The differential equation is \(x^2y''-5xy'+9y=0\) and one solution is \(y_1=x^3\)

Now, let the new solution be \(y=vy_1=vx^3\) where v is again a function of x.

Then,

\(y=vx^3\)

\(y'=3vx^2+x^3v'\)

\(y''=3(2vx+x^2v')+x^3v''+3x^2v'\)

\(=6vx+3x^2v'+x^3v''+3x^2v'\)

\(=6vx+6x^2v'+x^3v''\)

Substitute the values of \(y, y'\) and \(y''\) in \(x^2y''-5xy'+9y=0\) and obtain the differential equation in terms of v.

\(x^2(6vx+6x^2v'+x^3v'')-5x(3vx^2+x^3v')+9vx^3=0\)

\(6vx^3+6x^4v'+x^5v''-15x^3v-5x^4v'+9vx^3=0\)

\(x^5v''+x^4v'=0\)

Now let \(w=v'\).Then the equation becomes \(x^5w'+x^4w=0\)

Solve the equation \(x^5w'+x^4w=0\)

\(x^5\frac{dw}{dx}+x^4w=0\)

\(x^5\frac{dw}{dx}=-x^4w\)

\(\frac{dw}{w}=\frac{-1}{x}dx\)

\(\int\frac{dw}{w}=-\int\frac{1}{x}dx\)

\(\ln|w|=-\ln|x|+c\)

\(e^{\ln|w|}=e^{-\ln|x|+c}\)

\(w=c_1x^{-1}\)

\(v'=c_1x^{-1}\)

Integrate \(v'=c_1x^{-1}\) and obtain the value of v.

\(v=\int c_1 x^{-1}\)

\(=c_1\ln|x|+c_2\)

Therefore, the general solution is \(y=c_1x^3\ln x +c_2 x^3\)

The second solution is \(y_2=c_1x^3\ln|x|\)

Now, let the new solution be \(y=vy_1=vx^3\) where v is again a function of x.

Then,

\(y=vx^3\)

\(y'=3vx^2+x^3v'\)

\(y''=3(2vx+x^2v')+x^3v''+3x^2v'\)

\(=6vx+3x^2v'+x^3v''+3x^2v'\)

\(=6vx+6x^2v'+x^3v''\)

Substitute the values of \(y, y'\) and \(y''\) in \(x^2y''-5xy'+9y=0\) and obtain the differential equation in terms of v.

\(x^2(6vx+6x^2v'+x^3v'')-5x(3vx^2+x^3v')+9vx^3=0\)

\(6vx^3+6x^4v'+x^5v''-15x^3v-5x^4v'+9vx^3=0\)

\(x^5v''+x^4v'=0\)

Now let \(w=v'\).Then the equation becomes \(x^5w'+x^4w=0\)

Solve the equation \(x^5w'+x^4w=0\)

\(x^5\frac{dw}{dx}+x^4w=0\)

\(x^5\frac{dw}{dx}=-x^4w\)

\(\frac{dw}{w}=\frac{-1}{x}dx\)

\(\int\frac{dw}{w}=-\int\frac{1}{x}dx\)

\(\ln|w|=-\ln|x|+c\)

\(e^{\ln|w|}=e^{-\ln|x|+c}\)

\(w=c_1x^{-1}\)

\(v'=c_1x^{-1}\)

Integrate \(v'=c_1x^{-1}\) and obtain the value of v.

\(v=\int c_1 x^{-1}\)

\(=c_1\ln|x|+c_2\)

Therefore, the general solution is \(y=c_1x^3\ln x +c_2 x^3\)

The second solution is \(y_2=c_1x^3\ln|x|\)