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

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

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

For $x>0$ set $x=e^t$, i.e.\ $t=\ln x$. Alongside the usual two identities, a third-order Euler equation needs the third-derivative identity as well (dot denotes $d/dt$): \[x\,y' = \dot y, \qquad x^2y'' = \ddot y - \dot y, \qquad x^3y''' = \dddot y - 3\ddot y + 2\dot y.\]

Under this substitution the right-hand side $x^3+3x$ becomes $e^{3t}+3e^t$.

Step 2: Transformed equation

Substituting into $x^3y'''-x^2y''+2xy'-2y=x^3+3x$: $(\dddot y - 3\ddot y + 2\dot y) - (\ddot y - \dot y) + 2\dot y - 2y = e^{3t} + 3e^t$, $\dddot y - 4\ddot y + 5\dot y - 2y = e^{3t} + 3e^t$.

Step 3: Homogeneous solution

The characteristic equation is $r^3-4r^2+5r-2=0$. Testing $r=1$: $1-4+5-2=0. $\checkmark$$ Dividing out $(r-1)$: \[r^3-4r^2+5r-2 = (r-1)(r^2-3r+2) = (r-1)(r-1)(r-2) = (r-1)^2(r-2).\] So the roots are $r=1$ (double) and $r=2$, giving

\[ Y_h = (C_1+C_2t)e^{t} + C_3e^{2t}. \]

Step 4: Particular solution — superposition

Split $y^{*}=y_1^{*}+y_2^{*}$, where $y_1^{*}$ handles $e^{3t}$ and $y_2^{*}$ handles $3e^t$.

Part 1: $e^{3t}$. Here $\alpha=3$, not among the roots $\{1,1,2\}$, so $s=0$ and the trial is $y_1^{*}=Ae^{3t}$. Substituting into $\dddot y-4\ddot y+5\dot y-2y=e^{3t}$:

\[ 27A - 36A + 15A - 2A = 1 \implies 4A = 1 \implies A = \frac14. \] So $y_1^{*}=\dfrac{e^{3t}}{4}$.

Part 2: $3e^t$. Here $\alpha=1$ is a double root of the characteristic equation ($s=2$), so the trial must be multiplied by $t^2$: $y_2^{*}=Bt^2e^t$. Its derivatives are

\[ \dot y_2^{*} = B(2t+t^2)e^t, \quad \ddot y_2^{*} = B(2+4t+t^2)e^t, \quad \dddot y_2^{*} = B(6+6t+t^2)e^t. \]

Substituting into $\dddot y-4\ddot y+5\dot y-2y=3e^t$ and dividing through by $e^t$: $B(6+6t+t^2) - 4B(2+4t+t^2) + 5B(2t+t^2) - 2Bt^2 = 3.$

Collecting powers of $t$: the $t^2$ coefficient is $B(1-4+5-2)=0$, and the $t^1$ coefficient is $B(6-16+10)=0$ — both vanish identically, exactly as expected for a double-root resonance (the $t$ and $t^2$ terms must cancel, leaving only the constant term to carry the equation). The constant term gives

\[ B(6-8) = 3 \implies -2B = 3 \implies B = -\frac32. \] So $y_2^{*}=-\dfrac{3}{2}t^2e^t$.

Step 5: General solution in $t$, then return to $x$

\[ y = (C_1+C_2t)e^{t} + C_3e^{2t} + \frac{e^{3t}}{4} - \frac{3}{2}t^2e^{t}. \]

Since $e^t=x$, $e^{2t}=x^2$, $e^{3t}=x^3$, and $t=\ln x$, the general solution is $y = (C_1+C_2\ln x)\,x + C_3x^2 + \frac{x^3$.

Step 6: Verification

Substituting the full boxed expression back into the original equation $x^3y'''-x^2y''+2xy'-2y=x^3+3x$ confirms the identity term by term: the homogeneous pieces $x$, $x\ln x$, and $x^2$ annihilate the left side (they solve $x^3y'''-x^2y''+2xy'-2y=0$), while the particular piece $\tfrac14x^3-\tfrac32x(\ln x)^2$ reproduces exactly the forcing $x^3+3x$. By linearity, the boxed solution satisfies the original equation for all $C_1,C_2,C_3$. $\checkmark$