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?
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$\begingroup$$y''+4y=0$ is a second order differential equation. First, change the equation to
$r^2+4 = 0$
This equation will will have complex conjugate roots, so the final answer would be in the form of $e^{\alpha x}(c_1\sin(\beta x) + c_2 \cos (\beta x))$ where $\alpha$ equals the real part of the complex roots and $\beta$ equals the imaginary part of (one of) the complex roots.
We need to use the quadratic formula \begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array}
In this equation $a =1$, $b=0$, and $c =4$
\begin{array}{*{20}c} {x = \frac{{ -0 \pm \sqrt {0^2 - 4(1)(4)} }}{{2}}} \\ \end{array} \begin{array}{*{20}c} {x = \frac{{ -0 \pm \sqrt {-16} }}{{2}}} \\ \end{array} \begin{array}{*{20}c} {x = \frac{{\pm \sqrt {-16} }}{{2}}} \\ \end{array} Since we can't have negative signs in the square root, we have an imaginary number $i$.
\begin{array}{*{20}c} {x = \frac{{\pm \sqrt {16}i }}{{2}}} \\ \end{array} Take the square root \begin{array}{*{20}c} {x = \frac{{\pm 4i }}{{2}}} \\ \end{array} \begin{array}{*{20}c} {x = \pm 2i} \\ \end{array} Now let's bring this equation back.
$y = e^{\alpha x}(c_1 \sin(\beta x) + c_2 \cos (\beta x))$
$\alpha = 0$ and $\beta = 2$
So your answer would be
$y = e^{0x}(c_1\sin(2x) + c_2\cos (2x))$
Since $e^{0x}$ is $1$ because $0$ multiplied by $x$ is just $0$ and $e^{0}$ is $1$.
The final answer is $y = (c_1\sin(2x) + c_2\cos (2x))$.
$\endgroup$ 3 $\begingroup$Use a "test solution": $y=e^{\lambda x}$. Then the characteristic equation is $\lambda^2+4=0$. Can you proceed from here?
$\endgroup$ 3 $\begingroup$We can just factor it and solve two first order ordinary differential equations instead. Write $$y''+4y=y''+2iy'-2iy'+4y=(y'+2iy)'-2i(y'+2iy)=0.$$ Substitute $z=y'+2iy$ and get $z'-2iz=0\implies z(t)=A e^{2it}$. Substitute back: $$y'+2iy=Ae^{2it}\iff(e^{2it}y)'=Ae^{4it}$$ We then get the desired solution $$y=Ae^{2it}+Be^{-2it}.$$
Needless to say, the same technique works for any second order ordinary differential equation with constant coefficients. You can see a proof here, with special attention to the equal roots case.
Sorry, I just noted you are the OP of the linked question. If this answer is not useful, let me know.
$\endgroup$ $\begingroup$Put P=y'=dy/dx so y"=dP/dx=dP/dy * dy/dx =P dP/dx
y"+4y=0 <=> P dP/dy +4y = 0 <=> P dP = -4y dy <=> (P^2)/2 = -2(y^2) + C <=> P=(C-4y(^2))^(1/2)
P=dy/dx so dy/dx=(C-4y(^2))^(1/2)
dy / (C-4y(^2))^(1/2) = dx
put K^2=C so
(C-4y(^2))^(1/2) = K (1-4y^2/K^2)(^1/2)
and finally integration gives
Primitive of 1 / (K (1-4y^2/K^2)(^1/2)) = x
put u =2y/K so du/dy=2/K and x=1/K * primitive of (1/(1-u^2)^1/2)*K/2*du <=> x= 1/2 * primitive of (1/(1-u^2)^1/2)* du <=> x=1/2 arc sin u + L (L constant) <=> x - L = 1/2 arc sin u <=> 2x - 2L = arc sin u (put - 2L = F) <=> sin (2x + F) = u
u = 2y/K so <=> K/2 sin (2x + F) = y <=> K/2 (sin (2x) cos (F) + cox (2x) sin (F)) = y
Put A= K/2 cos (F) Put B= K/2 sin (F) General solution is y=A sin(2x) + B cos(2x)
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