SECOND-ORDER DIFFERENTIAL EQUATIONS - DIRECT METHOD

EDITOR: B. SOMANATHAN NAIR

We have already seen the method for solving the first-order differential and difference equations. We now consider the methods for solving the second-order differential and difference equations. First we consider second-order differential equations of the form

                                                (d²y/dt²) + p(dy/dt) + qy = 0  (1)

where we have assumed the forcing function to be zero. The characteristic equation of this differential equation can be written as                                                                                                          

                                                   (D² + pD + q)y = 0  (2)

where we have made the substitution D = d/dt. The quadratic given in the brackets has the roots

                                                          D₁, D₂ = α ± β  (3)                                       

where α = p/2 and β = (½)√(p²‒4q). It may be noted that α, called attenuation constant, introduces attenuation in the output wave form and β, called phase constant, produces phase delay in it. Depending on the values of α and β, we have four independent solutions of Eq. (2).

Solution 1: Both a and b are real and unequal

When both α and β are real, the roots from Eq. (2) are (α+β) and (α‒β), respectively. These roots can be seen to be real and are unequal. Using them, the solution can now be written as

                        y = Ae^(α+β)t + Be^(α‒β)t   (4)                                 

where A and B are the constants of integration. Equation (4) represents a damped single vibration, as shown in Fig. 1. Here α represents the damping in the vibration produced and β produces the delaying action in it.

Solution 2: Only a exists; b = 0

In this condition, both the roots are real and equal. This means that the system output has only attenuation and there is no propagation. The solution may be written in the form

                                    y = e^αt (A+ Bt)   (5)                                                  

which represents a single damped vibration of duration shorter than that shown in Fig. 1. The duration of the vibration becomes shorter since there is no propagation as b = 0. The resulting waveform, called critically damped waveform, is shown in Fig. 2.

 

Solution 3: Both a and b exist; a real, b imaginary

The roots are a +jb and a ‒jb, respectively. So, the solution can be written in a form similar to that given in Eq. (4), which yields

                                    y = Ae^(α+jβ)t + Be^(α‒β)t   (6)                              

                                     

Equation (6) may be expressed in the form           

            y = e^αt(Ae^jβt +Be^‒jβt)= e^αt(C cosbt + D sinbt)  (7)                       

where we define the new constants of integration  as C = A+jB and D = A‒jB. We find that Eq. (6) represents a damped oscillation, as shown in Fig. 3.

 

Solution 4: a = 0 and only b exists

Both the roots in this case can be seen to be purely imaginary, and are equal to ±jb. The solution, therefore, can be written by modifying Eq. (6), which yields

                                     y = Ae^jβt + Be^‒jβt = C cosbt + D sinbt  (8)

                       

This represents undamped pure sinusoidal oscillations (see Fig. 4).