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

x²y"+2xy'-3y=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"+2xy'-3y=0

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

L[y](x)=x²y"+2xy'-3y

Let's see,

w(r,x)=x^r

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

L[y](x)=x²(x^r )"+2x(x^r)'-3(x^r)

=x² (r(r-1)) x^r-2+2x (r) x^r-1-3 (x^r)

=(r²-r) x^r+2rx^r-3x^r

=(r²+r-3) x^r

Solving the indicial equation

r²+r-3=0

r= [-1±√(1+12)]/2

r= -1/2±√13/2

There are two distinct roots,

r₁= -1/2+√13/2 and r₂= -1/2-√13/2

Thus there are two linearly independent solutions given by,

y₁=c₁x^-1/2+√13/2 and y₂=c₂x^-1/2-√13/2

Hence, the general solution for the given equation will be,

y=c₁x^-1/2+√13/2 +c₂x^-1/2-√13/2