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

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

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

With $x=e^t$ ($t=\ln x$, $x>0$) and dots denoting $d/dt$, the chain rule gives the usual reduction identities

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

Step 2: Transformed equation and characteristic equation

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

The characteristic equation is

\[ r^2 + 2r + 1 = 0 \implies (r+1)^2 = 0 \implies r = -1 \ \text{(double root)}. \]

Step 3: General solution in $t$

For the repeated root $r=-1$, the fundamental system is $e^{-t}$ and $te^{-t}$, so

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

Step 4: Return to the variable $x$

Since $t=\ln x$, we have $e^{-t}=1/x$, so $y = \frac{C_1 + C_2\ln x$.

Step 5: Verification

Check the two basis functions separately. For $y=1/x$: $y'=-1/x^2$, $y''=2/x^3$, so

\[ x^2\!\left(\frac{2}{x^3}\right) + 3x\!\left(-\frac{1}{x^2}\right) + \frac1x = \frac2x - \frac3x + \frac1x = 0.\ \checkmark \]

For $y=(\ln x)/x$: $y' = \dfrac{1-\ln x}{x^2}$, $y'' = \dfrac{2\ln x - 3}{x^3}$, so

\[ x^2\!\left(\frac{2\ln x-3}{x^3}\right) + 3x\!\left(\frac{1-\ln x}{x^2}\right) + \frac{\ln x}{x} = \frac{(2\ln x-3)+(3-3\ln x)+\ln x}{x} = 0.\ \checkmark \]

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