Find the general solution of x^2 y'' + x y' + y = 0 , using…

Find the general solution of $x^2\,y'' + x\,y' + y = 0$, using homogeneity to reduce its order.

Step 1: Check homogeneity

Assign weights $[x]=1$, $[y]=k$. Every term $x^2y''$, $xy'$, $y$ has weight $k$, so the equation is homogeneous for any $k$: this is an Euler (equidimensional) equation.

Step 2: Substitution to reduce order

Use the standard Euler substitution $x = e^t$, $t = \ln x$ ($x>0$). Derivatives transform as:

\[ x\,y' = \dot y,\qquad x^2\,y'' = \ddot y - \dot y, \] where dots denote $d/dt$.

Step 3: Transform the equation

Substitute into $x^2y''+xy'+y=0$: \[(\ddot y - \dot y) + \dot y + y = 0 \implies \ddot y + y = 0. \tag{1}\]

Step 4: Solve the transformed equation

The characteristic equation $\lambda^2+1=0$ gives $\lambda=\pm i$, so: $y = C_1\cos t + C_2\sin t.$

Step 5: General solution

Substitute back $t=\ln x$: where $C_1$ and $C_2$ are arbitrary constants, valid for $x>0$.

Step 6: Verification

\[ y' = \frac{1}{x}\bigl(-C_1\sin(\ln x)+C_2\cos(\ln x)\bigr), \]

\[ y'' = \frac{1}{x^2}\bigl(C_1\sin(\ln x)-C_2\cos(\ln x)\bigr) - \frac{1}{x^2}\bigl(C_1\cos(\ln x)+C_2\sin(\ln x)\bigr). \]

Check $x^2y''+xy'+y$: each pair of sine/cosine terms cancels exactly, leaving $0$. $\checkmark$