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bayanhorani Group TitleBest ResponseYou've already chosen the best response.0
dw:1351014321172:dw
 2 years ago

bayanhorani Group TitleBest ResponseYou've already chosen the best response.0
dw:1351014635037:dw
 2 years ago

Raja99 Group TitleBest ResponseYou've already chosen the best response.0
use equation i cant understand plz
 2 years ago

er10 Group TitleBest ResponseYou've already chosen the best response.0
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 matrixvector multiplication on the far right handside, 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!
 2 years ago
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