Question

In Problems 11 and 12, use a substitution of the form to find a general solution to the given equation for x>c.

2(x-3)² y"+ 5(x-3)y'-2y=0

Short answer

The general solution for the given equation is y=c₁ (x-3)^1/2+c₂ (x-3)^-2.

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 an example of the Cauchy problem.

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

Find the general solution:

The given equation is

2(x-3)² y"+ 5(x-3)y'-2y=0

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

L[y] (x)= 2(x-3)² y"+ 5(x-3)y'-2y

Let set

w(r,x) = (x-3)^r

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

L[w] (x)= 2(x-3)² ((x-3)^r)"+ 5(x-3) ((x-3)^r)'-2 (x-3)^r

= 2(x-3)² (r(r-1))(x-3)^r-2+ 5(x-3)(r)((x-3)^r-1)-2 (x-3)^r

=2(r²-r)(x-3)^r+5r (x-3)^r-2(x-3)^r

=(2r²+3r-2) (x-3)^r

Solving the indicial equation,

2r²+3r-2=0

(r-1/2)(r+2)=0

There are two distinct roots,

r₁=1/2 and r₂= -2

Thus there are two linearly independent solutions given by,

y₁=c₁ (x-3)^1/2 and y₂=c₂ (x-3)^-2

Therefore, the general solution for the equation will be,

y=c₁ (x-3)^1/2+c₂ (x-3)^-2