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We have $$\frac{dT}{dt}=-a(T-T_{\infty})-\delta (t-1)$$ where $T_{\infty}$ is a constant and $\delta$ is the Dirac delta.

Determine the jump (discontinuity) condition for $T$ at $t=1$ and hence find $T(t)$ for $t>1$.

I'm just a little puzzled as this is first order. I have only ever done second order before. Is $T$ continuous at $t=1$ here? What is the jump condition? I think I just need a run down of what is actually going on here, for me this is very much a method at the moment, I see these questions I have a procedure to solve them rather than actually comprehending everything that's going on. My method doesn't work here do I can't do it.

Any help is appreciated. Thank you.

2 Answers 2

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Let $u=L\{\frac{dT}{dt}\}$, where $L$ is Laplace Transform. Then after taking Laplace Transform for both sides of the equation, one has $$ L\{\frac{dT}{dt}\}=-aL\{(T-T_{\infty})\}-L\{\delta (t-1)\}$$ or $$ su(s)-T(0)=-a(u(s)-\frac{T_\infty}{s})-e^{-s}. $$ Thus $$ u(s)=\frac{e^{-s}}{s+a}+\frac{aT_\infty}{s(s+a)}+\frac{T(0)}{s+a}.$$ So $$ T(t)=L^{-1}(u(s))=e^{-a t}(-e^a \theta (t-1)+T_\infty (e^{a t}-1)+T(0))$$ where $\theta$ is Heaviside Theta. So one has, for $t<1$, $$ T(t)=e^{-a t} (T_\infty \left(e^{a t}-1\right)+T(0))$$ and for $t\ge 1$, $$ T(t)=e^{-a t}(-e^a+T_\infty (e^{a t}-1)+T(0)).$$

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As usual, employ the exponential integrating factor to get \begin{align} \frac{d}{dt}(e^{at}(T(t)-T_∞))&=e^{at}\,δ(t−1)=e^{a·1}\,δ(t−1) \\[1em] e^{at}(T(t)-T_∞)-(T(0)-T_∞) &=-e^{a}\,\int_0^tδ(s−1)\,ds \\ &=-e^a·H(t-1) \end{align} where $H$ is the Heaviside jump function.