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### 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.d2y/dx2=5/x dy/dx-13/x2 y

The general solution for the given equation is y=c1x3 cos (2 lnx)+c2x3 sin(2 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

d2y/dx2=5/x dy/dx-13/x2 y

This can be re-written in the required form as

x2y"-5xy'+13y=0

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

L[y](x) = x2y"-5xy'+13y

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 )"-5x(xr)'+13(xr)

=x2 (r(r-1)) xr-2 -5x(r) xr-1 +13xr

=(r2-r) xr -5rxr+13xr

=(r2-6r+13) xr

Solving the indicial equation,

r2-6r+13=0

r=[2± √(36-4×13)]/2

=(6± 4i)/2

=3±2i

There are two complex conjugates,

r1=3+2i and r2=3-2i

Thus there are two linearly independent solutions given by,

x3+2i=x3 cos(2 lnx)+ix3 sin (2 lnx)

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

y1=x3 cos(2 lnx) and y2=x3 sin(2 lnx)

Therefore, the general solution for the equation will be,

y=c1 x3 cos(2 lnx)+c2 x3 sin(2 lnx)