ENGI 2422 Engineering Mathematics 2
Additional tutorial examples for
Chapter 5 (Laplace transforms)

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Find the complete solution to each of the following initial value problems

1.  

   

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   then the Laplace transform of the initial value problem is
   
   
   
   Therefore the complete solution is

   

   We can check that this solution is correct:
   
   Substituting into the left-hand side of the ODE:
   Ö
   Check the initial conditions:
   Ö

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   The alternative method [from Chapter 4] for the solution of this question is much longer:
   
   
   Using the method of undetermined coefficients to find the particular solution:
   
   
   This equation must be true for all values of t
   
   The general solution is
   
   Imposing the initial conditions,
   
   
   
   Therefore the complete solution is

   

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2. [Modification of example 4.3.3]

   

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   then the Laplace transform of the initial value problem is
   
   
   
   
   
   
   
   
   Therefore the complete solution is

   

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3.  

   

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   then the Laplace transform of the initial value problem is
   
   
   
   [Note:   one should always check for common factors between the numerator and denominator, just before resolving   Y(s)   into partial fractions.   There are no common factors here (unlike question 1 above).]
   All three denominators are non-repeated linear terms.
   Therefore the cover-up rule may be used to find all three values   a, b and c.
   
   
   
   
   Therefore the complete solution is

   

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4. [There was insufficient time to demonstrate this example during the tutorial.]

   

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   then the Laplace transform of the initial value problem is
   
   
   
   Note that this form of partial fractions is used, because
   
   The cover-up rule may be used to find a only.
   
   Clearing the denominators:
   
   Matching coefficients of powers of s:
   
   
   
   Therefore the complete solution is

   

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   Alternative solution:

   

   
   
   Using the method of undetermined coefficients to find the particular solution:
   
   
   The general solution is
   
   Imposing the initial conditions,
   
   
   
   Therefore the complete solution is