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Найдем базисные решения в виде
![x = Ae^{lambda t}quad y=Be^{lambda t}\\
left{
begin{aligned}
&lambda A = 6A+3B\
&lambda B = -8A-5B
end{aligned}
right.\\\
det left[begin{array}{ccc}6-lambda&3\-8&-5-lambdaend{array}right] =0\\
(lambda+5)(lambda-6)+24 = 0\
lambda^2 -lambda-6 = 0\
lambda_1 = 3\
lambda_2 = -2](https://tex.z-dn.net/?f=x+%3D+Ae%5E%7Blambda+t%7Dquad+y%3DBe%5E%7Blambda+t%7D%5C%5C%0Aleft%7B%0Abegin%7Baligned%7D%0A%26amp%3Blambda+A+%3D+6A%2B3B%5C%0A%26amp%3Blambda+B+%3D+-8A-5B%0Aend%7Baligned%7D%0Aright.%5C%5C%5C%0A+det+left%5Bbegin%7Barray%7D%7Bccc%7D6-lambda%26amp%3B3%5C-8%26amp%3B-5-lambdaend%7Barray%7Dright%5D+%3D0%5C%5C%0A%28lambda%2B5%29%28lambda-6%29%2B24+%3D+0%5C%0Alambda%5E2+-lambda-6+%3D+0%5C%0Alambda_1+%3D+3%5C%0Alambda_2+%3D+-2%0A "x = Ae^{lambda t}quad y=Be^{lambda t}\\
left{
begin{aligned}
&lambda A = 6A+3B\
&lambda B = -8A-5B
end{aligned}
right.\\\
det left[begin{array}{ccc}6-lambda&3\-8&-5-lambdaend{array}right] =0\\
(lambda+5)(lambda-6)+24 = 0\
lambda^2 -lambda-6 = 0\
lambda_1 = 3\
lambda_2 = -2")
Нашли собственные значения, теперь собственные вектора
![1)\
left[begin{array}{cc}3&3\-8&-8end{array}right]left[begin{array}{c}A\Bend{array}right] =0\\
A = -B = C_1
2)\
left[begin{array}{cc}8&3\-8&-3end{array}right]left[begin{array}{c}A\Bend{array}right] =0\\
A = 3C_2quad B=-8C_2\\
x(t) = C_1e^{3t}+3C_2e^{-2t}\
y(t) = -C_1e^{3t}-8C_2e^{-2t}](https://tex.z-dn.net/?f=1%29%5C%0Aleft%5Bbegin%7Barray%7D%7Bcc%7D3%26amp%3B3%5C-8%26amp%3B-8end%7Barray%7Dright%5Dleft%5Bbegin%7Barray%7D%7Bc%7DA%5CBend%7Barray%7Dright%5D+%3D0%5C%5C%0AA+%3D+-B+%3D+C_1%0A%0A2%29%5C%0Aleft%5Bbegin%7Barray%7D%7Bcc%7D8%26amp%3B3%5C-8%26amp%3B-3end%7Barray%7Dright%5Dleft%5Bbegin%7Barray%7D%7Bc%7DA%5CBend%7Barray%7Dright%5D+%3D0%5C%5C%0AA+%3D+3C_2quad+B%3D-8C_2%5C%5C%0Ax%28t%29+%3D+C_1e%5E%7B3t%7D%2B3C_2e%5E%7B-2t%7D%5C%0Ay%28t%29+%3D+-C_1e%5E%7B3t%7D-8C_2e%5E%7B-2t%7D "1)\
left[begin{array}{cc}3&3\-8&-8end{array}right]left[begin{array}{c}A\Bend{array}right] =0\\
A = -B = C_1
2)\
left[begin{array}{cc}8&3\-8&-3end{array}right]left[begin{array}{c}A\Bend{array}right] =0\\
A = 3C_2quad B=-8C_2\\
x(t) = C_1e^{3t}+3C_2e^{-2t}\
y(t) = -C_1e^{3t}-8C_2e^{-2t}")
Нашли собственные значения, теперь собственные вектора
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ничего не пойму
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какие - то каракули
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