SECOND-ORDER DIFFERENTIAL EQUATIONS CONTAINING FORCING FUNCTIONS - DIRECT METHOD

EDITOR: B. SOMANATHAN NAIR

The current flowing through a series RL circuit is given by

                                    di/dt + R/L = v(t)/L     (1)                           

where i = current flowing through the circuit, and v(t) is the applied (external) voltage. We know that the current stars flowing in the circuit only when the voltage v(t) is applied to it. Since the current flows only on the application of v(t), we call it as the forcing function of the circuit. This means that v(t) forces the  generation of the  transient in the circuit. However, it also forces the circuit to attain steady state.

            As another example, consider a thin galvanized-iron wire being stretched between and fastened to two vertical posts. Assume that the wire remains standstill at time t = 0. Now, if we pull the wire slightly upwards and then remove our hands, we find that the wire starts oscillating up and down. Here, we have forced the wire to oscillate by applying force using our hands. This transient situation (i.e., oscillation of the wire) can be stopped by applying pressure on the oscillating wire using our hands once again. This shows that the transient has been created and stopped by the forcing function, which is the pressure applied by our hands.

             Now, let us consider the second-order differential equation, given by

                      (d²y/dt²) + p(dy/dt)+qy = f(t)   (2)

In Eq. (2), f(t) is the forcing function. We find that f(t) can be a constant, or a function of time. We first consider the case when f(t) is a constant.

Case 1: f(t) is a Step (or Constant) Input

Equation (2) may now be written with f(t) = F, a constant force, as

                        (d²y/dt²) + p(dy/dt)+qy = F      (3)

Since the forcing function is a constant, a logical assumption is that in the steady state, the system has to deliver a constant output because it is the forcing function itself that produces the final steady state. Letting this solution to be the constant K, we find that K also must satisfy Eq. (3), as it must be a solution in both the transient and steady-state situations. Thus using y = K in Eq. (3), we get

                         (d²K/dt²) + p(dK/dt)+qK = F      (4)

But we find that since K is a constant, we have

                                                            d²K/dt² = dK/dt = 0    (5)

                                                                                             

Therefore, from Eq. (5), we find

                                                                        qK = F      (6)

                                                                                                       

or

                                               

K = F/q    (7)                                                          

This is the required solution.

Example 1:   Obtain the solution of the equation

               (d²v/dt²) + (RIL)(dy/dt)+ (1/LC) y = V/LC   (1)

                                                          

where V is a step input.

Solution: We first rewrite Eq. (1) as

                                                D² + (RIL)D + 1/LC)v = V/LC   (2)

                                                                                    

where we have chosen the operator D = d/dt, as before. As before, we have the total solution given by

                                                                        v = v_tr + v_ss­      (3)

where v_tr is the transient part and v_ss is the steady-state part of the solution. v_tr is obtained by solving the differential part of the equation, and we have already seen that there are four possible solutions to this (see earlier blog on solution of differential equation with forcing function equal to zero).

To get the steady-state part of the solution, since V is a step voltage, we can assume a constant as a solution, as in the case of the first-order equation. Thus, we use the relation

                                                            v = K, a constant 0 £ t £¥                           

                                                                           = 0, elsewhere      (4)                                                         

Using the relation from Eq. (4) into Eq. (1), we get

                                                D²K + (RIL)DK + 1/LC)K = V/LC   (5)

                                                                                   

 But, D²K = DK = 0, since the differential coefficient of a constant is zero. Therefore, Eq. (5) gives

                                       K = V    (6)                                                                             

This is the steady-state solution for a step input. The complete solution is therefore

                                                            v = v_tr + V                                                     

                                                                                                                               
Case 2: f(t) is an Impulse Input

When the input is a delta function, Eq. (2) may be written as

                          (d²y/dt²) + p(dy/dt)+qy = δ(t)   (8)                                               

Since the delta function is the derivative of the step function, we may obtain the solution of Eq. (8) by differentiating the solution of the equation with the forcing function a step input.

