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Expert-verified Found in: Page 450 ### 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 1-10, use a substitution y=xr to find the general solution to the given equation for x>0.x2y"+xy'(x)+17y=0

The general solution for the given equation is y=c1 x cos (4 lnx)+c2 x sin(4 lnx)..

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 both examples of Cauchy problems.

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

## Find the general solution:

The given equation is

x2y"+xy'(x)+17y=0

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

L[y](x)=x2y"+6xy'+6y

And let's set

w(r,x)=xr

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

L[y](x)=x2(xr)"-x(xr)'+17(xr)

=x2 (r(r-1)) xr-2-x(r)xr-1+17xr

=(r2-r) xr-rxr +17xr

=(r2-2r+17) xr

Solving the indicial equation

r2-2r+17=0

r=[2± √(4-4×17)]/2

=(2± 8i)/2

=1±4i

There are two complex conjugates,

r1=1+4i and r2=1-4i

Thus there are two linearly independent solutions given by

x1+4i=x1 cos(4 ln x)+ix1 sin(4 ln x)

You can write two linearly independent real-valued solutions as,

y1=x cos(4 ln x) and y2=x sin (4 ln x)

The general solution for the equation will be,

y=c1 x cos (4 lnx)+c2 x sin(4 lnx)

Therefore, the general solution for the given equation is y=c1 x cos (4 lnx)+c2 x sin(4 lnx). ### Want to see more solutions like these? 