Reduce to first order and solve: x^2y''-5xy'+9y=0 y_1=x^3

Reduce to first order and solve:
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Brighton
The differential equation is ${x}^{2}{y}^{″}-5x{y}^{\prime }+9y=0$ and one solution is ${y}_{1}={x}^{3}$
Now, let the new solution be $y=v{y}_{1}=v{x}^{3}$ where v is again a function of x.
Then,
$y=v{x}^{3}$
${y}^{\prime }=3v{x}^{2}+{x}^{3}{v}^{\prime }$
${y}^{″}=3\left(2vx+{x}^{2}{v}^{\prime }\right)+{x}^{3}{v}^{″}+3{x}^{2}{v}^{\prime }$
$=6vx+3{x}^{2}{v}^{\prime }+{x}^{3}{v}^{″}+3{x}^{2}{v}^{\prime }$
$=6vx+6{x}^{2}{v}^{\prime }+{x}^{3}{v}^{″}$
Substitute the values of $y,{y}^{\prime }$ and ${y}^{″}$ in ${x}^{2}{y}^{″}-5x{y}^{\prime }+9y=0$ and obtain the differential equation in terms of v.
${x}^{2}\left(6vx+6{x}^{2}{v}^{\prime }+{x}^{3}{v}^{″}\right)-5x\left(3v{x}^{2}+{x}^{3}{v}^{\prime }\right)+9v{x}^{3}=0$
$6v{x}^{3}+6{x}^{4}{v}^{\prime }+{x}^{5}{v}^{″}-15{x}^{3}v-5{x}^{4}{v}^{\prime }+9v{x}^{3}=0$
${x}^{5}{v}^{″}+{x}^{4}{v}^{\prime }=0$
Now let $w={v}^{\prime }$.Then the equation becomes ${x}^{5}{w}^{\prime }+{x}^{4}w=0$
Solve the equation ${x}^{5}{w}^{\prime }+{x}^{4}w=0$
${x}^{5}\frac{dw}{dx}+{x}^{4}w=0$
${x}^{5}\frac{dw}{dx}=-{x}^{4}w$
$\frac{dw}{w}=\frac{-1}{x}dx$
$\int \frac{dw}{w}=-\int \frac{1}{x}dx$
$\mathrm{ln}|w|=-\mathrm{ln}|x|+c$
${e}^{\mathrm{ln}|w|}={e}^{-\mathrm{ln}|x|+c}$
$w={c}_{1}{x}^{-1}$
${v}^{\prime }={c}_{1}{x}^{-1}$
Integrate ${v}^{\prime }={c}_{1}{x}^{-1}$ and obtain the value of v.
$v=\int {c}_{1}{x}^{-1}$
$={c}_{1}\mathrm{ln}|x|+{c}_{2}$
Therefore, the general solution is $y={c}_{1}{x}^{3}\mathrm{ln}x+{c}_{2}{x}^{3}$
The second solution is ${y}_{2}={c}_{1}{x}^{3}\mathrm{ln}|x|$