Find the general solution of the Euler equation x^2y''…

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

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

For $x>0$ set $x=e^t$, i.e.\ $t=\ln x$. The standard identities (dot denotes $d/dt$) are

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

Under this substitution the right-hand side $2x^3-x$ becomes $2e^{3t}-e^t$.

Step 2: Transformed equation

Substituting into $x^2y''-2xy'+2y=2x^3-x$: \[(\ddot y - \dot y) - 2\dot y + 2y = 2e^{3t} - e^t \implies \ddot y - 3\dot y + 2y = 2e^{3t} - e^t.\]

Step 3: Homogeneous solution

The characteristic equation is

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

\[ Y_h = C_1e^{t} + C_2e^{2t}. \]

Step 4: Particular solution

By the superposition principle, split $y^{*}=y_1^{*}+y_2^{*}$, where $y_1^{*}$ handles $2e^{3t}$ and $y_2^{*}$ handles $-e^t$.

Part 1: $2e^{3t}$. Here $\alpha=3$, which is not among the roots $\{1,2\}$, so $s=0$ and the trial is $y_1^{*}=Ae^{3t}$. Substituting:

\[ 9A - 9A + 2A = 2 \implies 2A = 2 \implies A = 1. \] So $y_1^{*}=e^{3t}$.

Part 2: $-e^t$. Here $\alpha=1$ IS a simple root of the characteristic equation ($s=1$), so the trial must be multiplied by $t$: $y_2^{*}=Bte^t$. Then

\[ \dot y_2^{*} = B(1+t)e^t, \qquad \ddot y_2^{*} = B(2+t)e^t. \]

Substituting into $\ddot y-3\dot y+2y=-e^t$: \[B(2+t)e^t - 3B(1+t)e^t + 2Bte^t = Be^t\bigl[(2+t)-3(1+t)+2t\bigr] = Be^t(-1) = -e^t.\] So $-B=-1 \implies B=1$, giving $y_2^{*}=te^t$.

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

\[ y = C_1e^{t} + C_2e^{2t} + e^{3t} + te^t. \]

Since $e^t=x$, $e^{2t}=x^2$, $e^{3t}=x^3$, and $t=\ln x$, the general solution is $y = C_1x + C_2x^2 + x^3 + x\ln x$.

Step 6: Verification

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

\[ x^2\!\left(\frac1x\right) - 2x(\ln x+1) + 2x\ln x = x - 2x\ln x - 2x + 2x\ln x = -x.\ \checkmark \]

For $y=x^3$: $x^2(6x)-2x(3x^2)+2x^3=6x^3-6x^3+2x^3=2x^3. $\checkmark$$ The homogeneous terms $x$ and $x^2$ satisfy the associated homogeneous equation identically (they are the $r=1,2$ solutions), so by linearity the full solution satisfies $x^2y''-2xy'+2y=2x^3-x$ for all $C_1,C_2$.