A community for students.
Here's the question you clicked on:
 0 viewing
bayanhorani
 2 years ago
i need to solve equation
bayanhorani
 2 years ago
i need to solve equation

This Question is Open

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

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

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

er10
 2 years ago
Best ResponseYou've already chosen the best response.0Here, $$\lambda$$ 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!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.