Solving the ODE $y''-4y=0$ I have $$y''-4y=0$$
I am trying to get to the solution which is $$y=c_1 e^{2x}+c_2 e^{-2x}$$
I am at the stage of $$\frac{1}{y}d^2y=4dx^2$$
Integrating both sides I get $$y(\ln(y)-1)=2x^2$$
Am I on the right lines?
Where do I go from here?
 A: Integrating $d^{2}y$
  and $dx^{2}$
  is unclear for me. I propose this redaction : let us denote by $y''$
  the second derivative of $y$
  with respect to $x$
 . Multiplying by $y'$
  the both sides of the equation yields to $$y'y''=4yy'.$$
 Formally integrating gives us$$\frac{\left(y'\right)^{2}}{2}=4\frac{y^{2}}{2}$$
 i.e.$$y'=\pm2y.$$
 Then we divide by $y$
  (where it is allowed!) and obtain $$\frac{y'}{y}=\pm2$$
 and after integration, we get $$\ln y=\pm2x+C$$
 with a real constant $C$
 , whence $$y\left(x\right)=\lambda e^{\pm2x}$$
 with a positive real constant $\lambda$
 .
Edit :
If you are seeking a solution of the form $y\left(x\right)=\lambda e^{rx}$
  where $\lambda\neq0$
  and $r\neq0$
  are real constants, then you obtain$$y'\left(x\right)=\lambda re^{rx}$$
 and $$y''\left(x\right)=\lambda r^{2}e^{rx}.$$
 Hence, the general equation $$ay''+by'+cy=0\,\,\,\,\,\left(*\right)$$
  becomes$$\left(a\lambda r^{2}+b\lambda r+c\lambda\right)e^{rx}=0$$
 and then you must satisfy to algebraic equation $$ar^{2}+br+c=0.$$
 The polynomial $ar^{2}+br+c$
  is the characteristic polynomial of the equation $\left(*\right)$ : it only depends on the coefficients $a,b,c$ of the differential equation.
A: Firstly, you must  get the characteristic equation which equal:
$$m^2-4=0$$
the two roots of the equation are
$$m_1=2$$ and $$m_2=-2$$
then the solution becomes
$$y=C_1e^{2x}+C_2e^{-2x}$$
A: Substitute $y=e^{mx}$ into the differential equation, this gives the charcteristic equation $$e^{mx}(m^2-4)=0.$$ Since $e^{mx} > 0$ for all $x \in \mathbb{R}$, we are left with $$m^2-4=0.$$ Solving for roots, we have $m_1=2$ and $m_2=-2$. The solution to your second-order differential equation takes the form $$y=c_1e^{m_1x}+c_2e^{m_2x}.$$
