Question

The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.2, y2 = y1(x e??p(x dx y 2 1 (x dx (5 as instructed, to find a second solution y2(x. xy'' y' = 0; y1 = ln x

Answer 1

Given a solution $y_1(x)=\ln x$, we can attempt to find a solution of the form $y_2(x)=v(x)y_1(x)$. We have derivatives

$y_2=v\ln x$
${y_2}'=v'\ln x+\dfrac vx$
${y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}$

Substituting into the ODE, we get

$v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0$
$v''x\ln x+(2+\ln x)v'=0$

Setting $w=v'$, we end up with the linear ODE

$w'x\ln x+(2+\ln x)w=0$

Multiplying both sides by $\ln x$, we have

$w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0$

and noting that

$\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x$

we can write the ODE as

$\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0$

Integrating both sides with respect to $x$, we get

$wx(\ln x)^2=C_1$
$w=\dfrac{C_1}{x(\ln x)^2}$

Now solve for $v$:

$v'=\dfrac{C_1}{x(\ln x)^2}$
$v=-\dfrac{C_1}{\ln x}+C_2$

So you have

$y_2=v\ln x=-C_1+C_2\ln x$

and given that $y_1=\ln x$, the second term in $y_2$ is already taken into account in the solution set, which means that $y_2=1$, i.e. any constant solution is in the solution set.