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The set of solutions of $(E): y' + a(x)y = 0$ ($a\,:\mathbb{R}\,\rightarrow\,\mathbb{R}$ continuous function) is a one-dimensional vector space.

If $f(x) = e^{-\int_0^x a(t)\,\mathrm{d}t}$ is solution of $(E)$, why the general solution of $(E)$ is of the form $Cf(x)$ ($C\,\in\,\mathbb{R}$)?

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Any nonzero element of a one-dimensional vector space spans the space... – Hans Lundmark Sep 16 '12 at 14:30
@Hans Lundmark: so the general solution is of the form $Cf(x)$ because $f(x)$ is in the vector space, and $Cf(x)$ spans it because it is a one-dimensional space? – David Robert Jones Sep 16 '12 at 15:12
@DavidRobertJones: The general solution is of the form $Cf(x)$ because it can span the space. It is a non zero vector as you found it above. – Babak S. Sep 16 '12 at 18:57

I am noting two important theorems which describe the formal form of a general solution of a homogenous linear nth-order differential equation. Let your equation is as: $$a_n(x)y^{(n)}+a_{n-1}(x)y^{(n-1)}+...+a_1(x)y'+a_0(x)y=0$$ where the functions $a_i(x), 1\leq i \leq n$ and $g(x)$ be continuous on an interval $I$ and $a_n(x)\neq0 $ over the interval. Then:

Theorem 1: There exists a linearly independent solutions, called fundamental set of solutions, for above equation on interval $I$ as: $$y_1,y_2,...y_n$$

Theorem 2. if $y_1,y_2,...y_n$ be a fundamental set of solutions of above linear and homogenous nth-order equation on interval $I$. Then the general solution of the equation on $I$ is defined to be: $$y=c_1y_1(x)+c_2y_2(x)+...+c_ny_n(x)$$ wherein $c_i, 1\leq i\leq n$ are arbitrary constants.

Now, I think you can conclude why the general solution of your first order linear and homogenous equation is as $$y=Cf(x)$$

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It's good you answered, lest the question go without one! +1 – amWhy Mar 18 '13 at 0:58

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