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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 453
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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Short Answer

Suppose r0 is a repeated root of the auxiliary equation ar2+br+c=0. Then, as we well know, is a solution to the equation ay"+by'+cy=0 where a, b, and c are constants. Use a derivation similar to the one given in this section for the case when the indicial equation has a repeated root to show that a second linearly independent solution is y2 (t)=tert .

The second solution for the differential equation is derived to be y2=ter0t where the first solution is given to be y1=er0t .

See the step by step solution

Step by Step Solution

Definition of Cauchy-Euler equation:

It has direct application to Fourier's approach in the study of partial differential equations, the Cauchy-Euler equation is significant in the theory of linear differential equations.

Prove that mixed partial derivatives are equal:

It is given that for the differential equation

ay"+by'+cy=0

Where a, b, c are constants, the characteristic equation is,

ar2+br+c=0

For which there is a repeated root at r0 and one solution of the equation is

y1=er0t

Using operator approach, if r0 is a repeated root, then

L[w] (et) = a (r-r0)2 ert

You observe that, the righthand side of the above equation has a factor (r-r0)2, so taking the partial derivative, w.r.t. r and setting r=r0 you will get 0,

𝛿 /𝛿r [L [w] (t)] |r=r0 = [a(r-r0)2 (rert)+2a (r-r0) ert]|r=r0 =0

You also note that w(r,et)=ert has continuous partial derivatives of all orders with respect to both and hence the mixed partial derivates are equal.

𝛿2w/𝛿r𝛿t2 = 𝛿2w/𝛿r2𝛿t ,

𝛿2w/𝛿r𝛿t = 𝛿2w/𝛿r𝛿t

Find the second solution:

Consequently, for a differential operator L, you have,

𝛿/𝛿r L[w] = 𝛿/𝛿r {a 𝛿2w/𝛿t2 +b 𝛿w/𝛿t+cw}

=a 𝛿3w/𝛿r𝛿t2 +b 𝛿2w/𝛿r𝛿t +c 𝛿w/𝛿r

=a 𝛿3w/𝛿t2𝛿r +b 𝛿2w/𝛿t𝛿r +c 𝛿w/𝛿r

=L [𝛿w/𝛿r]

From above, you can say that the operators 𝛿/𝛿r and L are commutative. Thus,

L [𝛿w/𝛿r]|r=r0 =0

Therefore, in the case of repeated roots at r0, the second linearly independent solution is,

y2(x) = 𝛿w/𝛿r (r0 , et)

=𝛿/𝛿r (ert) |r=r0

=ter0 t

Hence, you find the second solution for the differential equation is derived to be y2=ter0 t where the first solution is given to be y1=er0 t .

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