Find the general solution of the Euler equation x^3y''' +…

Find the general solution of the Euler equation $x^3y''' + xy' - y = 0$.

Step 1: The Euler substitution $x=e^t$

With $x=e^t$ ($t=\ln x$, $x>0$) and dots denoting $d/dt$, the reduction identities are

\[ x\,y' = \dot y, \qquad x^2y'' = \ddot y - \dot y, \qquad x^3y''' = \dddot y - 3\ddot y + 2\dot y. \]

Note that this equation has no $x^2y''$ term, so only the first and third identities are needed.

Step 2: Transformed equation and characteristic equation

Substituting into $x^3y'''+xy'-y=0$: \[(\dddot y - 3\ddot y + 2\dot y) + \dot y - y = 0 \implies \dddot y - 3\ddot y + 3\dot y - y = 0.\]

The characteristic equation is

\[ r^3 - 3r^2 + 3r - 1 = 0, \]

which is recognized as a perfect cube, $(r-1)^3=0$, so $p(r)=(r-1)^3$ has $r=1$ as a triple root. Indeed $p(1)=0$, $p'(r)=3(r-1)^2\Rightarrow p'(1)=0$, and $p''(r)=6(r-1)\Rightarrow p''(1)=0$.

Step 3: General solution in $t$

For a triple root $r$, the fundamental system is $e^{rt}, te^{rt}, t^2e^{rt}$. Since $p(1)=p'(1)=p''(1)=0$, each of $e^t,\ te^t,\ t^2e^t$ annihilates $p(D)=D^3-3D^2+3D-1$ (using $p(D)\bigl[t^ke^{rt}\bigr]=\sum_{j=0}^{k}\binom{k}{j}\frac{p^{(j)}(r)}{j!}\,\cdot j!\,t^{k-j}e^{rt}$, every term of which vanishes at $r=1$):

\[ y = \bigl(C_1 + C_2t + C_3t^2\bigr)e^{t}. \]

Step 4: Return to the variable $x$

Since $t=\ln x$, we have $e^t=x$, so $y = \bigl(C_1 + C_2\ln x + C_3\ln^2x\bigr)\,x$.

Step 5: Verification

Check the three basis functions directly. For $y=x$: $y'=1$, $y''=y'''=0$, so $x^3(0)+x(1)-x=0. $\checkmark$$

For $y=x\ln x$: $y'=\ln x+1$, $y''=1/x$, $y'''=-1/x^2$, so

\[ x^3\!\left(-\frac1{x^2}\right) + x(\ln x+1) - x\ln x = -x + x\ln x + x - x\ln x = 0.\ \checkmark \]

For $y=x\ln^2x$: $y' = \ln^2x + 2\ln x$, $y'' = \dfrac{2\ln x+2}{x}$, $y''' = \dfrac{-2\ln x}{x^2}$, so

\[ x^3\!\left(\frac{-2\ln x}{x^2}\right) + x\bigl(\ln^2x+2\ln x\bigr) - x\ln^2x = -2x\ln x + x\ln^2x+2x\ln x - x\ln^2x = 0.\ \checkmark \]

All three basis solutions satisfy the equation, so by linearity the boxed general solution does too, for all $C_1,C_2,C_3$.