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Q6E

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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 450
Fundamentals Of Differential Equations And Boundary Value Problems

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

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Short Answer

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

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)

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