Solving a System of First Order Differential Equations

**Belowzero78** · November 16th 2010, 06:08 PM

The system of first order differential equations:
$\frac{dx}{dt} = 0x - 1y$
$\frac{dy}{dt} = 1x +2y$
where $x(0) = -4, y(0) = -5$ has the solution $x(t) =$ and $y(t) =$

When i solved for the eigenvalues, i got one double root where $\lambda = 1$ . I know for a fact that given a double eigenvalue of the same value, there can only be one associated eigenvector. Therefore i must find a generalized eigenvector for the other root.

The resulting eigenvectors are: $V_{1} = \left[\begin{array}{cc}-1\\1\end{array}\right], V_{2} = \left[\begin{array}{cc}1\\0\end{array}\right]$

I finally get to my final answer equal to: $\left[\begin{array}{cc}-5e^{t}+9te^{t}\\-4e^{t}-9te^{t}\end{array}\right]$ where i found the constants $C_{1} = -5$ and $C_{2} = -9.$

I can't seem to get the correct answer! Anyone who contributes I thank them now.

**FernandoRevilla** · November 16th 2010, 10:08 PM

For

$A=\begin{bmatrix}{0}&{-1}\\{1}&{2}\end{bmatrix},\quad P=\begin{bmatrix}{-1}&{1}\\{1}&{0}\end{bmatrix},\quad J=\begin{bmatrix}{1}&{1}\\{0}&{1}\end{bmatrix}$

certainly $AP=PJ$ or equivalently $P^{-1}AP=J$ .

Then, $\begin{bmatrix}{x(t)}\\{y(t}\end{bmatrix}=e^{tA}\b egin{bmatrix}{-4}\\{-5}\end{bmatrix}=Pe^{tJ}P^{-1} \begin{bmatrix}{-4}\\{-5}\end{bmatrix}=\ldots$

I ignore your possible intermediate mistake.

Regards

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