end{array}

\right]\left[

\begin{array}{cc}

\cos (2 t) & 1-\sin (2 t) \\

1-\sin (2 t) & \cos (2 t) \\

\end{array}

\right]$

 

 

 

$=\left[

\begin{array}{cc}

(\cos (t)-\sin (t)) (-t \sin (t)+\sin (t)+t \cos (t)) & \left[t-\sin ^2(t)\right] \cos (2 t)-\sin (t) (\sin

(2 t)-1) \cos (t) \\

\left[t-\sin ^2(t)\right] \cos (2 t)-\sin (t) (\sin (2 t)-1) \cos (t) & (\cos (t)-\sin (t)) (-t \sin (t)+\sin

(t)+t \cos (t)) \\

\end{array}

\right]$

 

 

 

$\implies A(t)\cdot\displaystyle\int_0^tA(s)ds=\displaystyle\int_0^tA(s)ds\cdot A(t)$ olur ve bu durumda temel matris su sekilde verilir.

 

 

$\Phi(t)=e^{\displaystyle\int_0^tA(s)ds}=e^{\left[

\begin{array}{cc}

\sin (t) \cos (t) & t-\sin ^2(t) \\

t-\sin ^2(t) & \sin (t) \cos (t) \\

\end{array}

\right]}$

 

Eger matrisimiz kosegen olsaydi soyle bir kolayligimiz olurdu

 

$e^{\left[

\begin{array}{cc}

d_1& 0\\

0 & d_2 \\

\end{array}

\right]}=\left[

\begin{array}{cc}

e^{d_1}& 0\\

0 & e^{d_2} \\

\end{array}

\right]$

 
 

Sansliyiz ki $B=\left[

\begin{array}{cc}

\sin (t) \cos (t) & t-\sin ^2(t) \\

t-\sin ^2(t) & \sin (t) \cos (t) \\

\end{array}

\right]$ matrisimiz simetrik ve butun simetrik matrisler kosegenlestirilebilir. O zaman hadi kosegenlestirelim.

 Once ozdegerleri bulalim.

 

$\left|B-\lambda I

\right|=\left|\begin{array}{cc}

\sin (t) \cos (t) -\lambda& t-\sin ^2(t) \\

t-\sin ^2(t) & \sin (t) \cos (t)-\lambda \\

\end{array}

\right|=0$

 

$\implies\lambda ^2-2 \lambda \sin (t) \cos (t)-\sin ^4(t)+2 t \sin ^2(t)+\sin ^2(t) \cos ^2(t)-t^2=0$

 

$\lambda _1=\frac{1}{2} (-2 t+\sin (2 t)-\cos (2 t)+1),\,\, \lambda_2= \frac{1}{2} (2

t+\sin (2 t)+\cos (2 t)-1)$

 

$\lambda _1=\frac{1}{2} (-2 t+\sin (2 t)-\cos (2 t)+1): \implies (B-\lambda_1I)V=0$

 

$\left[

\begin{array}{cc}

\frac{1}{2} (2 t+\cos (2 t)-1) & t-\sin ^2(t) \\

t-\sin ^2(t) & \frac{1}{2} (2 t+\cos (2 t)-1) \\

\end{array}

\right]\left[

\begin{array}{c}

v_1 \\

v_2\\

\end{array}

\right]=\left[

\begin{array}{c}

0 \\

0\\

\end{array}

\right]$

 

 

 

$\implies \left[

\begin{array}{c}

v_1 \\

v_2\\

\end{array}

\right]=\left[

\begin{array}{c}

-1 \\

1\\

\end{array}

\right]$

 

$\lambda_2= \frac{1}{2} (2 t+\sin (2 t)+\cos (2 t)-1): \implies (B-\lambda_2I)V=0$

 

$\left[

\begin{array}{cc}

\frac{1}{2} (-2 t-\cos (2 t)+1) & t-\sin ^2(t) \\

t-\sin ^2(t) & \frac{1}{2} (-2 t-\cos (2 t)+1) \\

\end{array}

\right]\left[

\begin{array}{c}

v_3 \\

v_4\\

\end{array}

\right]=\left[

\begin{array}{c}

0 \\

0\\

\end{array}

\right]$

 
 

$\implies \left[

\begin{array}{c}

v_3 \\

v_4\\

\end{array}

\right]=\left[

\begin{array}{c}

1 \\

1\\

\end{array}

\right]$

 

 

Simdi $P=\left[

\begin{array}{cc}

v_1&v_3 \\

v_2 &v_4 \\

\end{array}

\right]=\left[

\begin{array}{cc}

-1&1 \\

1 &1 \\

\end{array}

\right]$ olarak tanimlayalim ve

 

 

$P^{-1}=\dfrac12\left[

\begin{array}{cc}

1&1 \\

-1 &1 \\

\end{array}

\right]$ olur.

