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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. x3y"'+4x2y"+10xy'-10y=0

The general solution for the given equation is y=c1x+c2x-1 cos(3 ln x)+c3 x-1 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

x3y"'+4x2y"+10xy'-10y=0

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

L[y](x)=x3y"'+4x2y"+10xy'-10y

And let's set

w(r,x)=xr

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

L[y](x)=x3(xr)"'+4x2(xr)"+10x(xr)'-10(xr)

=x3 (r(r-1)(r-2)) xr-3 +4x2 (r(r-1)) xr-2 +10x(r) xr-1-10xr

=(r3-3r2+2r) xr+4 (r2-r) xr +10r xr-10xr

=(r3+r2+8r-10) xr

Solving the indicial equation,

r3+r2+8r-10=0

By substitution, we can identify that r=1 is a solution of the equation,

(r-1) (r2+2r+10)=(r-1) [r- (-2± √(4-4×10)]/2)]

=(r-1) [r- (-2±6i)/2]

=(r-1) [r-(-1±3i)]

There is one real root and two complex conjugates,

r1=1

r2= -1+3i

r3= -1-3i

For complex roots there are two linearly independent solutions given by,

x1+3i = x-1 cos (3 ln x)+ix-1 sin(3 ln x)

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

y1=x,

y2=x-1 cos(3 ln x)

y3=x-1 sin(3 ln x)

Therefore, the general solution for the equation will be,

y=c1x+c2x-1 cos(3 ln x)+c3 x-1 sin(3 ln x) ### Want to see more solutions like these? 