Eigenvectors of first-order differential equation

[jeff1evesque]

Suppose that f is an eigenvector of T with corresponding eigenvalue \lambda. Then f' = T(f) = \lambdaf. This is a first-order differential equation whose solutions are of the form f(t) = ce^(\lambda*t) for some scalar constant c. Consequently, every real number \lambda is an eigenvalue of T, and \lambda corresponds to eigenvectors of the form ce^(\lambda*t ) for c not equal to 0. Note that for \lambda = 0, the eigenvectors are the nonzero constant functions.

Question: I was wondering if someone could explain the following sentence from above: "This is a first-order differential equation whose solutions are of the form f(t) = ce^(\lambda * t) for some scalar constant c. "

Work:
f ' = T(F) = \lambdaf.
Therefore, f ' = \lambda f ==> f' / f = \lambda
So, [Integral] f ' / f dt = [Integral] \lambda dt
Therefore, ln(f) = \lambdat + c
<==> e^( ln(f) ) = e^( \lambdat + c )
= e^l^n^|^f^| = e ^ ( \lambdat + c )
= e^ce^( \lambda * t )
<==> f = e^ce^( \lambda * t ) which does not equal ce^( lambda * t ) ??

Question: I'm kind of rusty with calculus but I didn't think [Integral] f &#039; / f dt = ln(f)
I thought it would be [Integral] f &#039; / f dt = [Integral] 1 / f dt = t / f

 

[tiny-tim]

Hi jeff1evesque!

(have a lambda: λ and an integral: ∫ and try using the X² tag just above the Reply box )

… f = e^ce^( \lambda * t ) which does not equal ce^( lambda * t ) ??

Yes it does … the constant e^c has simply been renamed c

Question: I'm kind of rusty with calculus but I didn't think [Integral] f &#039; / f dt = ln(f)
I thought it would be [Integral] f &#039; / f dt = [Integral] 1 / f dt = t / f

∫f/'f dt = ln(f) + c because d/dt (ln(t)) = 1/t, so by the chain rule d/dt (ln(f(t))) = f'(t)/f(t)

 

[jeff1evesque]

Wow, thanks a lot Tim.

Hi jeff1evesque!

(have a lambda: λ and an integral: ∫ and try using the X² tag just above the Reply box )


Yes it does … the constant e^c has simply been renamed c


∫f/'f dt = ln(f) + c because d/dt (ln(t)) = 1/t, so by the chain rule d/dt (ln(f(t))) = f'(t)/f(t)