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

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

Step 1: Check homogeneity

Assign weights $[x]=1$, $[y]=k$. Every term $xyy''$, $x(y')^2$, $yy'$ has weight $2k-1$, so the equation is homogeneous for any $k$.

Step 2: Logarithmic substitution

Divide by $y^2$ and set $z=\ln y$, so $z'=y'/y$, $z''=(yy''-(y')^2)/y^2$. The equation becomes: \[x\,z'' = z'. \tag{1}\]

Step 3: Reduce the order

Let $p=z'$, so $p'=z''$. Equation (1) becomes the separable ODE: \[x\,p' = p. \tag{2}\]

Step 4: Solve equation (2)

\[ \frac{dp}{p} = \frac{dx}{x} \implies p = C_1\,x. \]

Thus: \[z' = C_1\,x. \tag{3}\]

Step 5: Second integration

\[ z = \frac{C_1}{2}\,x^2 + C_2. \]

Step 6: General solution

Since $z=\ln y$: where $A=e^{C_2}$ and $C_1$ are arbitrary constants.

Step 7: Verification

Let $y = Ae^{C_1x^2/2}$. Then $y' = C_1xy$ and $y'' = C_1y + C_1x\,y' = C_1y(1+C_1x^2)$. Check $xyy''-x(y')^2-yy'$: \[C_1xy^2(1+C_1x^2) - C_1^2x^3y^2 - C_1xy^2 = C_1xy^2 + C_1^2x^3y^2 - C_1^2x^3y^2 - C_1xy^2 = 0.\ \checkmark\]