Fourier Transform of Newton's Law of Cooling

I am attempting to solve Newton's Law of Cooling differential equation with Fourier Transforms for a high school math report. Can Fourier Transforms be used to solve first-order ODEs? The equation is: $\frac{dT}{dt} = -kT + kT_a$, where $k$ and $T_a$ are constants. Since the Fourier Transform is a linear operator, I took the transform of each of the three terms: $$F\{T'\} = F\{-kT\} + F\{kT_a\}$$ $$F\{T'\} = -kF\{T\} + kT_aF\{1\}$$ The Fourier transform of 1 should be $\delta(s)$, so: $$F\{T'\} = -kF\{T\} + kT_a\delta(s)$$ The transform of the derivative of a function, using $s$ instead of $\omega$ in the formulas, is $F\{T'\} = 2\pi isF\{T\}$. Substituting this into the original equation, $$2\pi isF\{T\} = -kF\{T\} + kT_a\delta(s)$$ Adding the terms with the transform should yield $$2\pi isF\{T\} +kF\{T\} = kT_a\delta(s)$$ $$F\{T\}(2\pi is +k) = kT_a\delta(s)$$ $$F\{T\} = \frac{kT_a\delta(s)}{2\pi is + k}$$. The formula for the Inverse Fourier Transform is $f(t) = \int_{-\infty}^{\infty} e^{2\pi ist} F(s) ds$. Thus $$T = \int_{-\infty}^{\infty} \frac{e^{2\pi ist} \delta(s) kT_a}{2\pi is + k}ds$$, which I evaluated to $T_a$, dropping the integral and plugging in 0 for $s$ due to the $\delta(s)$ function...however, the actual solution to the differential equation is $T_a + Ae^{-kt}$, where A is a constant. Where did I go wrong?

• The reason is that $F[T-T_a] = F[e^{-kt}] = \int e^{2\pi i s - kt} ds$ does not exist (it is infinite) so your equation $2\pi i sF[T] = -kF[T] + kT_a\delta(s)$ reads $\infty = \infty$. – Winther Nov 23 '14 at 0:38

Fourier Transform can certainly be used to solve linear differential equations. Note that if you apply Fourier Transform of its ordinary form with respect to the variable $t$, you implicitly assume that $- \infty < t < + \infty$, i.e. you have no boundary. This is the reason why you cannot include boundary or initial conditions. You can get $T_a + A e^{-kt}$ if you apply Laplace Transform though.