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

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^([tex]\lambda[/tex] * t) for some scalar constant c. "

Work:

f ' = T(F) = [tex]\lambda[/tex]f.

Therefore, f ' = [tex]\lambda[/tex] f ==> [tex]f' / f [/tex] = [tex]\lambda[/tex]

So, [Integral] [tex]f ' / f [/tex] dt = [Integral] [tex]\lambda[/tex] dt

Therefore, ln(f) = [tex]\lambda[/tex]t + c

<==> e^( ln(f) ) = e^( [tex]\lambda[/tex]t + c )

= [tex]e^l^n^|^f^|[/tex] = e ^ ( [tex]\lambda[/tex]t + c )

= [tex]e^c[/tex]e^( [tex]\lambda[/tex] * t )

<==> f = [tex]e^c[/tex]e^( [tex]\lambda[/tex] * t ) which does not equal ce^( lambda * t ) ??

Question:I'm kind of rusty with calculus but I didn't think [Integral] [tex]f ' / f [/tex] dt = ln(f)

I thought it would be [Integral] [tex]f ' / f [/tex] dt = [Integral] [tex] 1 / f [/tex] dt = [tex]t / f[/tex]

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# Eigenvectors of first-order differential equation

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