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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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    @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.2012-09-16

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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! +12013-03-18