Birinchidan, tizim matritsa shaklida berilmaganiga e'tibor bering. Buning sababi shundaki, tizim matritsa shaklida hal etilmaydi. Shuni ham yodda tutingki, tizim bir hil emas.

Biz xuddi Laplasdan foydalanganimizdek boshlaymiz yagona differensial tenglamalarni yechish uchun aylantiradi. Ikkala differentsial tenglamaning konvertatsiyasini olamiz.

\[\begin{align*}s{X_1}\left( s \right) - {x_1}\left( 0 \right) & = 3{X_1}\left( s \right) - 3{X_2}\left( s \right) + \frac{2}{s}\\ s{X_2}\left( s \right) - {x_2}\left( 0 \right) & = - 6{X_1}\left( s \right) - \frac{1}{{{s^2}}}\end{align*}\]

Endi dastlabki holatni ulang va narsalarni biroz soddalashtiring.

\[\begin{align*}\left( {s - 3} \right){X_1}\left( s \right) + 3{X_2}\left( s \right) & = \frac{2}{s} + 1 = \frac{{2 + s}}{s}\\ 6{X_1}\left( s \right) + s{X_2}\left( s \right) & = - \frac{1}{{{s^2}}} - 1 = - \frac{{{s^2} + 1}}{{{s^2}}}\end{align*}\]

Endi biz buni transformatsiyalardan biri uchun hal qilishimiz kerak. Buni yuqoridagi tenglamani \(s\) ga, pastki qismini esa -3 ga ko‘paytirib, keyin qo‘shish orqali qilamiz. Bu beradi,

\[\left( {{s^2} - 3s - 18} \right){X_1}\left( s \right) = 2 + s + \frac{{3{s^2} + 3}}{{{s^2}}}\]

\(X_{1}\) ni hal qilish,

\[{X_1}\left( s \right) = \frac{{{s^3} + 5{s^2} + 3}}{{{s^2}\left( {s + 3} \right)\left( {s - 6} \right)}}\]

Qisman fraktsiyalash beradi,

\[{X_1}\left( s \right) = \frac{1}{{108}}\left( {\frac{{133}}{{s - 6}} - \frac{{28}}{{s + 3}} + \frac{3}{s} - \frac{{18}}{{{s^2}}}} \right)\]

Teskari konvertatsiya qilish bizga birinchi yechimni beradi,

\[{x_1}\left( t \right) = \frac{1}{{108}}\left( {133{{\bf{e}}^{6t}} - 28{{\bf{e}}^{ - 3t}} + 3 - 18t} \right)\]

Endi ikkinchi yechimni topish uchun biz orqaga qaytishimiz va \(X_{2}\) uchun konvertatsiyani topish uchun \(X_{1}\)-ni yo'q qilishimiz mumkin va ba'zida buni qilishimiz kerak bo'ladi. Biroq, bu holda ikkinchi differentsial tenglamaga e'tibor bering,

\[{x'_2} = - 6{x_1} - t\hspace{0.25in} \Rightarrow \hspace{0.25in}{x_2} = \int{{ - 6{x_1} - t}}\,dt\]

Shunday qilib, birinchi yechimni ulash va integratsiya qilish,

\[\begin{align*}{x_2}\left( t \right) & = - \frac{1}{{18}}\int{{133{{\bf{e}}^{6t}} - 28{{\bf{e}}^{ - 3t}} + 3}}\,dt\\ & = - \frac{1}{{108}}\left( {133{{\bf{e}}^{6t}} + 56{{\bf{e}}^{ - 3t}} + 18t} \right) + c\end{align*}\]

Endi, integratsiya doimiyligini olish uchun ikkinchi boshlang'ich shartni qayta qo'llash beradi

\[ - 1 = - \frac{1}{{108}}\left( {133 + 56} \right) + c\hspace{0.25in} \Rightarrow \hspace{0.25in}c = \frac{3}{4}\]

Ikkinchi yechim esa,

\[{x_2}\left( t \right) = - \frac{1}{{108}}\left( {133{{\bf{e}}^{6t}} + 56{{\bf{e}}^{ - 3t}} + 18t - 81} \right)\]

Shunday qilib, bularning barchasini birlashtirib, tizimga yechim topadi:

\[\begin{align*}{x_1}\left( t \right) & = \frac{1}{{108}}\left( {133{{\bf{e}}^{6t}} - 28{{\bf{e}}^{ - 3t}} + 3 - 18t} \right)\\ {x_2}\left( t \right) & = - \frac{1}{{108}}\left( {133{{\bf{e}}^{6t}} + 56{{\bf{e}}^{ - 3t}} + 18t - 81} \right)\end{align*}\]