Case 3: f(t) is an Ramp Input

When the input is a ramp function, Eq. (2) may be written as

                              (d²y/dt²) + p(dy/dt)+qy = r(t)   (9)                                                         

where r(t) is a ramp function. Just as the delta function is obtained by differentiating the step function, we know that the ramp function is obtained by integrating the step function. Therefore, we obtain the solution of Eq. (9) by differentiating the solution of the equation with a step input as the forcing function. This is illustrated in Example 2.

 Case 4: f(t) is an Exponential Input

When the input is an exponential function, Eq. (2) may be written as

                        (d²y/dt²) + p(dy/dt)+qy = Fe^‒mt   (10)

                               

When we apply an exponentially decaying function as input, we find that in the steady state the output will also be exponentially decaying. So the response must have the same exponent m, but its magnitude will be less. This is also illustrated in Example 2.

Example 2: Obtain the solution of the equation

(d²v/dt²) + 6(dv/dt)+5v = V  (1)                                                                  

when (a) V = 4 u(t) (b) V = 4 d (t) (c) V = 4 r(t) (d) V = 20 e^-4t .

Solution:

(a) Step input: We first rewrite Eq. (1) in the form

          D² + 6D + 5v = 0  (2)

                                   

The transient solution has the roots ‒5, ‒1. the transient solution therefore is

                                 v = Ae^‒5t + Be^‒t      (3)

                                                                                     

            Now, for the steady-state solution, we know that V = 4. As per our earlier theory (see Case 1), we assume a constant K as the solution and find that the steady-state solution is

                                      K = V = 4      (4)                                                         

Thus the complete solution is

                                    v = Ae^‒5t + Be^‒t + 4     (5)

                                                                                        

The constants A and B can be evaluated using the boundary conditions (i) v = 0 at t = 0, and (ii) dv/dt = 0. Solving, we get A = 1, B = -5. Then the final solution with the forcing function as step input is

                                   v_step = e^‒5t ‒ 5e^‒t + 4     (6)                                                   

(b) Impulse input: As stated above, we get the solution for the impulse input by differentiating Eq. (7). Thus, we find the final solution with the forcing as impulse input is

                        v_impulse = (dV_step/dt) =(dV_step/dt) (e^‒5t ‒ 5e^‒t + 4) = 5(e^‒t ‒ e^‒5t )  (7)

                          

(c) Ramp input: Here we get the solution by integrating Eq. (7). Thus, we have the final solution with the forcing as ramp input is

                        v_ramp = ∫(e^‒5t ‒ 5e^‒t  + 4)dt = (‒1/5) e^‒5t + 5e^‒t + 4t + A     (7)

                                                                

where A is the constant of integration. To evaluate A, we use the boundary condition that at t = 0, v_ramp = 0. Solving, we get A = -5 + 1/5. The final solution then becomes

v_ramp = ‒(1/5) e^‒5t + 5e^‒t + 4t + (1/5)‒5

          = (1/5)(1‒ e^‒5t) ‒ 5(1‒e^‒t) + 4t       (8)

                  

                                                           

(c) Exponential input: To get the solution for the exponential input, as suggested above we assume v to be given by

                                                            V = Fe^‒4t  (9)

Then using Eq. (9) in Eq. (2), we get

      D²(Fe^‒4t) + 6D(Fe^‒4t) + 5 Fe^‒4t  = 20e^‒4t   (10)

                                                                                                           

Carrying the differentiations in Eq. (10) yields

                                                16 Fe^‒4t ‒24 Fe^‒4t + 5Fe^‒4t = 20e^‒4t (11)

Canceling common term e^‒4t and reducing the resultant expression, we get

                                                            F = -20/3  (12)                                                                     

Thus the complete solution is

                        v_exponential = Ae^‒5t + Be^‒t ‒ (20/3)e^‒4t(13)                           

The constants A and B can be evaluated by using the boundary conditions that (i) v = 0 at t = 0, and (ii) dv/dt = 0.

