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.