I have a differential equation and I'm trying to understand the solution printed in the back of the book. I will specify what part I'm struggling to understand:

Problem statement: Solve the following equation. $y(x)=\int_0^xy(t)dt + x + 1$.

Verbatim solution from back of the book that I'm trying to understand:
$y(x)=\int_0^xy(t)dt + x + 1 ⇒ \frac{dy}{dx} = y(x) + 1 ⇒ \frac{dy}{dx}- y = 1 ⇒ u(x) = e^{\int-1dx} = e^{-x} ⇒ \frac{d}{dx}[e^{-x}y] = e^{-x} ⇒ e^{-x}y = \int e^{-x}dx = -e^{-x} + C ⇒ y = Ce^{x} - 1.$

But there is an implied initial condition here: $y(0)=1.$ Therefore, $1=y(0)=Ce^{0}-1⇒C=2$, so that the solution is $y=2e^{x}-1.$ [Try any other value of C and see that the function doesn't satisfy the original equation.]

I have 2 questions:

  1. Why is $y(x)=\int_0^xy(t)dt = y(x)$, and not $y(x)-y(0)$?

  2. Why is there an implied initial condition when there is no implied condition stated in the problem?


1) This is the direct result of the first part of the Fundamental Theorem of Calculus.

2) You can see in the original equation, when $x=0$ the integral vanishes, as does $x$, leaving you the condition $y(0)=1$. This is certainly true, so this condition has to be satisfied.

  • 1
    $\begingroup$ Nice edits, thank you. I was being lazy. $\endgroup$ Sep 10 '13 at 13:49

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