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

Find the general solution of the Euler equation $x^2y'' - 2y = \sin(\ln x)$.

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

This is an Euler (equidimensional) equation, so set $x=e^t$, i.e. $t=\ln x$, and write $\dot y = dy/dt$. The standard identities for this substitution are

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

Substituting into $x^2y''-2y=\sin(\ln x)$ turns it into a constant-coefficient linear equation in $t$: $\ddot y - \dot y - 2y = \sin t$.

Step 2: Homogeneous solution

The characteristic equation is

\[ r^2 - r - 2 = (r-2)(r+1) = 0 \;\Longrightarrow\; r=2,\ -1, \]

so $y_h(t) = C_1e^{2t} + C_2e^{-t}$. Returning to $x=e^t$ (so $e^{2t}=x^2$ and $e^{-t}=1/x$), $y_h = C_1x^2 + \frac{C_2}{x}$.

Step 3: Particular solution

Since $r=\pm i$ is not a root of the characteristic equation, try $y_p = A\sin t + B\cos t$. Then $\dot y_p = A\cos t - B\sin t$ and $\ddot y_p = -A\sin t - B\cos t$, so

\[ \ddot y_p - \dot y_p - 2y_p = (-3A+B)\sin t + (-A-3B)\cos t. \]

Matching this to $\sin t$ gives the linear system

\[ -3A+B = 1, \qquad -A-3B = 0. \]

The second equation gives $A=-3B$; substituting into the first, $9B+B=1$, so $B=\tfrac1{10}$ and $A=-\tfrac{3}{10}$. Hence

\[ y_p(t) = -\frac{3}{10}\sin t + \frac{1}{10}\cos t \;\Longrightarrow\; y_p(x) = -\frac{3}{10}\sin(\ln x) + \frac{1}{10}\cos(\ln x). \]

Step 4: General solution

The solution is recorded in the result below.

Step 5: Verification

It is quickest to verify $y_p$ directly in $x$. With $y_p=-\tfrac{3}{10}\sin(\ln x)+\tfrac1{10}\cos(\ln x)$, the chain rule ($\tfrac{d}{dx}\ln x = 1/x$) gives

\[ y_p' = \frac1x\left[-\frac{3}{10}\cos(\ln x) - \frac{1}{10}\sin(\ln x)\right]. \]

Differentiating once more (quotient rule, using $\tfrac{d}{dx}$ of the bracket $=\tfrac1x\bigl[\tfrac3{10}\sin(\ln x)-\tfrac1{10}\cos(\ln x)\bigr]$): \[y_p'' = \frac{1}{x^2}\left[\frac{2}{5}\sin(\ln x) + \frac{1}{5}\cos(\ln x)\right].\]

So $x^2y_p'' = \tfrac25\sin(\ln x)+\tfrac15\cos(\ln x)$, while $2y_p = -\tfrac35\sin(\ln x)+\tfrac15\cos(\ln x)$. Subtracting, \[x^2y_p'' - 2y_p = \left(\frac25+\frac35\right)\sin(\ln x) + \left(\frac15-\frac15\right)\cos(\ln x) = \sin(\ln x).\ \checkmark\]

The homogeneous terms check trivially: $x^2(x^2)''-2x^2 = 2x^2-2x^2=0$, and $x^2(1/x)''-2(1/x)=x^2\cdot\tfrac{2}{x^3}-\tfrac2x=\tfrac2x-\tfrac2x=0$, so both satisfy the homogeneous Euler equation, confirming the boxed general solution.