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Solve the IVP with initial values x(0)=(1,0,-2)

$$A = \begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$$ So far My work consists of me finding the eigenvectors and eigenvalues first.

$$det(A-LI) = det(\begin{bmatrix} 1-L & 0 & 1 \\ 1 & 1-L & 0 \\ 0 & 0 & 1-L\end{bmatrix})=0$$

L = (1,1,1)

(0,0,1;1,0,0;0,0,0)(x;y;z)=(0;0;0) $$A-1 = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$ * $$\begin{bmatrix} x\\y\\z\end{bmatrix}$$

(0;1;0)=v1

(0,0,1;1,0,0;0,0,0)(x1;y1;z1)=(0;1;0)

(0;0;1)=v2

(0,0,1;1,0,0;0,0,0)(x2;y2;z2)=(0;1;1)

(0;0;0)=v3

using the general solution I get x(t)=c1(0;1;0)e^t+c2((0;1;0)te^t+(0;0;1)e^t)+c3(t^2(0;1;0)e^t+t(0;0;1)e^t+(0;0;0)e^t) Is this right so far? I can't find many online resources for 3 repeated eigenvalues.

MathematicalAnomaly
  • 1,021

1 Answers1

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first find the eigen values. when you take the determinant you get solve for landa and get landa = 0,1,-1 and you use these values to find the eigen vectors when you row reduce

HamkinsStudent
  • 641