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Expert-verified Found in: Page 453 ### 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 # 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

## 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 . ### Want to see more solutions like these? 