Question

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

Answer 1

Standard reduction of order procedure: suppose there is a second solution of the form $y_2(x)=v(x)y_1(x)$, which has derivatives

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

Substitute these terms into the ODE:

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

and replacing $v'=w$, we have an ODE linear in $w$:

$w'x^6+wx^5=0$

Divide both sides by $x^5$, giving

$w'x+w=0$

and noting that the left hand side is a derivative of a product, namely

$\dfrac{\mathrm d}{\mathrm dx}[wx]=0$

we can then integrate both sides to obtain

$wx=C_1$
$w=\dfrac{C_1}x$

Solve for $v$:

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

Now

$y=C_1x^4\ln|x|+C_2x^4$

where the second term is already accounted for by $y_1$, which means $y_2=x^4\ln x$, and the above is the general solution for the ODE.