= Logarithmic-derivative reduction of a second-order equation
{title2=$y''=(y')^2/y+a(x)y'\Longrightarrow (y'/y)'=a(x)(y'/y)$}
Where $y\ne0$, its <logarithmic derivative> $q=y'/y$ changes this nonlinear second-order <ordinary differential equation> into the <first-order linear differential equation> $q'=a(x)q$. For continuous $a$, initial values $y(x_0)=y_0\ne0$ and $y'(x_0)=v_0$ therefore give
$$
q(x)=\frac{v_0}{y_0}\exp\left(\int_{x_0}^x a(t)\,dt\right),\qquad
y(x)=y_0\exp\left(\int_{x_0}^x q(t)\,dt\right).
$$
Differentiation proves both the equation and the initial data. The exponential preserves the sign of $y_0$, so the solution never reaches the singular value zero on any finite interval where $a$ is continuous. For $a(x)=2x$ and $y_0=v_0=1$, the expression reduces to $y(x)=\exp(\int_0^x e^{t^2}\,dt)$.
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