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Q18E

Expert-verifiedFound in: Page 453

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**Suppose r _{0}**

The second solution for the differential equation is derived to be y_{2}=te^{r0}^{t} where the first solution is given to be y_{1}=e^{r0}^{t} .

**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.**

It is given that for the differential equation

ay"+by'+cy=0

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

ar^{2}+br+c=0

For which there is a repeated root at r_{0} and one solution of the equation is

y_{1}=e^{r0}^{t}

Using operator approach, if r_{0} is a repeated root, then

L[w] (e^{t}) = a (r-r_{0})^{2} e^{rt}

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

𝛿 /𝛿r [L [w] (t)] |_{r=r0 }= [a(r-r_{0})^{2} (re^{rt})+2a (r-r_{0}) e^{rt}]|_{r=r0 }=0

You also note that w(r,e^{t})=e^{rt} has continuous partial derivatives of all orders with respect to both and hence the mixed partial derivates are equal.

𝛿^{2}w/𝛿r𝛿t^{2} = 𝛿^{2}w/𝛿r^{2}𝛿t ,

𝛿^{2}w/𝛿r𝛿t = 𝛿^{2}w/𝛿r𝛿t

Consequently, for a differential operator L, you have,

𝛿/𝛿r L[w] = 𝛿/𝛿r {a 𝛿^{2}w/𝛿t^{2} +b 𝛿w/𝛿t+cw}

=a 𝛿^{3}w/𝛿r𝛿t^{2} +b 𝛿^{2}w/𝛿r𝛿t +c 𝛿w/𝛿r

=a 𝛿^{3}w/𝛿t^{2}𝛿r +b 𝛿^{2}w/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 r_{0}, the second linearly independent solution is,

y_{2}(x) = 𝛿w/𝛿r (r_{0} , e^{t})

=𝛿/𝛿r (e^{rt}) |_{r=r0}

=te^{r0 }^{t}

Hence, you find the second solution for the differential equation is derived to be y_{2}=te^{r0 }^{t} where the first solution is given to be y_{1}=e^{r0 }^{t} .

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