Solution
a. If \(y=ux\), then \(y'=u'x+u\) and \(y''=u''x+2u'\), so
\[\begin{aligned} x^2y''+xy'-y&=x^2(u''x+2u')+x(u'x+u)-ux\\ &=x^3u''+3x^2u'.\end{aligned}\]
Therefore \(y=ux\) is a solution of Equation \ref{eq:5.6.12} if and only if
\[x^3u''+3x^2u'=x^2+1, \nonumber\]
which is a first order equation in \(u'\). We rewrite it as
\[\label{eq:5.6.13} u''+{3\over x}u'={1\over x}+{1\over x^3}.\]
To focus on how we apply variation of parameters to this equation, we temporarily write \(z=u'\), so that Equation \ref{eq:5.6.13} becomes
\[\label{eq:5.6.14} z'+{3\over x}z={1\over x}+{1\over x^3}.\]
We leave it to you to show by separation of variables that \(z_1=1/x^3\) is a solution of the complementary equation
\[z'+{3\over x}z=0 \nonumber\]
for Equation \ref{eq:5.6.14}. By variation of parameters, every solution of Equation \ref{eq:5.6.14} is of the form
\[z={v\over x^3} \quad \text{where} \quad {v'\over x^3}={1\over x}+{1\over x^3}, \quad \text{so} \quad v'=x^2+1 \quad \text{and} \quad v={x^3\over 3}+x+C_1. \nonumber\]
Since \(u'=z=v/x^3\), \(u\) is a solution of Equation \ref{eq:5.6.14} if and only if
\[u'={v\over x^3}={1\over3}+{1\over x^2}+{C_1\over x^3}.\nonumber\]
Integrating this yields
\[u={x\over 3}-{1\over x}-{C_1\over2x^2}+C_2.\nonumber\]
Therefore the general solution of Equation \ref{eq:5.6.12} is
\[\label{eq:5.6.15} y=ux={x^2\over 3}-1-{C_1\over2x}+C_2x.\]
Reasoning as in the solution of Example \(\PageIndex{1a}\), we conclude that \(y_1=x\) and \(y_2=1/x\) form a fundamental set of solutions for Equation \ref{eq:5.6.11}.
As we explained above, we rename the constants in Equation \ref{eq:5.6.15} and rewrite it as
\[\label{eq:5.6.16} y={x^2\over3}-1+c_1x+{c_2\over x}.\]
b. Differentiating Equation \ref{eq:5.6.16} yields
\[\label{eq:5.6.17} y'={2x\over 3}+c_1-{c_2\over x^2}.\]
Setting \(x=1\) in Equation \ref{eq:5.6.16} and Equation \ref{eq:5.6.17} and imposing the initial conditions \(y(1)=2\) and \(y'(1)=-3\) yields
\[\begin{aligned} c_1+c_2&= \phantom{-}{8\over 3} \\ c_1-c_2&= -{11\over 3}.\end{aligned}\]
Solving these equations yields \(c_1=-1/2\), \(c_2=19/6\). Therefore the solution of Equation \ref{eq:5.6.12} is
\[y={x^2\over 3}-1-{x\over 2}+{19\over 6x}.\nonumber\]