Question

3 points -5 3 7 Consider the initital value problem - x, x(0) 2 -10 0 The smaller eigenvalue is וג- The bigger eigenvalue is X2- The general solution can be written as xC1 The solution of the Initial Value Problem is eit+ x

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Answer #1

Eigenvalues are given by

det( ー10 (t 5) (t 10) - 60

Smaller eigenvalue is : -11

Larger eigenvalue is: -4

eigenvectors are given by

-5-101-tl)u =0

t=-11

-5 3 2-10| +111)o = 0 ー10

​​​​

t=-4

-5 3 -10 41)

v=egin{bmatrix} 3 1 end{bmatrix}

General solution is

x=C_1egin{bmatrix} -1 2 end{bmatrix}e^{lambda_1t}+C_2egin{bmatrix} 3 1 end{bmatrix}e^{lambda_2t}

x(0)=egin{bmatrix} 7 0 end{bmatrix}=C_1egin{bmatrix} -1 2 end{bmatrix}+C_2egin{bmatrix} 3 1 end{bmatrix} 7=-C_1+3C_2 0=2C_1+C_2 C_1=-1,C_2=2

x=-egin{bmatrix} -1 2 end{bmatrix}e^{lambda_1t}+2egin{bmatrix} 3 1 end{bmatrix}e^{lambda_2t}

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