How do I solve $y''+4y=0$? This problem is in Penney's Elementary Differential Equations, listed as a reducible, 2nd-order DE.  The chapter has taught two techniques to be used, which are for when either $x$ or $y(x)$ is missing.  It didn't show how to solve when $y'(x)$ is missing.
I checked with WolframAlpha, and it suggests starting by assuming that $y$ is proportional to $e^{\lambda x}$.  
If I didn't know to assume this, how could I solve this otherwise?

$y''+4y=0$ is a second order differential equation.
First, change the equation to
$r^2+4 = 0$
This equation will will have complex conjugate roots, so the final answer would be in the form of $e^{\alpha x}(c_1\sin(\beta x) + c_2 \cos (\beta x))$
where $\alpha$ equals the real part of the complex roots and $\beta$ equals the imaginary part of (one of) the complex roots. 
We need to use the quadratic formula 
\begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array}
In this equation $a =1$, $b=0$, and $c =4$
\begin{array}{*{20}c} {x = \frac{{ -0 \pm \sqrt {0^2 - 4(1)(4)} }}{{2}}} \\ \end{array}
\begin{array}{*{20}c} {x = \frac{{ -0 \pm \sqrt {-16} }}{{2}}} \\ \end{array}
\begin{array}{*{20}c} {x = \frac{{\pm \sqrt {-16} }}{{2}}} \\ \end{array}
Since we can't have negative signs in the square root, we have an imaginary number $i$.
\begin{array}{*{20}c} {x = \frac{{\pm \sqrt {16}i }}{{2}}} \\ \end{array}
Take the square root
\begin{array}{*{20}c} {x = \frac{{\pm 4i }}{{2}}} \\ \end{array}
\begin{array}{*{20}c} {x = \pm 2i} \\ \end{array}
Now let's bring this equation back. 
$y = e^{\alpha x}(c_1 \sin(\beta x) + c_2 \cos (\beta x))$
$\alpha = 0$ and $\beta = 2$
So your answer would be
$y = e^{0x}(c_1\sin(2x) + c_2\cos (2x))$
Since $e^{0x}$ is $1$ because $0$ multiplied by $x$ is just $0$ and $e^{0}$ is $1$. 
The final answer is $y = (c_1\sin(2x) + c_2\cos (2x))$.
