Question

X ^ (2) y '' - 7xy '+ 16y = 0, y1 = x ^ 4

Answer 1

Given a solution $y_1(x)=x^4$, we can attempt to find another via reduction of order of the form $y_2(x)=x^4v(x)$. This has derivatives

${y_2}'=4x^3v+x^4v'$
${y_2}''=12x^2v+8x^3v'+x^4v''$

Substituting into the ODE yields

$x^2(x^4v''+8x^3v'+12x^2v)-7x(x^4v'+4x^3v)+16x^4v=0$
$x^6v''+(8x^5-7x^5)v'+(12x^4-28x^4+16x^4)v=0$
$x^6v''+x^5v'=0$

Now letting $u(x)=v'(x)$, so that $u'(x)=v''(x)$, you end up with the ODE linear in $u$

$x^6u'+x^5u=0$

Assuming $x\neq0$ (which is reasonable, since $x=0$ is a singular point), you can divide through by $x^5$ and end up with

$xu'+u=(xu)'=0$

and integrating both sides with respect to $x$ gives

$xu=C_1\implies u=\dfrac{C_1}x$

Back-substitute to solve for $v$:

$v'=\dfrac{C_1}x\implies v=C_1\ln|x|+C_2$

and again to solve for $y$:

$y=x^4v\implies \dfrac y{x^4}=C_1\ln|x|+C_2$
$\implies y=C_1\underbrace{x^4\ln|x|}_{y_2}+C_2\underbrace{x^4}_{y_1}$

Alternatively, you can solve this ODE from scratch by employing the Euler substitution (which works because this equation is of the Cauchy-Euler type), $t=\ln x$. You'll arrive at the same solution, but it doesn't hurt to know there's more than one way to solve this.