Differential system with initial value problem 2nd order

Question

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.

Answer 1

Use Laplace transform:

$$ \begin{cases} y''_1(t)=6y_2(t)−10y_1(t)\\ y''_2(t)=6y_1(t)−10y_2(t) \end{cases}\Longleftrightarrow $$ $$ \begin{cases} \mathcal{L}_t\left[y''_1(t)\right]_{(s)}=\mathcal{L}_t\left[6y_2(t)−10y_1(t)\right]_{(s)}\\ \mathcal{L}_t\left[y''_2(t)\right]_{(s)}=\mathcal{L}_t\left[6y_1(t)−10y_2(t)\right]_{(s)} \end{cases}\Longleftrightarrow $$

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Use:

• $$\mathcal{L}_t\left[1\right]_{(s)}=\frac{1}{s}$$
• $$\mathcal{L}_t\left[y_n(t)\right]_{(s)}=\text{Y}_n(s)$$
• $$\mathcal{L}_t\left[y''_n(t)\right]_{(s)}=s^2\text{Y}_n(s)-sy_n(0)-y'_n(0)$$

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

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

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

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 $y_2\left(\frac{\pi}{2}\right)$:

$$y_2(t)=\frac{\cos(2t)-\cos(4t)}{2}\to y_2\left(\frac{\pi}{2}\right)=\frac{\cos(\pi)-\cos(2\pi)}{2}=-1$$

Answer 2

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

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 $\ds{\,\vec{\mathrm{r}}\,''\pars{x} - \,\mathsf{A}\,\vec{\mathrm{r}}\pars{x} = \vec{0}}$.

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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.

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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}

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