1 Answer

Matt Austin · Answer 1 · 2020-08-02T05:21:37+0000

For the corresponding homogeneous equation,

x^2y''+xy'+y=0

we can look for a solution of the form
y=x^m , with derivatives
y'=mx^(m-1) and
y''=m(m-1)x^(m-2) . Substituting these into the ODE gives

$m(m-1)x^m+mx^m+x^m=0\implies m^2+1=0\implies m=\pm i$

which admits two solutions,
y_1=x^i and
y_2=x^(-i) , which we can write as

$x^i=e^(\ln x^i)=e^(i\ln x)=\cos(\ln x)+i\sin(\ln x)$

and by the same token,

$x^(-i)=\cos(\ln x)-i\sin(\ln x)$

so we see two independent solutions that make up the characteristic solution,

$y_c=C_1\cos(\ln x)+C_2\sin(\ln x)$

For the non-homogeneous ODE, we make the substitution

$x=e^t\iff t=\ln x$

so that by the chain rule, the first derivative becomes

$(\mathrm dy)/(\mathrm dx)=(\mathrm dy)/(\mathrm dt)(\mathrm dt)/(\mathrm dx)=(\mathrm dy)/(\mathrm dt)\frac1x$

$(\mathrm dy)/(\mathrm dx)=e^(-t)(\mathrm dy)/(\mathrm dt)$

Let
$f(t)=(\mathrm dy)/(\mathrm dx)$ . Then the second derivative becomes

$(\mathrm d^2y)/(\mathrm dx^2)=(\mathrm df)/(\mathrm dx)=(\mathrm df)/(\mathrm dt)(\mathrm dt)/(\mathrm dx)=\left(-e^(-t)(\mathrm dy)/(\mathrm dt)+e^(-t)(\mathrm d^2y)/(\mathrm dt^2)\right)\frac1x$

$(\mathrm d^2y)/(\mathrm dx^2)=e^(-2t)\left((\mathrm d^2y)/(\mathrm dt^2)-(\mathrm dy)/(\mathrm dt)\right)$

Substituting these into the ODE gives

$e^(2t)\left(e^(-2t)\left((\mathrm d^2y)/(\mathrm dt^2)-(\mathrm dy)/(\mathrm dt)\right)\right)+e^t\left(e^(-t)(\mathrm dy)/(\mathrm dt)\right)+y=t^2+2e^t$