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

Find the general solution of the Euler equation $x^2y'' - 4xy' + 6y = x$.

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 $x$ becomes $e^t$.

Step 2: Transformed equation

Substituting into $x^2y''-4xy'+6y=x$: \[(\ddot y - \dot y) - 4\dot y + 6y = e^t \implies \ddot y - 5\dot y + 6y = e^t.\]

Step 3: Homogeneous solution

The characteristic equation is

\[ r^2 - 5r + 6 = 0 \implies (r-2)(r-3)=0 \implies r=2,\,3, \]

\[ Y_h = C_1e^{2t} + C_2e^{3t}. \]

Step 4: Particular solution

The forcing term is $e^t$, so $\alpha=1$. Since $\alpha=1$ is not a root of the characteristic equation, $s=0$ and the trial is $y^{*}=Ae^t$. Substituting $\dot y^{*}=\ddot y^{*}=Ae^t$ into $\ddot y-5\dot y+6y=e^t$:

\[ A - 5A + 6A = e^t \cdot 1 \implies 2A = 1 \implies A = \frac12. \] So $y^{*}=\dfrac{e^t}{2}$.

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

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

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

Step 6: Verification

For $y=\dfrac{x}{2}$: $y'=\dfrac12$, $y''=0$, so $x^2(0)-4x\!\left(\dfrac12\right)+6\!\left(\dfrac{x}{2}\right)=-2x+3x=x. $\checkmark$$ For $y=x^2$: $x^2(2)-4x(2x)+6x^2=2x^2-8x^2+6x^2=0$, and for $y=x^3$: $x^2(6x)-4x(3x^2)+6x^3=6x^3-12x^3+6x^3=0$, so both homogeneous terms satisfy the associated homogeneous equation. By linearity, the full solution satisfies $x^2y''-4xy'+6y=x$ for all $C_1,C_2$.