Find a general solution to the given Cauchy-Euler equation.

(a) \(4{x^3}{y^{\prime \prime \prime }} + 8{x^2}{y^{\prime \prime }} - x{y^\prime } + y = 0,\quad x > 0\)

(b) \({x^3}{y^{\prime \prime \prime }} + 2{x^2}{y^{\prime \prime }} + 2x{y^\prime } + 4y = 0,\quad x > 0\)

Consider the given equation:

\(4{x^3}{y^{\prime \prime \prime }} + 8{x^2}{y^{\prime \prime }} - x{y^\prime } + y = 0\)

To solve this equation, we insert \(y(x) = {x^r}\) in this equation, but first, we find the derivatives.

\(\begin{array}{c}{y^\prime }(x) = r{x^{r - 1}}\\{y^{\prime \prime }}(x) = r(r - 1){x^{r - 2}}\\{y^{\prime \prime \prime }}(x) = r(r - 1)(r - 2){x^{r - 3}}\end{array}\)

Now we substitute this into the given equation and get,

\(\begin{array}{c}4{x^3}r(r - 1)(r - 2){x^{r - 3}} + 8{x^2}r(r - 1){x^{r - 2}} - xr{x^{r - 1}} + {x^r} = 0\\4r(r - 1)(r - 2){x^r} + 8r(r - 1){x^r} - r{x^r} + {x^r} = 0\end{array}\)

By dividing the last equation with \({{\bf{x}}^r}\) we get the characteristic equation,

\(\begin{array}{c}4r(r - 1)(r - 2) + 8r(r - 1) - r + 1 = 0\\4{r^3} - 4{r^2} - r + 1 = 0\\4{r^2}(r - 1) - (r - 1) = 0\\(r - 1)\left( {4{r^2} - 1} \right) = 0\\(r - 1)(2r - 1)(2r + 1) = 0.\end{array}\)

The roots of the characteristic equation are \({r_1} = 1,{r_2} = 1/2\)and \({r_3} = - 1/2\). Therefore, a solution to the given homogeneous equation is

\(y(x) = {c_1}x + {c_2}{x^{\frac{1}{2}}} + {c_3}{x^{ - \frac{1}{2}}}.\)

Consider the given equation,

\({x^3}{y^{\prime \prime \prime }} + 2{x^2}{y^{\prime \prime }} + 2x{y^\prime } + 4y = 0\)

To solve this equation, we insert \(y(x) = {x^r}\) in this equation, but first, we find the derivatives:

\(\begin{array}{c}{y^\prime }(x) = r{x^{r - 1}}{y^{\prime \prime }}(x)\\ = r(r - 1){x^{r - 2}}{y^{\prime \prime \prime }}(x)\\ = r(r - 1)(r - 2){x^{r - 3}}\end{array}\)

Now we substitute this into the given equation and get:

\(\begin{array}{*{20}{r}}{{x^3}r(r - 1)(r - 2){x^{r - 3}} + 2{x^2}r(r - 1){x^{r - 2}} + 2xr{x^{r - 1}} + 4{x^r} = 0}\\{r(r - 1)(r - 2){x^r} + 2r(r - 1){x^r} + 2r{x^r} + 4{x^r} = 0}\end{array}\)

By dividing the last equation with \({x^r}\) we get the characteristic equation,

\(\begin{array}{*{20}{r}}{r(r - 1)(r - 2) + 2r(r - 1) + 2r + 4 = 0}\\{{r^3} - {r^2} + 2r + 4 = 0}\end{array}\)

By inspection, we find that \(r = - 1\) is a root, and using polynomial division we get:

\(\begin{array}{*{20}{r}}{(r + 1)\left( {{r^2} - 2r + 4} \right) = 0}\\{(r + 1){{(r - 2)}^2} = 0}\end{array}\)

The roots of the characteristic equation are \({r_1} = - 1\)and\({r_2} = {r_3} = 2\). Therefore, a solution to the given homogeneous equation is:

\(y(x) = {c_1}{x^{ - 1}} + {c_2}{x^2} + {c_3}{x^2}\ln x\)