SOLUTIONS OF PROBLEMS IN SECOND-ORDER DIFFERNCE EQUATIONS WITH FORCING FUNCTIONS EQUAL TO ZERO

EDITOR: B. SOMANATHAN NAIR

Example 1:  Obtain the solution of the equation

                                                y(n) + 4y(n‒1) + 3y(n‒2) = 0  (1)

                                                                                                             

Solution: The transient part of the solution can be obtained from the characteristic equation

                                     (C² + 4C + 3)y = 0    (2)                                            

The roots of Eq. (2) are ‒3, and ‒1, which are real and unequal. Hence, we use Solution I:

                                    y(n) = A(‒3)ⁿ + B(‒1)ⁿ  (3)                                          

            The constants A and B are evaluated with initial conditions. Let the initial conditions be given as y(0) = 1, and y(1) = 0. With n = 0, we get

                                    y(0) = A + B = 1  (4)                                                   

With n = 1, from Eq. (3), we get

                                y(1) = ‒3A ‒ B = 0 (5)                                                    

Solving, we find A = -1/2, B = 3/2. So the solution is

                        y(n) = (‒1/2)(‒3)ⁿ + (3/2)(‒1)ⁿ  (6)

                                      

The solution can be checked by substituting the value of y(n) from Eq. (6) into Eq. (1). For this, we determine y(0),  y(1), and y(2) from Eq. (6). Thus at n = 0, we get

                                   

                        y(0) = ‒1/2 + (3/2) = 1  (7)

                       

With n = 1, Eq. (6) gives

                                                y(1) = (‒1/2)( ‒3) + (3/2)( ‒1) = 0 (8)

                                                                         

When n = 2, Eq. (6) gives

                                            y(1) = (‒1/2)( ‒3)² + (3/2)( ‒1)² = 3 (9)

                                                                       

Now consider Eq. (1) again, which states that

                            y(n) + 4y(n‒1) + 3y(n‒2) = 0      

Now, substituting for y(0),  y(1), and y(2) from Eqs. (7), (8), and (9), respectively, we get

                                                y(2) + 4y(1) + 3y(0) = ‒3+0+3 = 0              

The obtained value of y(n) thus agrees with the given equation. Hence it is a valid solution.

Example 2:  Obtain the solution of the equation

                                                y(n) + 2y(n‒1) + y(n‒2) = 0    (1)

Solution: The characteristic equation is

                               (C² + 2C + 1)y = 0            (2)                                                                

and the roots of the characteristic equation are

           

                           C₁, C₂ = ‒1 and ‒1            (3)                                             

Therefore, the solution is

                                                            y(n) = (‒1)ⁿ (A + Bn)    (4)

                                                                                               

As before, to evaluate A and B, let y(0) = 1, and y(1) = 0. Then with n = 0, we get

                                                                        1 = A                    (5)

and with y(1) = 0, we find

                                                            0 = (‒)(A + B)

                                   

which gives

           

                                    B = -A = -1  (6)                                                                     

Thus the complete solution is therefore

                                    y(n) = (‒1)ⁿ (1‒n)    (7)                                                      

Example 3:  Obtain the solution of the equation

                                                 y(n) + 4y(n‒1) + 8y(n‒2) = 0    (1)

                                                      

Solution: The characteristic equation is

                                    (C² + 4C + 8)y = 0          (2)                                   

           

and the roots of the characteristic equation are

                                                C₁, C₂ = ‒2‒2j, ‒2+2j             (3)

We get, with α ± jβ = ‒ 2 ± j2, we choose Solution III. For this, we find

­                        rⁿ = √(α² + β²) =√(2² + 2²) = 2√2   (4)

                   θ = tan^‒1 (α/β) = tan^‒1 (2/2) = 3π/4  (5)

                                                                        

Using these, the solution can now be written as

            y(n) = (2√2)ⁿ [A cos (3nπ/4) + B sin (3nπ/4)]  (6)

                       

Example 4: Obtain the solution of the equation

                                                    y(n) +y(n‒2) = 0    (1)

                                                      

                                                              

Solution: The characteristic equation is

                                    (C²  + 1)y = 0  (2)                                                      

and the roots of the characteristic equation are

                                    C₁, C₂ = ±j    (3)                                                                     

