# How does the integral $\int y''\cdot y'\text{ d}x$ work in the context of differential equations?

In this question, the accepted answer starts off solving the diffeq $$(y'/y)'=-ay$$ by substituting $$u = \log y$$ to get $$u'' = -a e^{u}$$ and then multiplying both sides with $$u'$$ and integrating both sides to get $$\frac{(u')^{2}}{2} = -a e^{u} + \frac{c_{1}}{2}$$ I am baffled at how to get $$\int u'' \cdot u' \text{ d}x=\frac{(u')^{2}}{2}$$ through the context of differentials. Typically, I can understand integrating say $$u'(x)$$ through differentials, where we have $$\int \frac{du}{dx}\text{ d}x = \int \text{ d}u = u+C$$ How do I get the same thing for integrating $$u'' u'$$?

Obviously, I can see this through the chain rule where $$\frac{d}{dx}\left[\frac12 y'^2\right] = \frac12\cdot 2y'\cdot y'' = y'' y'$$ but I cannot reason how how the integral would work $$\int \frac{d^2u}{dx^2}\cdot \frac{du}{dx}\text{ d}x = \int \frac{d^2u}{dx^2} \text{ d}u?????$$

Help is appreciated! :)

• Let $y = u'$ and rinse repeat the same logic above.. Commented Mar 5, 2023 at 23:23
• To answer the question in the title: $[(y’)^2]’=2y’y”\Rightarrow \int{y’y”dx}=\frac 12(y’)^2+C$. Commented Mar 5, 2023 at 23:49

Let $$z=y'$$ ; thus $$\hspace{3pt} dz/dx=y''$$, that is, by law of substitution, $$dz = y''dx$$
$$I = \int y'' y' dx = \int y' (y'' dx) = \int z dz = z^2/2 + c= (y')^2/2+c$$
I can understand integrating say $$u′(x)$$ through differentials, where we have $$\int \dfrac {du}{dx} dx=\int du=u+C$$
You can use the same trick: $$\int u'' \cdot u' \text{ d}x=\int u' \dfrac {du'}{dx}\text{ d}x=\int u' \text{ d}u'=\frac{(u')^{2}}{2}+C$$