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Differential Equations
The Geometric View of ODEs
01Direction Fields and the Geometric View of ODEs02Integral Curves and the Isocline Method03Perpendicular Isoclines: The Circles of y' = -x/y04Isocline Corridors and a Trap for Solutions05Existence and Uniqueness: One Curve Through a Point06Where Existence and Uniqueness FailProblem set0/10Problem set 20/10MIT problem set0/3Practice∞
Numerical Methods
01Euler's Method and Its Three Equations02Euler's Method by Hand: Building the Table03Convexity and the Sign of Euler's Error04First Order Error: Halve the Step, Halve the Error05Improved Euler (RK2): Averaging Two Slopes06Runge-Kutta 4 and the Price of Accuracy07Singular Points: Where Every Method FailsProblem set0/10Problem set 20/10MIT problem set0/1Practice∞
First-Order Linear Equations
01First Order Linear: Standard Form and the Sign of p02The Conduction-Diffusion Model: Newton's Law of Cooling03The Integrating Factor and the Four-Step Method04Integrating Factors: Two Mistakes and a Trick Question05Constant Coefficients: Steady State and TransientProblem set0/10Problem set 20/10MIT problem set0/1Practice∞

Constant Coefficients: Steady State and Transient

Why does the initial temperature $T_0$ stop mattering once $e^{-kt}$ decays, leaving only the steady-state integral?


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    1. 1

      Standard form dTdt+kT=kTe\frac{dT}{dt} + kT = kT_edtdT​+kT=kTe​

      Write the conduction equation in standard form with kTekT_ekTe​ on the right rather than TeT_eTe​, and carry out the units comparison of the two sides.

    2. 2

      The integrating factor ekte^{kt}ekt

      Integrate k dtk\,dtkdt, multiply through, write the compact form (ektT)′=kTe(t)ekt(e^{kt}T)' = kT_e(t)e^{kt}(ektT)′=kTe​(t)ekt, and integrate to T=e−kt∫kTe(t)ekt dt+Ce−ktT = e^{-kt}\int kT_e(t)e^{kt}\,dt + Ce^{-kt}T=e−kt∫kTe​(t)ektdt+Ce−kt.

    3. 3

      Definite-integral solution with T(0)=T0T(0) = T_0T(0)=T0​

      Redo the integral from 000 to ttt in the dummy variable t1t_1t1​, then set t=0t = 0t=0 and record the resulting value C=T0C = T_0C=T0​.

    4. 4

      Steady-state term and transient T0e−ktT_0 e^{-kt}T0​e−kt

      Name the two terms of T=e−kt∫0tkTe(t1)ekt1 dt1+T0e−ktT = e^{-kt}\int_0^t kT_e(t_1)e^{kt_1}\,dt_1 + T_0 e^{-kt}T=e−kt∫0t​kTe​(t1​)ekt1​dt1​+T0​e−kt and write the t→∞t \to \inftyt→∞ behaviour of each for k>0k > 0k>0.

    Attempt 1 of 2