Find the general solution of the generalized Euler equation…

Find the general solution of the generalized Euler equation $(1+x)^2y'' - 3(1+x)y' + 4y = (1+x)^3$.

Step 1: A shifted Euler substitution

This equation is not homogeneous in $x$ itself, but it is homogeneous in $u=1+x$: every coefficient is proportional to a power of $(1+x)$. So the same reduction applies with $x$ replaced by $u=1+x$. Set $u=e^t$, i.e.\ $t=\ln(1+x)$ (for $x>-1$). Since $du=dx$, the chain rule gives exactly the usual Euler identities with $u$ in place of $x$:

\[ (1+x)\,y' = \dot y, \qquad (1+x)^2y'' = \ddot y - \dot y, \] where the dot denotes $d/dt$. Under this substitution the right-hand side $(1+x)^3$ becomes $e^{3t}$.

Step 2: Transformed equation

Substituting into $(1+x)^2y''-3(1+x)y'+4y=(1+x)^3$: \[(\ddot y - \dot y) - 3\dot y + 4y = e^{3t} \implies \ddot y - 4\dot y + 4y = e^{3t}.\]

Step 3: Homogeneous solution

The characteristic equation is

\[ r^2 - 4r + 4 = 0 \implies (r-2)^2 = 0 \implies r = 2\ (\text{double}), \]

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

Remark: this is exactly the characteristic equation that would arise from the homogeneous Euler equation $x^2y''-3xy'+4y=0$ — the same repeated root $r=2$ shows up here simply with $x$ replaced by $1+x$.

Step 4: Particular solution

The forcing term is $e^{3t}$, so $\alpha=3$. Since $\alpha=3$ is not a root of $(r-2)^2=0$, $s=0$ and the trial is $y^{*}=Ae^{3t}$. Substituting $\dot y^{*}=3Ae^{3t}$, $\ddot y^{*}=9Ae^{3t}$ into $\ddot y-4\dot y+4y=e^{3t}$:

\[ 9A - 12A + 4A = 1 \implies A = 1. \] So $y^{*}=e^{3t}$.

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

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

Since $t=\ln(1+x)$, we have $e^t=1+x$, $e^{2t}=(1+x)^2$, $e^{3t}=(1+x)^3$, so the general solution is $y = \bigl(C_1+C_2\ln(1+x)\bigr)(1+x)^2 + (1+x)^3$.

Step 6: Verification

For $y=(1+x)^3$: $y'=3(1+x)^2$, $y''=6(1+x)$, so

\[ (1+x)^2\bigl[6(1+x)\bigr] - 3(1+x)\bigl[3(1+x)^2\bigr] + 4(1+x)^3 = 6(1+x)^3 - 9(1+x)^3 + 4(1+x)^3 = (1+x)^3.\ \checkmark \]

For $y=(1+x)^2\ln(1+x)$: using $u=1+x$, $y=u^2\ln u$ satisfies the homogeneous equation $u^2y''-3uy'+4y=0$ by construction (it is the second solution at the double root $r=2$), and likewise $y=(1+x)^2$ does. By linearity, the boxed solution satisfies $(1+x)^2y''-3(1+x)y'+4y=(1+x)^3$ for all $C_1,C_2$.