# Solving differential equations by definite integration

I'm used to solving differential equations through indefinite integration and then making use of the boundary constraints to find the constants. Now I've learned they can also be solved by using definite integrals, but I don't quite understand the approach yet. Take e. g. the ODE $$C\frac{d^2u(x)}{dx^2}+p=0$$ where the domain for $$x$$ is $$[0;l]$$ and the BCs are $$u(l)=u_1$$ and $$C\frac{du(0)}{dx}:=Cu'(0)=-N$$. ($$C$$, $$p$$, $$l$$, $$u_1$$ and $$N$$ are all constants). Could anyone explain how to solve the problem using definite integration? (I have the solution here, but I have no idea how it's done ;).

I’m on my phone so I won’t go too in depth. The TLDR is to just integrate from the IC of your independent variable (e.g. $$x_0$$), to your dependent variable (e.g. $$x$$).

$$C\frac{d^2u}{dx^2}+p=0$$

$$\int_{x_0}^x(C\frac{d^2u}{dx^2}+p)dx= \int_{x_0}^x0dx$$

$$(C\frac{du}{dx}+px)\Big|_{x_0}^x=c\Big|_{x_0}^x$$

$$C(\frac{du}{dx}-\frac{du}{dx}\big|_{x_0}) +p(x-x_0)=0$$

$$C(\frac{du}{dx}-u’_0) +p(x-x_0)=0$$

$$\int_{x_0}^x(C(\frac{du}{dx}-u’_0) +p(x-x_0))dx= \int_{x_0}^x 0dx$$

$$(C(u(x)-u’_0x)+p(\frac{1}{2}x^2-x_0x)) \Big|_{x_0}^x=0 \Big|_{x_0}^x$$

$$C(u(x)-u_0-u’_0(x-x_0))+p(\frac{1}{2}(x^2-x_0^2)-x_0(x-x_0))=0$$.

You can solve for $$u$$ as it takes a while to LaTeX on the phone. And you’ll have to find someone else on using definite integrals to solve boundary condition problems (e.g. $$x(0)=x_0, x(1)=x_1$$) as I’m not sure on that front.

• Ah, ok. To be clean you should use an auxiliary variable in the integral (i.e $\int_{x_1}^x f(y)dy$) since $\int_{x_1}^x f(x)dx$ is a bit "sloppy" (though we all have done it ;). The BCs can then be inserted readily since evaluation of the definite integral (in case of a first order ODE) yields $f(x)-f(x_0)$, where $f(x_0)$ is our known BC value. For a second order ODE (as described above), we have to play the game twice (after the first integration, we can insert our $u'(0)=-N$ BC and after the second integration we plug in the $u(l)=u_1$ BC). Thanx Jan 3, 2021 at 11:31