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I'm learning about how to solve a first-order linear differnetial equation, say $$y’(x)+p(x)y=q(x)$$. Where $p(x)$ and $q(x)$ are continuous over interval $I$.

See I already know that, in order to solve this equation, multiply both sides or the equation through by $\exp\int p(x)dx$, then we will have $$d\left(\exp \left[\int p(x)dx\right]y(x)\right)=d\left(\exp\left[\int p(x)dx\right]q(x)dx\right).$$ Then the solution will come without any extra effort:$$y(x)=\exp\left[-\int p(x)dx\right]\left(C+\int q(x)\exp\left[\int p(x)dx\right]dx\right).$$ I have no problem understanding this. However, I'm not so sure how to re-write the solution in the form of definite integrals. In my textbook, the solution can be written like this $$y(x)=\exp\left[-\int_{x_0}^xp(t)dt\right]\cdot\left(C+\int_{x_0}^xq(s)\exp\left[\int_{x_0}^sp(t)dt\right]ds\right) \quad (x_0\in I).$$The following are my questions.

  1. What is $x_0$ here? Is it arbitrarily chosen over $I$?
  2. How do I get the "definite integral form" from the "indefinite integral form"?
  3. Why do we need the "definite integral form"? Is it kind of like the period of finding the potential function of a conservative vector field?

I'm really confused about this "difinite integral form". If you feel it is necessary to add some important things which are not included in my questions&statements, just be free to talk. Any help will be appreciated!

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  • $\begingroup$ Check it should be $y'+p(x)y=q(x)$ (mind prime). $\endgroup$
    – Z Ahmed
    Oct 9, 2020 at 8:43
  • $\begingroup$ Sorry about that.... $\endgroup$
    – Scanners
    Oct 9, 2020 at 9:38
  • $\begingroup$ Sorry about that.... it’s $y’+p(x)y=q(x)$, thanks for reminding me of that. $\endgroup$
    – Scanners
    Oct 9, 2020 at 9:45

1 Answer 1

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1,3) When solving differential equations, you will almost, if not always get what is called a family of solutions. This is simply because you can add or multiply by constants and have the derivatives unchanged. For example, $(f(x)+c)'\equiv f'(x)$.

However, in real-world situations, there is only one real answer to our equation. In order to achieve it, we must subject our DE to "initial conditions" which reflect this real world situation.

In this situation, we are using arbitrary initial conditions. They would most likely be written as $y(x_0)=y_0$. As you surmised, it is arbitrary here, but would be picked specifically in real world situations.

  1. Exactly as shown in your question:

$$F(x)+C=\int f(x) dx =\int_{x_0}^x f(t) dt=F(x)-F(x_0)$$


Explaining initial conditions with a basic example.

Let's suppose you're solving $$y'=y$$

This is a rather simple example, and the solution family $y(x)=Ce^x, C\in \Bbb R$

However, without any more information, we can't pinpoint what $C$ is exactly. But, if I gave you the following IVP:

$$\begin{cases} y'=y \\ y(0)=1 \end{cases}$$

we may apply this initial condition $y(0)=1$ to our solution family and see $1=C(1)\implies C=1$ and so we have the unique solution $y(x)=e^x$


I carefully re-read your question and spotted an error:

$$d\left(\exp \left[\int p(x)dx\right]y(x)\right)=\color{red}{d}\left(\exp\left[\int p(x)dx\right]q(x)dx\right)$$ there is no differential on the RHS. What you do after this is integrate both sides between $x_0$ and $x$ which automatically gets rid of any integration constants.

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  • $\begingroup$ Thanks a lot! Well, I do understand things about "innitial conditions". I just don't understand why this expression $$y(x)=e^{-\int_{x_0}^xp(t)dt}\left(C+\int_{x_0}^xq(s)e^{\int_{x_0}^sp(t)dt}ds\right)$$can solve the Cauchy problem $y'+p(x)y=q(x),\quad y(x_0)=y_0$. Isn't there still uncertain constant $C$ in this expression? $\endgroup$
    – Scanners
    Oct 9, 2020 at 10:11
  • $\begingroup$ Edited to account $\endgroup$ Oct 9, 2020 at 23:44

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