In Problems 1-10, use the substitution y=x^rto find a general solution to the given equation for x>0.

x³y"'+9x²y"+19xy'+8y=0

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.

The given equation is

x³y"'+9x²y"+19xy'+8y=0

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

L [y] (x)= x³y"'+9x²y"+19xy'+8y

Let set

w(r,x) = x^r

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

L[w] (x)= x³(x^r )"'+9x²(x^r )"+19x(x^r ) '+8(x^r )

= x³(x^r )"'+9x²(x^r )"+19x(x^r)'+8(x^r )

=x³[r(r-1)(r-2)]( x^r-3 )+9x^2[r (r-1)] (x^r-2)+19x(r)(x^r-1 )+8(x^r )

=(r^3-3r²+2r) x^r+9(r²-r) x^r +15r x^r+8x^r

=(r³++6r²+12r+8) x^r

Solving the indicial equation,

r³++6r²+12r+8=0

By substitution , you can identify that r= -2 is a solution of the equation,

(r+2) (r²+4r+4)=(r+2)(r+2)²

(r+2)³= r= -2, -2, -2

There are three simultaneous roots at x=-2, thus you can write three linearly independent real-valued solutions as,

y₁=x^-2

y₂=x^-2 (lnx)

y₃=x^-2 (lnx)²

Therefore, the general solution for the equation will be,

y= c₁x+c₂x^-2 cos(lnx) +c₃x^-2 (lnx)²