Find the general solution of x^2 y y'' = (y - x y')^2…

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

Step 1: Check homogeneity

Assign weights $[x]=1$, $[y]=k$. Both $[x^2yy'']$ and $[(y-xy')^2]$ equal $2k$, so the equation is homogeneous for any $k$.

Step 2: Substitution

Set $y = x\,v(x)$, so $y' = v+xv'$ and $y-xy' = -x^2v'$; also $y''=2v'+xv''$. Substitute into $x^2yy''=(y-xy')^2$: \[x^2\cdot xv\cdot(2v'+xv'') = x^4(v')^2 \implies x\,v\,v'' + 2v\,v' - x(v')^2 = 0. \tag{1}\]

Step 3: Reduce the order

Divide (1) by $v^2$ and set $w = v'/v$, so $v'=wv$, $v''=v(w'+w^2)$: \[x(w'+w^2) + 2w - xw^2 = 0 \implies x\,w' + 2w = 0. \tag{2}\]

Step 4: Solve equation (2)

\[ \frac{dw}{w} = -\frac{2\,dx}{x} \implies w = \frac{C_1}{x^2}. \]

Recall $w = v'/v$: \[\frac{v'}{v} = \frac{C_1}{x^2}. \tag{3}\]

Step 5: Second integration

\[ \frac{dv}{v} = \frac{C_1\,dx}{x^2} \implies \ln|v| = -\frac{C_1}{x} + C_2' \implies v = A\,e^{-C_1/x}. \]

Step 6: General solution

Since $y = xv$: where $C_1$ and $C_2$ are arbitrary constants, valid for $x \neq 0$.

Step 7: Verification

Let $y = A\,e^{-\alpha/x}x$ ($A=C_2$, $\alpha=C_1$). Then $y' = Ae^{-\alpha/x}\!\left(1+\dfrac{\alpha}{x}\right)$ and $y'' = Ae^{-\alpha/x}\dfrac{\alpha^2}{x^3}$. Also $y-xy' = -\alpha A e^{-\alpha/x}$. Check:

\[ x^2yy'' = x^2\cdot Axe^{-\alpha/x}\cdot Ae^{-\alpha/x}\frac{\alpha^2}{x^3} = A^2\alpha^2 e^{-2\alpha/x} = \left(-\alpha Ae^{-\alpha/x}\right)^2 = (y-xy')^2.\ \checkmark \]