Find the general solution of y y'' - 2y y'ln y = (y')^2
Find the general solution of $y\,y'' - 2y\,y'\ln y = (y')^2$.
Explanation of Method
Reduce via $z=\ln y$, then $p=z'$ treated as a function of $z$; three cases by sign.
Step 1: Logarithmic substitution
Divide by $y^2$ and recognize the left side as a derivative. Set $z=\ln y$, so $z'=y'/y$, $z''=(yy''-(y')^2)/y^2$. The equation becomes: \[z'' = 2z'\,z. \tag{1}\]
Step 2: Second substitution
Let $p = z'$; since $x$ is absent, treat $p$ as a function of $z$: $p' = p\,dp/dz$. Equation (1) becomes: \[p\,\frac{dp}{dz} = 2zp. \tag{2}\]
Step 3: Separate variables
Case 1 · p = 0.
Then $z' = 0$, so $z = \mathrm{const}$, i.e. $y = \mathrm{const}$.
Case 2 · p ≠ 0.
Divide (2) by $p$ and integrate: $\frac{dp}{dz} = 2z \implies p = z^2 + C_1.$
Thus: \[z' = z^2 + C_1. \tag{3}\]
Step 4: Integrate equation (3)
This is separable; the sign of $C_1$ determines the form.
Case A · C₁ = 0.
\[ \frac{dz}{z^2} = dx \implies z = -\frac{1}{x+C_2}. \]
Case B · C₁ = k² > 0.
\[ \frac{dz}{z^2+k^2} = dx \implies z = k\tan\!\left(k(x+C_2)\right). \]
Case C · C₁ = −k² < 0.
\[ \frac{dz}{z^2-k^2} = dx \implies z = -k\tanh\!\left(k(x+C_2)\right)\ \text{ or }\ z = -k\coth\!\left(k(x+C_2)\right). \]
Step 5: General solution
Since $y = e^z$, for each case:
Step 6: Verification
Since $z' = z^2+C_1$, differentiate: $z'' = 2z\,z' = 2z(z^2+C_1)$, which equals $2z'z$ as required by (1). And since $z''=\dfrac{yy''-(y')^2}{y^2}$, $z'=\dfrac{y'}{y}$, the original equation $yy''-2yy'\ln y-(y')^2 = y^2(z''-2z'z)$ is satisfied identically. $\checkmark$
Result
Final answer:
\[\boxed{\,y = e^{-1/(x+C_2)}\quad(C_1=0);\qquad y = e^{\,k\tan(k(x+C_2))}\quad(C_1=k^2>0);\,}\]