Numerical Analysis Using MATLAB and Spreadsheets

5.4: Solution of the Homogeneous ODE

5.4 Solution of the Homogeneous ODE

Let the solutions of the homogeneous ODE


be of the form


Then, by substitution of (5.20) into (5.19) we get


or


We observe that (5.21) can be satisfied when


but the only meaningful solution is the quantity enclosed in parentheses since the latter two yield trivial (meaningless) solutions. We, therefore, accept the expression inside the parentheses as the only meaningful solution and this is referred to as the characteristic (auxiliary) equation, that is,


`

Since the characteristic equation is an algebraic equation of an nth-power polynomial, its solutions are s 1, s 2, s 3,..., s n and thus the solutions of the homogeneous ODE are:


Case I - Distinct Roots

If the roots of the characteristic equation are distinct (different from each another), the n solutions of (5.23) are independent and the most general solution is:


Case II - Repeated Roots

If two or more roots of the characteristic equation are repeated (same roots), then some of the terms of (5.24) are not independent and therefore (5.25) does not represent the most general solution. If, for example, s 1 = s 2, then,


and we see that one term of (5.25) is lost. In this case, we express one of the terms of (5.25), say k 2e s 1t as k 2te s 1t. These two represent two independent solutions and therefore the most general solution...

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