Differential system with initial value problem 2nd order i got a problem solving this Diff. system with initial value problem 2nd order.
$$ y''_1=−10y_1+6y_2 $$
$$y''_2=6y_1−10y_2$$
$$y_1(0)=1,y_2(0)=0,y_1'(0)=0,y_2'(0)=0 $$
i need the value for: $$ y_2(\pi/2)= ? $$
according my math script, i got 2 different values and i dont know which one is right or both wrong.
my values are: $-1$ and $0$
I hope someone can tell me which one is right or if both are wrong which value would be right, so i can look over and try to find my mistake.
Thx.
 A: Use Laplace transform:
$$
\begin{cases}
y''_1(t)=6y_2(t)−10y_1(t)\\
y''_2(t)=6y_1(t)−10y_2(t)
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
\mathcal{L}_t\left[y''_1(t)\right]_{(s)}=\mathcal{L}_t\left[6y_2(t)−10y_1(t)\right]_{(s)}\\
\mathcal{L}_t\left[y''_2(t)\right]_{(s)}=\mathcal{L}_t\left[6y_1(t)−10y_2(t)\right]_{(s)}
\end{cases}\Longleftrightarrow
$$

Use:


*

*$$\mathcal{L}_t\left[1\right]_{(s)}=\frac{1}{s}$$

*$$\mathcal{L}_t\left[y_n(t)\right]_{(s)}=\text{Y}_n(s)$$

*$$\mathcal{L}_t\left[y''_n(t)\right]_{(s)}=s^2\text{Y}_n(s)-sy_n(0)-y'_n(0)$$


$$
\begin{cases}
s^2\text{Y}_1(s)-sy_1(0)-y'_1(0)=6\text{Y}_2(s)−10\text{Y}_1(s)\\
s^2\text{Y}_2(s)-sy_2(0)-y'_2(0)=6\text{Y}_1(s)−10\text{Y}_2(s)
\end{cases}\Longleftrightarrow
$$

Use the initial conditions $y_1(0)=1,y_2(0)=0,y′_1(0)=0,y′_2(0)=0$:

$$
\begin{cases}
s^2\text{Y}_1(s)-s=6\text{Y}_2(s)−10\text{Y}_1(s)\\
s^2\text{Y}_2(s)=6\text{Y}_1(s)−10\text{Y}_2(s)
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
s^2\text{Y}_1(s)+10\text{Y}_1(s)=6\text{Y}_2(s)+s\\
s^2\text{Y}_2(s)+10\text{Y}_2(s)=6\text{Y}_1(s)
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
\text{Y}_1(s)\left[s^2+10\right]=6\text{Y}_2(s)+s\\
\text{Y}_2(s)\left[s^2+10\right]=6\text{Y}_1(s)
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
\text{Y}_1(s)=\frac{6\text{Y}_2(s)+s}{s^2+10}\\
\text{Y}_2(s)=\frac{6\text{Y}_1(s)}{s^2+10}
\end{cases}
$$
Now, using substitution:


*

*$$\text{Y}_1(s)=\frac{s(10+s^2)}{s^4+20s^2+64}$$

*$$\text{Y}_2(s)=\frac{6s}{s^4+20s^2+64}$$


With inverse Laplace transform:


*

*$$y_1(t)=\frac{\cos(2t)+\cos(4t)}{2}$$

*$$y_2(t)=\frac{\cos(2t)-\cos(4t)}{2}$$


So, for $y_2\left(\frac{\pi}{2}\right)$:
$$y_2(t)=\frac{\cos(2t)-\cos(4t)}{2}\to y_2\left(\frac{\pi}{2}\right)=\frac{\cos(\pi)-\cos(2\pi)}{2}=-1$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Lets $\ds{\,\vec{\mathrm{r}}\pars{x} \equiv
{\,\mathrm{y}_{1}\pars{x} \choose \,\mathrm{y}_{2}\pars{x}}}$ and
  $\ds{\,\mathsf{A} \equiv
\pars{\begin{array}{rr}
\ds{-10} & \ds{6}
\\
\ds{6} & \ds{-10}
\end{array}}}$ such that $\ds{\,\vec{\mathrm{r}}\,''\pars{x} - \,\mathsf{A}\,\vec{\mathrm{r}}\pars{x} = \vec{0}}$. 


Note that $\ds{\,\mathsf{k}^{2} = \,\mathsf{A}}$ where
$\ds{\quad\,\mathsf{k} \equiv
\pars{\begin{array}{rr}
\ds{3} & \ds{-1}
\\
\ds{-1} & \ds{3}
\end{array}}\ic\quad}$ with eigenvalues $\ds{\pars{4\ic, 2\ic}}$ and orthonormalized eigenvectors
$\ds{\braces{\vphantom{\huge A^{A^{A^{A}}}}
{1 \over \root{2}}{-1 \choose \phantom{-}1},\
{1 \over \root{2}}{1 \choose 1}}}$, respectively.

 Then,
\begin{align}
\,\vec{\mathrm{r}}\pars{x} & =
\cosh\pars{\,\mathsf{k}x}{1 \choose 0}
\\[4mm] & =
\cosh\pars{4\ic x}{1 \over \root{2}}{-1 \choose 1}
{1 \over \root{2}}\pars{-1\quad 1}{1 \choose 0} +
\cosh\pars{2\ic x}{1 \over \root{2}}{1 \choose 1}
{1 \over \root{2}}\pars{1\quad 1}{1 \choose 0}
\\[4mm] & =
\half\cos\pars{4x}
\pars{\begin{array}{rr}
\ds{1} & \ds{-1}
\\
\ds{-1} & \ds{1}
\end{array}}{1 \choose 0} +
\half\cos\pars{2x}
\pars{\begin{array}{rr}
\ds{1} & \ds{1}
\\
\ds{1} & \ds{1}
\end{array}}{1 \choose 0}
\\[4mm] & =
\half\pars{\begin{array}{r}
\ds{\cos\pars{4x} + \cos\pars{2x}}
\\[1mm]
\ds{-\cos\pars{4x} + \cos\pars{2x}}
\end{array}}\
\imp\
\left\lbrace\begin{array}{rcl}
\ds{\,\mathrm{y}_{1}\pars{x}} & \ds{=} &
\ds{\half\bracks{\cos\pars{2x} + \cos\pars{4x}}}
\\[2mm]
\ds{\,\mathrm{y}_{2}\pars{x}} & \ds{=} &
\ds{\half\bracks{\cos\pars{2x} - \cos\pars{4x}}}
\end{array}\right.
\end{align}

$$
\color{#f00}{\,\mathrm{y}_{2}\pars{\pi \over 2}} =
\half\bracks{\cos\pars{2\,{\pi \over 2}} - \cos\pars{4\,{\pi \over 2}}} =
\color{#f00}{-1}
$$
