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MxwllsPersuasns

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## Homework Statement

We assume from ODE theory that given a smooth A: I → gl(n;R) there exists a

unique smooth solution F : I → gl(n;R), defined on the same interval I on which

A is defined, of the initial value problem F' = FA and F(t

_{0}) = F

_{0}∈ gl(n;R) given.

(i) Show that two solutions F

_{i}: I → GL(n;R) of the ODE F' = FA satisfy

F2 = CF1 for a constant invertible matrix C ∈ GL(n;R).

(ii) Show that for A a constant matrix F(t) = exp(tA) is a solution of F' = FA

(iii) If A: I → gl(n;R) is not constant, why is F(t) = exp(∫A(s)ds [from t

_{0}to t]) not

solving F' = FA, or is it? Explain.

## Homework Equations

F' = FA then...

|F|' = tr(A)*|F| where |...| signifies the determinant

## The Attempt at a Solution

**i)**I recall once I needed to show the uniqueness of the solutions for a standard complex valued ODE and the solution was given as something like z(t) = z

_{0}exp(of an integral) and I wanted to show that w(t) was also a solution. If I recall correctly someone told me to differentiate the ratio of the two (z(t)/w(t)) and I found that it ended up equaling zero or something like that and that showed there was only a constant difference between them or something like that? Can anyone help here.. I'm not sure if this is the right way to show F

_{1}and F

_{2}are actually separate solutions.

**ii)**For this part I would assume that since we are given the solution we can just plug it into the DE and show that it works, correct?

**iii)**Would this be because when A is constant it can be taken out of the exp in a nice and easy way when exp(...) is differentiated? I have a feeling there's more to this and ii) but I can't think of anything more complex than just trying the solution.