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

Find the general solution of the Euler equation $x^2y'' - 3xy' + 4y = 0$.

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

This is an Euler (equidimensional) equation: the coefficient of $y^{(k)}$ is proportional to $x^k$. Such equations are reduced to constant-coefficient linear equations by the substitution $x=e^t$, i.e.\ $t=\ln x$ (for $x>0$). Writing dots for $d/dt$, the chain rule gives the standard identities

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

Step 2: Transformed equation and characteristic equation

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

The characteristic equation is

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

Step 3: General solution in $t$

For a repeated root $r$, the fundamental system is $e^{rt}$ and $te^{rt}$, so

\[ y = (C_1 + C_2 t)\,e^{2t}. \]

Step 4: Return to the variable $x$

Since $t=\ln x$, we have $e^{2t}=x^2$ and $t=\ln x$, so $y = x^2\bigl(C_1 + C_2\ln x\bigr)$.

Step 5: Verification

Check the two basis functions separately. For $y=x^2$: $y'=2x$, $y''=2$, so

\[ x^2(2) - 3x(2x) + 4x^2 = 2x^2 - 6x^2 + 4x^2 = 0.\ \checkmark \]

For $y=x^2\ln x$: $y' = 2x\ln x + x$, $y'' = 2\ln x + 3$, so

\[ x^2(2\ln x+3) - 3x(2x\ln x+x) + 4x^2\ln x = (2x^2\ln x+3x^2) - (6x^2\ln x+3x^2) + 4x^2\ln x = 0.\ \checkmark \]

Both basis solutions satisfy the equation, so by linearity the boxed general solution does too, for all $C_1,C_2$.