What kind of vector space does the set of (complex/real) solutions of the following differential equation form? I'm doing a multiple choice exercise and I'm stuck at a question, where there are two possible answers (may be both or none correct).

Consider the differential equation
$$y''+y'+y=0$$
for $y(x)$. Which of the following statements are true?
  
  
*
  
*The set of complex solutions forms a two-dimensional complex vector space.
  
*The set of real solutions forms a two-dimensional real vector space.

I'm really not comfortable with vector spaces and the differentiation of real and complex vector space makes me even more unconfident. I solved the differential equation with the ansatz $y(x) = e^{\lambda x}$ and got to $\lambda^2 + \lambda + 1 = 0$, an equation which does not possess any real solutions. However, when I proceeded, I figured that I could rewrite the exponential function (with complex exponent) with the help of the sine and the cosine, so in the end I got a general solution with no complex coefficients.
What kind of vector space is this?
 A: Both statements are true.
A: As Huy says, both statements are true.
Well, if you are a hammer, all the problems seem nails to you, but since this is a linear differential equation, Laplace transform again gives us the general solution in both, real and complex, cases.
$$
{\cal L}[y''] + {\cal L}[y'] + {\cal L}[y] = 0 \ \Longleftrightarrow \ (s^2 F(s) -sy(0) -y'(0) ) + (sF(s) -y(0)) + F(s) = 0 \ ,
$$
where $F(s) = {\cal L}[y]$. Now, put $a= y(0)$ and $b= y'(0)$ and solve this equation for $F(s)$:
$$
F(s) = \dfrac{as + (a+b)}{s^2+s+1} \ .
$$
Since the polynomial $s^2+s+1$ has no real roots, the distinction between the real and complex cases arrives at this point.
In the complex case, you write 
$$
s^2 + s + 1 = (s-\alpha)(s-\beta)  \ ,
$$
where $\alpha = \dfrac{-1+i\sqrt{3}}{2}$ and $\beta = \dfrac{-1-i\sqrt{3}}{2}$ are the roots of your characteristic polynomial. Hence the solution of your differential equation is
$$
y(t) = A{\cal L}^{-1} \left[\dfrac{1}{s-\alpha} \right] + B {\cal L}^{-1} \left[\dfrac{1}{s-\beta} \right] = A e^{\alpha t} + B e^{\beta t} \ ,
$$
where $A$ and $B$ are complex numbers, depending on $a , b$, which again I let you the pleasure to compute  :-) .
Anyway, the fact is that the solution of your differential equation is the $\mathbb{C}$-linear span of two linearly independent complex functions 
$$
e^{\alpha t} \qquad \text{and} \qquad e^{\beta t} \ .
$$
Thus a $\mathbb{C}$-vector space of dimension two.
In the real case, you write
$$
s^2 + s + 1 = \left(s + \frac{1}{2} \right)^2 + \frac{3}{4}
$$
and this time, the solution set of your differential equation is
$$
y(t) = 
A {\cal L}^{-1} \left[ \frac{s + \frac{1}{2}}{ \left( s+ \frac{1}{2} \right)^2+ \frac{3}{4} }  \right] +
B {\cal L}^{-1} \left[ \frac{ \frac{\sqrt{3}}{2}}{ \left( s+ \frac{1}{2} \right)^2+ \frac{3}{4} }  \right] 
= A e^{-\frac{t}{2}}\cos \left( \frac{\sqrt{3}}{2}t\right) +
B e^{-\frac{t}{2}}\sin \left( \frac{\sqrt{3}}{2}t\right)  \ ,
$$
where $A$ and $B$ are real numbers, depending on $a,b$, which... Anyway again: the solution set is the $\mathbb{R}$-linear span of two linearly independent real functions 
$$
e^{-\frac{t}{2}}\cos \left( \frac{\sqrt{3}}{2} t \right) \qquad \text{and} \qquad e^{-\frac{t}{2}}\sin \left( \frac{\sqrt{3}}{2} t \right)  \ .
$$
Thus, a real vector space of dimension two.
Alternatively, you could read what Wikipedia says about linear differential equations.
