Seen as I have written lots in comments already I would like to make an answer. I saw Hagen's edited answer pop up while writing, but I'm not sure I understand it.
Your idea is wrong. A solution of a linear differential equation of order $n$ (with constant coefficients) has $n$ degrees of freedom.
I don't know if you have had linear algebra, but I will try to give an explanation of how it is used in the theory of linear differential equations (and hope that this is correct, if not any corrections are appreciated.)
First of all you need to know that certain sets of functions, with a natural addition and scalar multiplication are vector spaces, i.e. they satisfy the axioms of a vector space. This means we will be able to apply theorems established in linear algebra to solve problems regarding these sets of functions. In particular the set we will consider is $\mathsf C^\infty$, the set of all functions from $\mathbb R\to \mathbb C$ with derivatives of all orders.
It turns out that $\mathsf{D}:\mathsf C^\infty\to\mathsf C^\infty,f(x)\mapsto f'(x)$ is a linear operator and so is any sum, scalar multiple or composition of $\mathsf D$. This means we can apply theorems we know about linear transformations to $\mathsf D$. It also is the case that when you have a differential operator $p(\mathsf D)$, basically a polynomial in $\mathsf D$, it can be factored just like the characteristic polynomial of the differential equation. This can be used to solve the homogeneous equation, example: $f''(x)+3f'(x)+2f(x)=0$, can be written as $\mathsf D^2(f)+3\mathsf D(f)+2\mathrm{id_{\mathsf C^\infty}}(f)=0$ and just as the characteristic polynomial is factored, this is $(\mathsf D+2)(\mathsf D +1)(f)=\\(\mathsf D+2)(f'(x)+f(x))\\f''(x)+f'(x)+2f'(x)+2f(x)=\\f''(x)+3f'(x)+2f(x)=0$ which we see is the same as the equation we started from. This can be used in a similar way as factoring polynomials to say that $f''(x)+2f'(x)+f(x)=0\iff f\in\mathrm{ker}(\mathsf D+2)+\mathrm{ker}(\mathsf D+1)$ (you might expect it to lie in the union of the kernels, but this is not the case.) This works analogously for higher orders and by the similarity between the factorization of the differential operator and the characteristic polynomial we have that any $f$ which satisfies the homogeneous equation lies in the sum of these kernels, each with dimension $1$, so we end up with a solution space with the same dimension as the order of the equation. This is a bit of a hand waving argument and here I have assumed there aren't any repeated roots, but it works in general.
Now I will explain the way I see it for the case of an inhomogeneous equation (I hope it is right). You probably know the idea of a line defined by a support vector and a direction vector. I will try to show the same sort of idea applies here. Any function in the kernel which we saw in the previous paragraph can be added to a solution of a differential equation of the form $f''(x)+af'(x)+bf(x)=g(x)$. In this way we can get the set of solutions if we know one solution $f_p$ to the equation above and we know the solution space of the homogeneous equation $\mathsf K_p$ (by the method above), that is $f_p+\mathsf K_H$. So if you think of it geometrically, all solutions to $f''(x)+af'(x)+bf(x)=g(x)$ will lie in a "plane" which is parallel to the "plane" we get from the homogeneous equation and which is supported by a (not unique) support vector which you can find for example by educated guess. It follows (think of the plane or line, where the support vector adds no degree of freedom) that there are just as many degrees of freedom in the solution to the inhomogeneous equation as in the solution to the homogeneous equation, which we have seen to be the order of the equation.
I have written quite a lot and hope it is at least in some way helpful and correct. Else, just post a comment.