Since j corresponds to 90^º or π/2, we use Solution IV. Thus

                         y(n) = A cos (nπ/2) + B sin (nπ/2) (4)                              

CHECK: The solution given by Eq. (4) may now be checked as follows. First, as before, n = 0, we have

                                       y(0) = A       (5)                                 

When n = 2

                                            y(2) = A cos (2π/2) + B sin 2nπ/2)

                                                 = A cos (π) = ‒A (6)                      

Substituting for y(0) and y(2) in (4), we get        

                                    y(2) + y(0) =‒A+ A = 0         (7)                 

The result we find in Eq. (7) shows the accuracy of the solution for y(n).

SOLUTIONS OF PROBLEMS IN SECOND-ORDER DIFFERENTIAL EQUATIONS WITH FORCING FUNCTION EQUAL TO ZERO

EDITOR: B. SOMANATHAN NAIR

Example 1: Solve the differential equation

                                           (d²y/dt²) + 4(dy/dt) + 3y = 0  (1)

The boundary conditions given are: at  t = 0, y(0) = 1 and dy/dt = 0.

Solution: The characteristic equation of this differential equation is                                                                                                                       

                                                   (D² + 4D + 3)y = 0  (2)

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

                                                 D₁, D₂ = ‒2 ± 1 = ‒3, ‒1  (3)                                           

where α = ‒4/2 and β = (½)√(16‒12). Since the roots are real and unequal, we choose Solution I. y therefore is:

                               y = Ae^‒3t + Be^‒t   (4)                                 

                   

where A and B are the constants of integration. To evaluate the constants of integration, we now use the boundary conditions y(0) = 1 and dy/dt = 0 at t = 0. Using the first condition in Eq. (4), we get

                                                           1 = A + B (5)

Now, differentiating Eq. (4), and substituting the second condition, we obtain

                                                  dy/dt = 0 = ‒3A ‒ B   (6)

From Eq. (6), we have

                                                  3A = ‒ B    (7)

Solving Eqs. (5) and (7) yields A = ‒½ and B = 3/2. The complete solution, therefore, is

y = ‒ (1/2) e^‒3t + (3/2)e^‒t   (8)

Note: Reason for using the boundary conditions

Differential equations are equations governing the operation of systems under working conditions. As such they represent continuously varying (i.e., dynamic) functions. So we can not use these variable quantities to evaluate constants of integration; we have to have stable values to evaluate them. Therefore we use boundary conditions related to the system under consideration to evaluate them. This is because at boundaries of the system, values are constant and yet the differential equation will be satisfied at these boundaries also as they are part of the system.  

Example 2: Solve the differential equation

                                       (d²y/dt²) + 4(dy/dt) + 4y = 0  (1)

Solution:  The characteristic equation is

(D² + 4D + 4)y = 0  (2)

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

                                                          D₁, D₂ = ‒2, ‒2  (3)                                     

where α = ‒2 and β = 0. Since the roots are real and equal, we choose Solution II. y therefore is:

                                    y = e^‒2t (A+Bt)   (4)                              

where A and B are the constants of integration.

Example 3: Solve the differential equation

                                         (d²y/dt²) + 4(dy/dt) + 8y = 0  (1)                                                                          

Solution:  The characteristic equation of this equation can be written as

                                                   (D² + 4D + 8)y = 0  (2)

The roots are ‒2 + 2j and ‒2 ‒ 2j, respectively. So, we choose solution III, which yields

                        y = e^‒2t(Ae^j2t+Be^‒j2t)= e^‒2t(C cos 2t + D sin2t)  (3)        

                                                                                                  

Example 4:  Solve the differential equation

                                                 (d²y/dt²) + 4y = 0  (1)

Solution:  The characteristic equation is

                     

          (D² + 4)y = 0  (2)

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

                                                          D₁, D₂ = ‒2j, 2j  (3)                                     

where α = 0   and β = ‒2j, 2j. Since the roots are imaginary and equal, we choose Solution IV. y therefore is:

                                                y = A cos 2t + B sin2t  (4) 

                                       

where A and B are the constants of integration.