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

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

Step 1: Check homogeneity

Assign weights $[x]=1$, $[y]=k$. Every term has weight $2k$, so the equation is homogeneous for any $k$.

Step 2: Substitution

Set $y = xv$, so $y'=v+xv'$, $y''=2v'+xv''$. Substituting and simplifying (dividing by $x^3$): \[2x\,v\,v'' - x(v')^2 + 2v\,v' = 0. \tag{1}\]

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

This is a Bernoulli equation in $w(x)$.

Step 3: Solve the Bernoulli equation

Rewrite (2) as $w' + w/x = -w^2/2$, a Bernoulli equation with $n=2$. Set $u = 1/w$, so $u' = -w'/w^2$; dividing (2) by $w^2$ gives: \[u' - \frac{u}{x} = \frac{1}{2}. \tag{3}\]

Step 4: Solve the linear ODE (3)

The integrating factor is $1/x$. Multiply (3) by $1/x$ and integrate: \[\frac{d}{dx}\!\left(\frac{u}{x}\right) = \frac{1}{2x} \implies \frac{u}{x} = \frac{1}{2}\ln|x| + C_1 \implies u = x\!\left(\frac{\ln|x|}{2}+C_1\right).\]

Since $u = v/v'$: \[\frac{v'}{v} = \frac{2}{x(\ln x + 2C_1)}. \tag{4}\]

Step 5: Integrate equation (4)

With $t = \ln x + 2C_1$ (relabelled $2C_1 \to C_1$): \[\int\frac{dv}{v} = \int\frac{2\,dt}{t} = 2\ln|t| \implies v = C_2(\ln x + C_1)^2.\]

Step 6: General solution

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

Step 7: Verification

Let $y = A\,x\,L^2$ ($A=C_2$, $L=\ln x+C_1$). Then $y' = A(L^2+2L) = AL(L+2)$ and $y''=\dfrac{2A(L+1)}{x}$. Compute: $(y')^2 - 2yy'' = A^2L^2\!\left[(L+2)^2-4(L+1)\right] = A^2L^4.$

Since $y^2 = A^2x^2L^4$, check $x^2\bigl((y')^2-2yy''\bigr)-y^2$: $x^2A^2L^4 - A^2x^2L^4 = 0.$ $\checkmark$