## bayanhorani 2 years ago i need to solve equation

• This Question is Open
1. bayanhorani

|dw:1351014321172:dw|

2. bayanhorani

|dw:1351014635037:dw|

3. Raja99

use equation i cant understand plz

4. er10

Here, $$\lambda$$ is a repeated eigenvalue for the linear system $$\frac{d\,\mathbf{X}}{dt} = A\mathbf{X}$$ so if $$\mathbf{X}_1(t) = (\mathbf{w} + t\mathbf{v})e^{\lambda t}$$ satisfies the linear system of equations, then $\frac{d\, \mathbf{X}_1}{d t} = \frac{d\, }{dt} \left( (\mathbf{w} + t\mathbf{v})e^{\lambda t} \right) = \frac{d\, }{dt} \left( \mathbf{w}e^{\lambda t} + t\mathbf{v}e^{\lambda t} \right) = \lambda\mathbf{w}e^{\lambda t} + \mathbf{v}e^{\lambda t} + \lambda t \mathbf{v}e^{\lambda t} = A\mathbf{X}_1$ So, by evaluating the matrix-vector multiplication on the far right hand-side, and factoring, we find $e^{\lambda t}(\lambda \mathbf{w} + \mathbf{v}) + te^{\lambda t}(\lambda \mathbf{v}) = A(\mathbf{w} + t\mathbf{v})e^{\lambda t} = e^{\lambda t}(A\mathbf{w}) + te^{\lambda t}(A\mathbf{v})$ by equating like terms, $\lambda \mathbf{w} + \mathbf{v} = A\mathbf{w} \text{ and } \lambda \mathbf{v} = A\mathbf{v}.$ For the first equation we can solve for the desired $\mathbf{v} = A\mathbf{w} - \lambda \mathbf{w} = \mathbf{w}(A - \lambda I),$ so $\mathbf{v} = \mathbf{w}(A - \lambda I)$ as desired. I hope this helps!