# Eigenvectors of first-order differential equation

1. May 17, 2009

### jeff1evesque

Suppose that f is an eigenvector of T with corresponding eigenvalue $$\lambda$$. Then f' = T(f) = $$\lambda$$f. 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) = $$\lambda$$f.
Therefore, f ' = $$\lambda$$ f ==> $$f' / f$$ = $$\lambda$$
So, [Integral] $$f ' / f$$ dt = [Integral] $$\lambda$$ dt
Therefore, ln(f) = $$\lambda$$t + c
<==> e^( ln(f) ) = e^( $$\lambda$$t + c )
= $$e^l^n^|^f^|$$ = e ^ ( $$\lambda$$t + c )
= $$e^c$$e^( $$\lambda$$ * t )
<==> f = $$e^c$$e^( $$\lambda$$ * t ) which does not equal ce^( lambda * t ) ??

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

Last edited: May 17, 2009
2. May 18, 2009

### tiny-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 ec 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)

3. May 18, 2009

### jeff1evesque

Wow, thanks a lot Tim.