1.5: Solving SecondOrder ODE models
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A secondorder ODE model can be similarly solved by applying the Laplace transform to both sides of the differential equation. We consider a general secondorder ODE model, described as: \[\ddot{y}(t)\; +\; a_{1} \dot{y}(t)\; +\; a_{2} y(t)=b_{1} \dot{u}(t)+b_{0} u(t)\] We assume the following initial conditions for the ODE model: \(y\left(0\right)=y_0,\ \dot{y}\left(0\right)={\dot{y}}_0,\ \dot{u}\left(0\right)=u_0.\) The application of the Laplace transform gives an algebraic equation: \[\left(s^{2} +a_{1} s+a_{2} \right)\; y(s)\left(s+a_{1} \right)y_{0} \dot{y}_{0} =\left(b_{1} s+b_{2} \right)u(s)b_{1} u_{0}\] which can then be solved for \(y(s)\) to obtain: \[y\left(s\right)=\frac{1}{s^2+a_1s+a_2}\left[\left(s+a_1\right)y_0+{\dot{y}}_0+\left(b_1s+b_2\right)u\left(s\right)b_1u_0\right]\] The denominator term in the above expression represents the characteristic polynomial of the ODE model. Further, the characteristics of the timeresponse, \(y\left(t\right)\), obtained from the application of inverse Laplace transform to \(y\left(s\right)\), are determined by the roots of the characteristic equation: \(s^{2} +a_{1} s+a_{2} =0\), as illustrated in the following example.
Example 1.9: The mass–spring–damper system
We consider the massspringdamper system model (Example 1.8), in the presence of a unitstep input function, \(u\left(t\right)\), assuming zero initial conditions; then, application of the Laplace transform produces the following algebraic equation: \[\left(ms^{2} +bs+k\right)x(s)=f(s)\] Next, let \(f\left(s\right)=1/s\), and consider two cases based on the assumed system parameter values that results in either real or complex roots of the characteristic equation: \(ms^{2} +bs+k=0\).
Case I (Real Roots). Let the parameter values be: \(m=1,\ k=2,\ b=3\); then, the characteristic equation is given as: \(s^2+3s+2=0\). The equaiton has real roots at: \(s=1,\ 2\).
The algebraic equation describing the system output is: \(\left(s^2+3s+2\right)x\left(s\right)=f\left(s\right)\). Assuming a unitstep input, the output variable is solved as: \(x\left(s\right)=\frac{1}{s\left(s+1\right)\left(s+2\right)}\).
In order to apply the inverse Laplace transform, we need to carry out partial fraction expansion (PFE) of the output. We may use online SimboLab partial fraction calculator for this purpose (https://www.symbolab.com/solver/part...nscalculator/). The result is given as: \[x(s)=\frac{1}{2s}\frac{1}{s+1}+\frac{1}{2(s+2)}\] By applying the inverse Laplace transform, the time response of the springmassdamper system is obtained as (Figure 10a): \[x\left(t\right)=\left(\frac{1}{2}e^{t}+\frac{1}{2}e^{2t}\right)u(t)\] where \(u\left(t\right)\) represents the unitstep function.
Case II (Complex Roots). Alternatively, let the parameter values be: \(m=1,\ k=2,\ b=2\); then, the characteristic equation is given as: \(s^2+2s+2=0\). The equaiton has complex roots at: \(s=1\pm j1\).
The algebraic equation describing the system output is: \(\left(s^2+2s+2\right)x\left(s\right)=f\left(s\right)\). Assuming a unitstep input, the output variable is solved as: \(x\left(s\right)=\frac{1}{s\left(s^2+2s+2\right)}\). Taking help from SimboLab for partial fractions, the output is given as: \[x(s)=\frac{1}{2s}\frac{s+2}{2(s^2+2s+2)}\] The quadratic factor is alternately expressed as: \({\left(s+1\right)}^2+1^2\), which helps when applying the inverse Laplace transform. Also, the quadratic term is split to write: \[x\left(s\right)=\frac{1}{2s}\frac{s+1}{2\left[{\left(s+1\right)}^2+1^2\right]}\frac{1}{2\left[{\left(s+1\right)}^2+1^2\right]}\] By applying the inverse Laplace transform, the time response of the springmassdamper system is obtained as (Figure 10b): \[x\left(t\right)=\frac{1}{2}\left(1e^{t}{\mathrm{cos} t\ }e^{t}{\mathrm{sin} t\ }\right)u\left(t\right)\\] where \(u\left(t\right)\) represents the unitstep function.
A comparison of the time response in the two cases reveals that:

The steadystate response in both cases settles at \(x_{\infty }=\frac{1}{2}\).
image12 The time response in the case of real roots resembles that of a firstorder systems. Whereas, the time response in the case of complex roots has an overshoot that causes it to be oscillatory.
Figure 10: Time response of secondorder system models: characteristic equation with real roots (left); with complex roots (right).