Find a general solution. Check your answer by substitution ODEs of this kind have important applications to be discussed in Secs. 2.4, 2.7, and 2.9. 4y''-25y=0

${\color{#c34632} {4y''-25y=0}}\;\;\;\; \Rightarrow \;\; \textbf{Homogeneous 2nd Order D.E , With Constant Coeff.}$

$\Rightarrow \textbf{Form :}\;\; \boxed{\;ay''+by'+cy=0\;}$

$Where\; : \;a\;,\;b\;,\;c \;\; \Rightarrow Constants$

$In\; order\; to\; solve\; this\; D.E \;, \;we\; need \;to\; get\; the\; "characterestic\; equation\;" by\; replacing$

$\boxed{y'' \;\; \Rightarrow \;\; m^{2}\;\;\;,\;\;\; y'\;\; \Rightarrow \;\; m\;\;\;,\;\;\;y\;\; \Rightarrow \;\; 1}$

$\textbf{Then Characterestic Eq :}\;\; \Rightarrow \;\;{\color{#c34632} {4m^{2}-25=0}}$

$(2m-5)(2m+5)=0$

$m_{1}=\dfrac{5}{2}\;\;\;,\;\;\;m_{2}=\dfrac{-5}{2}$

$\therefore \;\; \textbf{Solution is : }\; {\color{#c34632} {y=C_{1}e^{m_{1}x}+C_{2}e^{m_{2}x}}}$

$$\therefore \boxed{\;\color{#4257b2} {y=C_{1}e^{(5/2)x}+C_{2}e^{(-5/2)x}}\;}$$

The characteristic equation is

$$4\lambda ^2 - 25 = 0.$$

Let's find the roots of the characteristic equation.

$$\begin{align*} (2 \lambda) ^2 - 5^2 = 0 &\iff \lambda (2 \lambda - 5)(2 \lambda + 5) = 0 \\ & \iff 2 \lambda - 5 = 0 \quad \vee \quad 2 \lambda + 5 = 0 \\ & \iff \lambda = \frac{5}{2} \quad \vee \quad \lambda = -\frac{5}{2} \end{align*}$$

So, it has the distinct real roots:

$$\lambda _1 = \dfrac{5}{2} \quad \wedge \quad \lambda _2 = -\dfrac{5}{2}$$

A basis is:

$$\begin{align*} \begin{cases} & \phi _1 (x) = e^{\lambda _1 x} = e^{\frac{5x}{2}} \\ & \phi _2 (x)= e^{\lambda _2 x} = e^{-\frac{5x}{2}} \end{cases} \end{align*}$$

The general solution is:

$$\begin{align*} y &= c_1 \phi _1 (x) + c_2 \phi _2 (x) \\ & = \boxed{{\color{#4257b2}{ c_1e^{\frac{5x}{2}} + c_2e^{-\frac{5x}{2}} }}} \end{align*}$$