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

x²y"+xy'(x)+17y=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

x²y"+xy'(x)+17y=0

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

L[y](x)=x²y"+6xy'+6y

And let's set

w(r,x)=x^r

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

L[y](x)=x²(x^r)"-x(x^r)'+17(x^r)

=x²(r(r-1)) x^r-2-x(r)x^r-1+17x^r

=(r²-r) x^r-rx^r +17x^r

=(r²-2r+17) x^r

Solving the indicial equation

r²-2r+17=0

r=[2± √(4-4×17)]/2

=(2± 8i)/2

=1±4i

There are two complex conjugates,

r₁=1+4i and r₂=1-4i

Thus there are two linearly independent solutions given by

x¹⁺⁴ⁱ=x¹ cos(4 ln x)+ix¹ sin(4 ln x)

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

y₁=x cos(4 ln x) and y₂=x sin (4 ln x)

The general solution for the equation will be,

y=c₁ x cos (4 lnx)+c₂ x sin(4 lnx)

Therefore, the general solution for the given equation is y=c₁ x cos (4 lnx)+c₂ x sin(4 lnx).