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Vice-versa, if you choose to use the negative radical for the value of "m" in equation 11, then the

corresponding value of "n" in equation 12 uses the positive radical.

The multiplier form “m” and “n” arise in the rework of the left side of equation 8 which is,

again:

(D - m)(D - n)y

(13)

So the m and n are interchangeable as far as the two parenthesized operators are concerned, and

you can use either set of m and n solutions in equation 13.

All of this means that equation 5 can be rewritten as:

(D m)(Dn)y = 0

(14)

in which m and n are the roots given by equation 11 and 12 of the quadratic equation 10. Since

equation 10 is controlled by the b and c coefficients of equation 5, we can write:

(D-r₁) (D-r₂)y = 0

(15)

where r and r₂ are the roots of the quadratic equation controlled by the “b” and “c” coefficients

of equation 5.

Most ODE textbooks glide by the extended analysis given on pages 2 and 3 by simply setting the

bracketed quantity on the left side of equation 8 to zero and by treating the D operator as an

unknown. We have:

(D² + b + c)y = 0

(16)/nENGG 614 2023

Seventh Class 10-17-23

Revised 10-16-23, 12:00 PM

whose roots r₁ and r₂ are:

2

Page 4 of 19

}

which are identical to the solutions for m and n on pages 2 and 3. Actually, this just recognizes

that the solution for y is e**, which when substituted in the ODE yields the characteristic

equation r + br + c = 0. The "D" in equation 16 simply replaces the r in the characteristic

equation.

So we can now write equation 5, the homogeneous complement of equation 2, in operator form

as:

(17)

(Dr.)(D = r₂)y = 0

Notice that equation 16, from which we deriver and r₂, is identical to the characteristic

equation we would have for equation 5, using the earlier "reasoned" approach of assuming a

solution of the form ex.

We now use equation 17 to solve equation 5. We have, following the operator procedure

described in the paragraphs below equation 8:

(Dr.)(Dy - r₂y) = 0

(18)

Fig: 1

Fig: 2