Find the general integral of yy'' + y'^2 = f(x) , where f(x)…

Find the general integral of $yy'' + y'^2 = f(x)$, where $f(x)$ is a given function.

Step 1: Recognize a perfect derivative

The key observation is that the left side is already an exact derivative: by the product rule, $\frac{d}{dx}(yy') = y'^2 + yy''.$ So the equation is simply

\[ \frac{d}{dx}(yy') = f(x). \]

Step 2: Integrate once

Integrating both sides directly, $yy' = F(x) + C_1, \qquad \text{where } F(x)=\int f(x)\,dx \text{ (any antiderivative).}$.

Step 3: Recognize a second perfect derivative

Since $yy' = \tfrac12\dfrac{d}{dx}(y^2)$, the equation becomes

\[ \frac12\frac{d}{dx}(y^2) = F(x) + C_1. \]

Step 4: Integrate again

Integrating once more, with $G(x)=\int F(x)\,dx$ an iterated antiderivative of $f$: $\frac{y^2}{2} = G(x) + C_1x + \frac{C_2}{2}.$ $y^2 = 2G(x) + 2C_1x + C_2, \qquad G(x)=\iint f(x)\,dx\,dx$.

with $C_1,C_2$ arbitrary constants; equivalently $y=\pm\sqrt{2G(x)+2C_1x+C_2}$ wherever the radicand is positive.

Step 5: Verification

Differentiating $y^2=2G(x)+2C_1x+C_2$ once: $2yy' = 2G'(x)+2C_1 = 2F(x)+2C_1$, i.e. $yy'=F(x)+C_1$, recovering Step 2. Differentiating again: $y'^2+yy'' = F'(x) = f(x)$ (since $G'=F$ and $F'=f$), which is exactly the original equation. $\checkmark$

Step 6: Illustrative special cases