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

Find the general solution of the Euler equation $x^3y''' + x^2y'' + 3xy' - 8y = 0$.

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

With $x=e^t$ ($t=\ln x$, $x>0$) and dots denoting $d/dt$, repeated application of the chain rule (equivalently, of the operator identity $x\dfrac{d}{dx}=\dfrac{d}{dt}$) gives the three reduction identities needed for a third-order Euler equation:

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

Step 2: Transformed equation and characteristic equation

Substituting into $x^3y'''+x^2y''+3xy'-8y=0$: \[(\dddot y - 3\ddot y + 2\dot y) + (\ddot y - \dot y) + 3\dot y - 8y = 0 \implies \dddot y - 2\ddot y + 4\dot y - 8y = 0.\]

The characteristic equation is

\[ r^3 - 2r^2 + 4r - 8 = 0. \]

Factor by grouping: \[r^2(r-2) + 4(r-2) = (r-2)(r^2+4) = 0 \implies r = 2, \ \ r = \pm 2i.\]

Step 3: General solution in $t$

The real root $r=2$ contributes $e^{2t}$, and the complex-conjugate pair $r=\pm2i$ contributes $\cos2t,\ \sin2t$: $y = C_1e^{2t} + C_2\cos2t + C_3\sin2t.$

Step 4: Return to the variable $x$

Since $t=\ln x$, we have $e^{2t}=x^2$, $\cos2t=\cos(2\ln x)$, $\sin2t=\sin(2\ln x)$, so $y = C_1x^2 + C_2\cos(2\ln x) + C_3\sin(2\ln x)$.

Step 5: Verification

For $y=x^2$: $y'=2x$, $y''=2$, $y'''=0$, so

\[ x^3(0) + x^2(2) + 3x(2x) - 8x^2 = 2x^2 + 6x^2 - 8x^2 = 0.\ \checkmark \]

For $y=\cos(2\ln x)$, write $u=\ln x$. Then $y'=-\dfrac{2\sin2u}{x}$, $y''=\dfrac{2\sin2u-4\cos2u}{x^2}$, $y'''=\dfrac{4\sin2u+12\cos2u}{x^3}$, and

\[ x^3y''' + x^2y'' + 3xy' - 8y = (4\sin2u+12\cos2u) + (2\sin2u-4\cos2u) - 6\sin2u - 8\cos2u = 0.\ \checkmark \]

The $\sin2u$ coefficients $4+2-6=0$ and the $\cos2u$ coefficients $12-4-8=0$ both vanish; the $\sin(2\ln x)$ term checks identically by the same computation. By linearity, the full boxed solution satisfies the original equation for all $C_1,C_2,C_3$.