 
 

$D=P^{-1}\left[\displaystyle\int_0^tA(s)ds\right]P$

 

 

$=\dfrac12\left[

\begin{array}{cc}

1&1 \\

-1 &1 \\

\end{array}

\right]\left[

\begin{array}{cc}

\sin (t) \cos (t) & t-\sin ^2(t) \\

t-\sin ^2(t) & \sin (t) \cos (t) \\

\end{array}

\right]\left[

\begin{array}{cc}

-1&1 \\

1 &1 \\

\end{array}

\right]$

 

 

$\implies D=\left[

\begin{array}{cc}

t-\sin ^2(t)+\sin (t) \cos (t) & 0 \\

0 & -t+\sin ^2(t)+\sin (t) \cos (t) \\

\end{array}

\right]$

 
 

$PDP^{-1}=\left[\displaystyle\int_0^tA(s)ds\right]\implies Pe^DP^{-1}=e^{\displaystyle\int_0^tA(s)ds}$

 
 

$\Phi(t)=e^{\displaystyle\int_0^tA(s)ds}=Pe^DP^{-1}$

 

 

$=Pe^{\left[

\begin{array}{cc}

t-\sin ^2(t)+\sin (t) \cos (t) & 0 \\

0 & -t+\sin ^2(t)+\sin (t) \cos (t) \\

\end{array}

\right]}P^{-1}$

 

 

$=\left[

\begin{array}{cc}

-1&1 \\

1 &1 \\

\end{array}

\right]\left[

\begin{array}{cc}

e^{t-\sin ^2(t)+\sin (t) \cos (t)} & 0 \\

0 &e^{ -t+\sin ^2(t)+\sin (t) \cos (t)} \\

\end{array}

\right]\dfrac12\left[

\begin{array}{cc}

1&1 \\

-1 &1 \\

\end{array}

\right]$

 

 

$\Phi(t)=\left[

\begin{array}{cc}

\frac{1}{2} \left[e^{2 t}+e^{2 \sin ^2(t)}\right] e^{-t-\sin ^2(t)+\sin (t) \cos (t)} & \frac{1}{2}

\left[e^{2 t}-e^{2 \sin ^2(t)}\right] e^{-t-\sin ^2(t)+\sin (t) \cos (t)} \\

\frac{1}{2}\left[e^{2 t}-e^{2 \sin ^2(t)}\right] e^{-t-\sin ^2(t)+\sin (t) \cos (t)} & \frac{1}{2}

\left[e^{2 t}+e^{2 \sin ^2(t)}\right] e^{-t-\sin ^2(t)+\sin (t) \cos (t)} \\

\end{array}

\right]$

 

 

$X(t)=\Phi(t)C$

 
 

$\left[

\begin{array}{c}

x_1(t)\\

x_2(t)\\

\end{array}

\right]=\left[

\begin{array}{cc}

\frac{1}{2} \left[e^{2 t}+e^{2 \sin ^2(t)}\right] e^{-t-\sin ^2(t)+\sin (t) \cos (t)} & \frac{1}{2}

\left[e^{2 t}-e^{2 \sin ^2(t)}\right] e^{-t-\sin ^2(t)+\sin (t) \cos (t)} \\

\frac{1}{2} \left[e^{2 t}-e^{2 \sin ^2(t)}\right] e^{-t-\sin ^2(t)+\sin (t) \cos (t)} & \frac{1}{2}

\left[e^{2 t}+e^{2 \sin ^2(t)}\right] e^{-t-\sin ^2(t)+\sin (t) \cos (t)} \\

\end{array}

\right]\left[

\begin{array}{c}

C_1 \\

C_2\\

\end{array}

\right]$

 

 

$\left[
\begin{array}{c}
C_1 \\
C_2\\
\end{array}

\right]=\left[
\begin{array}{c}
1 \\
2\\
\end{array}
\right]$ baslangic degerleri icin cozumler.

 

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