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

Find the general solution of $y\,y'' - (y')^2 = y^2\ln y$.

Step 1: Recognize a logarithmic substitution

Divide by $y^2$ and recognize the left side as a derivative: \[\frac{y\,y''-(y')^2}{y^2} = \ln y,\qquad \frac{d}{dx}\!\left(\frac{y'}{y}\right) = \frac{y''y-(y')^2}{y^2}.\]

Set $z = \ln y$, so $z' = \dfrac{y'}{y}$ and $z'' = \dfrac{yy''-(y')^2}{y^2}$. The equation becomes: \[z'' = z. \tag{1}\]

Step 2: Solve the linear ODE

Equation (1) is linear with constant coefficients. The characteristic equation $\lambda^2-1=0$ gives $\lambda = \pm 1$, so: \[z = C_1 e^x + C_2 e^{-x}. \tag{2}\]

Step 3: General solution

Since $z = \ln y$, we have $y = e^z$: where $C_1$ and $C_2$ are arbitrary constants.

Step 4: Verification

From $\ln y = C_1 e^x + C_2 e^{-x}$: \[z' = \frac{y'}{y} = C_1 e^x - C_2 e^{-x},\qquad z'' = \frac{yy''-(y')^2}{y^2} = C_1 e^x + C_2 e^{-x} = z = \ln y.\]

Thus: $yy'' - (y')^2 = y^2 z'' = y^2\ln y.$ $\checkmark$