Find the solution of the following Second Order Differential Equationy″−9y=0,y(0)=2,y′(0)=0

We have given
${\displaystyle y9y=0}$
Auxiliary equation of given differential equation
${\displaystyle m^2-9=0}$
${\displaystyle (m-3)(m+3)=0}$
${\displaystyle m=3,-3}$
Therefore, general solution of differential equation
${\displaystyle y=c_1{e}^{3x}+c_2{e}^{-3x}}$
We have ${\displaystyle y\left(0\right)=2}$
${\displaystyle \Rightarrow y\left(0\right)=c_1{e}^{3\left(0\right)}+c_2{e}^{-3\left(0\right)}}$
${\displaystyle 2=c_1+c_2}$ (i)
${\displaystyle y^{\prime }\left(x\right)=3c_1{e}^{3x}-3c_2{e}^{-3x}}$
We have ${\displaystyle y^{\prime }\left(0\right)=0}$
${\displaystyle \Rightarrow y^{\prime }\left(x\right)=3c_1{e}^{3\left(0\right)}-3c_2{e}^{-3\left(0\right)}}$
${\displaystyle 0=3c_1-3c_2}$
${\displaystyle 0=c_1-c_2}$ (2)
Now, solve equation (1) and (2) we get
${\displaystyle c_1=1,c_2=1}$
Therefore, particular solution is
${\displaystyle y=e^{3x}+e^{-3x}}$

${\displaystyle y9y=0,y\left(0\right)=2,y^{\prime }\left(0\right)=0}$
The auxiliary equation for this equation is:
${\displaystyle m^2-9=0}$
On solving this equation for "m".
${\displaystyle m^2-9=0}$
${\displaystyle m^2=9}$
${\displaystyle m=\pm \sqrt{9}}$
${\displaystyle m=\pm 3}$
${\displaystyle m=-3,3}$
Then, the solution of this given differential equation is:
${\displaystyle y=c_1{e}^{-3x}+c_2{e}^{3x}}$
Given that,
${\displaystyle y\left(0\right)=2}$
So, replace x with 0 and y with 2 in it.
${\displaystyle y=c_1{e}^{-3x}+c_2{e}^{3x}}$
${\displaystyle 2=c_1{e}^{-3\left(0\right)}+c_2{e}^{3\left(0\right)}}$
${\displaystyle 2=c_1+c_2}$ (1)
Now, differentiate the obtained solution with respect to x.
${\displaystyle y^{\prime }=-3c_1{e}^{-3x}+3c_2{e}^{3x}}$
And, also given that,
${\displaystyle y^{\prime }\left(0\right)=0}$
So, replace x with 0 and y with 0 in it.
${\displaystyle y^{\prime }=-3c_1{e}^{-3x}+3c_2{e}^{3x}}$
${\displaystyle 0=-3c_1{e}^{-3\left(0\right)}+3c_2{e}^{3\left(0\right)}}$
${\displaystyle 0=-3c_1+3c_2}$ (2)
Now, solve equations (1) and (2) by the elimination or substitution method.
From equation (2), ${\displaystyle c_1=c_2}$, put this value in equation (1), which gives ${\displaystyle c_1=c_2=1}$
Then, ${\displaystyle c_1=1,c_2=1}$
Thus, the solution of the given differential equation is:
${\displaystyle y=c_1{e}^{-3x}+c_2{e}^{3x}}$
${\displaystyle y=1e^{-3x}+1e^{3x}}$
${\displaystyle y=e^{-3x}+e^{3x}}$

$\begin{array}{}{y}^{″}-9y=0\\ m^2-9=0\\ (m-3)(m+3)=0\\ m=3,-3\\ y=c_1{e}^{3x}+c_2{e}^{-3x}\\ y(0)=2\\ \Rightarrow y(0)=c_1{e}^{3(0)}+c_2{e}^{-3(0)}\\ 2=c_1+c_2(i)\\ y^{\prime }(x)=3c_1{e}^{3x}-3c_2{e}^{-3x}\\ y^{\prime }(0)=0\\ \Rightarrow y^{\prime }(x)=3c_1{e}^{3(0)}-3c_2{e}^{-3(0)}\\ 0=3c_1-3c_2\\ 0=c_1-c_2(2)\\ c_1=1,c_2=1\\ y=e^{3x}+e^{-3x}\end{array}$