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Find the unit step response of the system $$\ddot x + 2 \dot x.$$ That is, find $x$ in the ODE $$\ddot x + 2 \dot x = H(t)$$ where $H$ is the Heaviside unit step function

If this is impossible, or the question not well-defined, prove, or at least explain, so.

Background: A similar question is to solve the ODE $$\ddot x + 2 \dot x = \delta,$$ where $\delta$ is the unit impulse function. That's easily solved as $$x = \frac 1 2 (1 - e^{-2t}), $$ $t \geq 0$. In contrast, the above question seems very hard to answer. I tried multiple approaches and got inconsistent results (e.g. $x = t/2$), none of which checked.

SRobertJames
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    The function $x = \frac 1 2 (1 - e^{-2t})$ satisfies $x''+2x'=0$, not $\delta$. – Anne Bauval Dec 04 '24 at 15:05
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    Have you tried using Laplace transforms? That's usually a good route for time-dependent linear ODEs which 'start' at time $t=0$. – Semiclassical Dec 04 '24 at 15:07
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    You can reduce this by substituting $y(t) = \dot{x}$ and using an integrating factor. You could also solve with Laplace transforms using suitable initial conditions. – Sean Roberson Dec 04 '24 at 15:07
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    Solve $y'+2y=H$ separately on $\Bbb R_+$ and on $\Bbb R_-$. – Anne Bauval Dec 04 '24 at 15:14
  • You are not new to this site and you know you are meant to provide your attempts, because we cannot guess your level and where you did a false move $-$ because your solution for the Dirac delta source is wrong $-$, even if we could show you several techniques out of the clear blue sky. – Abezhiko Dec 04 '24 at 16:27

2 Answers2

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We begin with taking the anti-derivative: $$ \dot x + 2 x = t H(t) + A, $$ where $A$ is some constant.

Then we multiply with the integrating factor $e^{2t}.$ This makes the left hand side become a derivative: $$ \frac{d}{dt}(e^{2t}x) = e^{2t}\dot x + e^{2t}2x = tH(t)e^{2t} + Ae^{2t}. $$

Taking anti-derivative again gives $$ e^{2t}x = \left( \frac14 (2t-1) e^{2t} + \frac14 \right) H(t) + \frac12 Ae^{2t} + B, $$ where $B$ is some constant.

Finally we multiply by $e^{-2t}$ and get $$ x = \left( \frac14 (2t-1) + \frac14 e^{-2t} \right) H(t) + \frac12 A + B e^{-2t} . $$

md2perpe
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$y=\dot{x}$ fulfills the differential equation $\dot{y}(t)+2y(t)=H(t)$. The general solution of the associated homogeneous equation is given by $y_h(t)=A e^{-2t}$. A particular solution of the inhomogeneous equation can be found by employing the ansatz $y_p(t)=C(t)e^{-2t}$, where the function $C(t)$ is determined by the differential equation $\dot{C}(t)=H(t) e^{2t}$ leading to $C(t)=[H(-t)+H(t)e^{2t}]/2$ (up to an irrelevant additive constant). The general solution for $y(t)$ is thus given by $$y(t)= A e^{-2t}+\frac{1}{2}H(-t)e^{-2t}+\frac{1}{2}H(t), $$ which can easily be checked by using $\dot{H}(t)=\delta(t)$ and $\delta(t) f(t)=\delta(t) f(0)$. Finding $x(t)$ by integrating $y(t)$ is left as an easy exercise.

Hyperon
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    @SRobertJames In your question, you wanted to find the solution of the ODE $\ddot{x}(t)+2 \dot{x}(t) =H(t)$. Defining $y(t)=\dot{x}(t)$, the function $y(t)$ obviously fulfills the ODE $\dot{y}(t)+2y(t)=H(t)$. Once you have found $y(t)$, $x(t)$ is obtained by a simple integration. – Hyperon Dec 04 '24 at 16:52
  • Got it, thank you for clarifying. – SRobertJames Dec 04 '24 at 19:30