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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. d2y/dx2=1/x dy/dx-4/x2 y

The general solution for the given equation is y=c1 x cos(√3 ln x) +c2 x sin (√3 ln x) .

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=1/x dy/dx-4/x2 y

This can be re-written in the required form as,

x2y"-xy'+4y=0

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

L[y](x) =x2y"-xy'+4y

And let's set

w(r,x)=xr

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

L[w](x)=x2(xr)"-x(xr)'+4(xr)

=x2 (r(r-1)) xr-2 -x(r) xr-1 +4xr

=(r2-r) xr-rxr +4xr

=(r2-2r+4) xr

Solving the indicial equation,

r2-2r+4=0

r=[2± √(4-4×4)]/2

=(2± 2√3 i)/2

=1±√3i

There are two complex conjugates,

r1=1+√3i and r2=1-√3i

Thus there are two linearly independent solutions given by,

x3+2i = x cos(√3 ln x)+ix sin(√3 ln x)

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

y1=x cos(√3 ln x) and y2=x sin (√3 ln x)

Therefore, the general solution for the equation will be,

y=c1 x cos(√3 ln x) +c2 x sin (√3 ln x)