Find the general solution of the Euler equation y'' +…

Find the general solution of the Euler equation $y'' + \frac{y'}{x} + \frac{y}{x^2} = 0$.

Step 1: Standard Euler form

Multiplying through by $x^2$ clears the fractions and puts the equation in the standard Euler (equidimensional) form, in which the coefficient of $y^{(k)}$ is proportional to $x^k$:

\[ x^2y'' + xy' + y = 0. \]

Step 2: The Euler substitution $x=e^t$

For $x>0$ set $x=e^t$, i.e.\ $t=\ln x$. By the chain rule, $x\,y'=\dot y$, and differentiating this identity again gives the standard second identity, where the dot denotes $d/dt$:

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

Step 3: Transformed equation

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

Step 4: Solving the constant-coefficient equation in $t$

The characteristic equation is

\[ r^2 + 1 = 0 \implies r = \pm i, \]

a purely imaginary conjugate pair, so the general solution in $t$ is

\[ y = C_1\cos t + C_2\sin t. \]

Step 5: Return to the variable $x$

Since $t=\ln x$, $y = C_1\cos(\ln x) + C_2\sin(\ln x)$.

Step 6: Verification

Take $y=\cos(\ln x)$ (the $\sin(\ln x)$ term checks identically by the same computation). By the chain rule, $y' = -\frac{\sin(\ln x)}{x}, \qquad y'' = \frac{-\cos(\ln x)+\sin(\ln x)}{x^2}.$

Then \[y'' + \frac{y'}{x} + \frac{y}{x^2} = \frac{-\cos(\ln x)+\sin(\ln x)}{x^2} + \frac{-\sin(\ln x)}{x^2} + \frac{\cos(\ln x)}{x^2} = 0.\ \checkmark\]

The $\sin(\ln x)$ contributions from the first two terms cancel, and the $\cos(\ln x)$ contributions from the first and third terms cancel, confirming the equation is satisfied identically.