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For a differential equation like this with real coefficients: $$\frac{d^2y}{dx^2}+p \frac{dy}{dx}+q y = 0$$ By solving $$\lambda ^2+p\lambda +q = 0$$ we obtain two eigenvalue $\lambda_1 = a+bi$ and $\lambda_2 = a-bi$, if the discriminant $\Delta$ is smaller than 0. then if we want to obtain two real function solutions linearly independent,

$e^{ax}\cos(bx), e^{ax}\sin(bx)$

Then question is for a differential equation with solution space of dimension $d$, is there any guarantee that the real sub-solution space is also a space of dimension $d$. Is it always possible to find $d$ linearly independent real solutions?

I have to add that the above is just an example. What I'm asking is the general nth order ODE with n linearly independent complex solutions. Is the dimension of its real solution space the same with the complex one?

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Are you considering $d$ to be an arbitrary positive integer?

Then the answer is yes. In fact, one can show the following is true:

For the homogeneous equation with real coefficients: $$\tag{1}\def\sss{} a_{\sss n} y^{\sss(\!n\!)} +a_{\sss n\!-\!1} y^{\sss(\!n\!-\!1\! )} +\cdots+a_{\sss1} y' +a_{\sss0} y = 0,\quad a_n\ne0 $$

The characteristic polynomial (c.p.) is $$ \tag{2}a_{\sss n}x^n+a_{\sss n\!-\!1}x^{n-1} +\cdots+a_{\sss1}x +a_{\sss0} . $$

Find the roots and their multiplicities of (2).

If $c$ is a real root of (2) with multiplicity $k $, then $k$-independent solutions of (1) are $$ e^{ct}, xe^{ct}, x^2 e^{ct}, \ldots, x^{k-1}e^{ct} $$

If $a+bi$ is a complex root of (2) with multiplicity $k$, then $2k$-independent solutions of (1) are $$ e^{at}\sin (bt), x e^{at}\sin (bt), \ldots, x^{k-1} e^{at}\sin (bt) $$ $$ e^{at}\cos (bt), xe^{at}\cos (bt) , \ldots, x^{k-1}e^{at}\cos (bt) $$

Moreover, the set of all solutions found from the above will be independent (for instance, one can compute the Wronskian) and there will be $n$ of them (this follows from the Fundamental Theorem of Algebra).

The general solution to (1) is

$$y_c=c_{\sss 1}y_{\sss1}+c_{\sss2}y_{\sss2}+\cdots+c_{\sss n}y_{\sss n}$$

where $y_1$, $y_2$, $\ldots\,$, $y_n$ are the $n$-solutions found above.

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Hello David, I read now your answer and I have a question... Is the dimension of the solution space of a differential equation always equal to the order of the differential equation? – Mary Star Nov 2 '15 at 11:46

If the differential equation is linear and all coefficients are real, then the real part and the imaginary part of a solution separately solve the equation, so in this case the dimension of the space of real solutions over $\mathbb R$ is the same as the dimension of the space of complex solutions over $\mathbb C$. However, if you make the coefficients complex, this is no longer the case, and you may get fewer real solutions.

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Hello joriki, I read now your answer and I have a question... When is the dimension of the solution space of a differential equation always equal to the order of the differential equation? – Mary Star Nov 2 '15 at 11:54
@MaryStar: That's quite a general question. Generally the set of solutions of a differential equation need not be a vector space or even a manifold (and thus in particular need not have a dimension). Are you perhaps thinking of a particular type of differential equations? Perhaps linear differential equations with constant coefficients, such as in this question? – joriki Nov 2 '15 at 12:40
Yes, such a type of differential equations. – Mary Star Nov 2 '15 at 12:48
@MaryStar: I think you'll find the answer to your question here:… (where "degree" is used synonymously with "order"): The solutions of a homogeneous linear differential equation of order $n$ always form a vector space of dimension $n$. – joriki Nov 2 '15 at 13:52
Ok... Does the same as in that answer stand also at the first case of the following question… ? In this case we have a non-homogeneous differential equation and we are working in a specific ring. – Mary Star Nov 2 '15 at 16:09

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