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

Find the general solution of the Euler equation $x^3y''' + 2x^2y'' - xy' + y = 0$.

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

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

Step 2: Transformed equation and characteristic equation

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

The characteristic equation is

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

Factor by grouping: \[r^2(r-1) - (r-1) = (r-1)(r^2-1) = (r-1)(r-1)(r+1) = (r-1)^2(r+1) = 0,\]

giving roots $r=1$ (double) and $r=-1$ (simple).

Step 3: General solution in $t$

The double root $r=1$ contributes $e^t$ and $te^t$; the simple root $r=-1$ contributes $e^{-t}$: $y = (C_1 + C_2t)\,e^{t} + C_3e^{-t}.$

Step 4: Return to the variable $x$

Since $t=\ln x$, we have $e^t=x$ and $e^{-t}=1/x$, so $y = (C_1 + C_2\ln x)\,x + \frac{C_3$.

Step 5: Verification

Check the three basis functions directly. For $y=x$: $y'=1$, $y''=y'''=0$, so $x^3(0)+2x^2(0)-x(1)+x=0. $\checkmark$$

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

\[ x^3\!\left(-\frac1{x^2}\right) + 2x^2\!\left(\frac1x\right) - x(\ln x+1) + x\ln x = -x + 2x - x\ln x - x + x\ln x = 0.\ \checkmark \]

For $y=1/x$: $y'=-1/x^2$, $y''=2/x^3$, $y'''=-6/x^4$, so

\[ x^3\!\left(-\frac6{x^4}\right) + 2x^2\!\left(\frac2{x^3}\right) - x\!\left(-\frac1{x^2}\right) + \frac1x = -\frac6x + \frac4x + \frac1x + \frac1x = 0.\ \checkmark \]

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