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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 # In Problems 11 and 12, use a substitution of the form to find a general solution to the given equation for x>c.2(x-3)2 y"+ 5(x-3)y'-2y=0

The general solution for the given equation is y=c1 (x-3)1/2+c2 (x-3)-2.

See the step by step solution

## Define Cauchy-Euler equations:

In mathematics, a Cauchy problem is one in which the solution to a partial differential equation must satisfy specific constraints specified on a hypersurface in the domain.

An initial value problem or a boundary value problem is an example of the Cauchy problem.

The equation will be in the form of, ax2y"+bxy'+cy=0.

## Find the general solution:

The given equation is

2(x-3)2 y"+ 5(x-3)y'-2y=0

Let L be the differential operator defined by the left-hand side of equation, that is

L[y] (x)= 2(x-3)2 y"+ 5(x-3)y'-2y

Let set

w(r,x) = (x-3)r

Substituting w(r,x) in place of y(x), you get

L[w] (x)= 2(x-3)2 ((x-3)r)"+ 5(x-3) ((x-3)r)'-2 (x-3)r

= 2(x-3)2 (r(r-1))(x-3)r-2+ 5(x-3)(r)((x-3)r-1)-2 (x-3)r

=2(r2-r)(x-3)r+5r (x-3)r-2(x-3)r

=(2r2+3r-2) (x-3)r

Solving the indicial equation,

2r2+3r-2=0

(r-1/2)(r+2)=0

There are two distinct roots,

r1=1/2 and r2= -2

Thus there are two linearly independent solutions given by,

y1=c1 (x-3)1/2 and y2=c2 (x-3)-2

Therefore, the general solution for the equation will be,

y=c1 (x-3)1/2+c2 (x-3)-2 ### Want to see more solutions like these? 