Inhomogenous Differential system

Question

This is the system,

$$ y'_1 =y_1+y_2+1 $$ $$ y'_2= -y_1+y_2+1 $$ initial value problem which fulfill: $$y_1(0)=1$$ $$y_2(0)=-1$$ value to find $$y_1(π)= \text{?}$$

Answer 1

Use Laplace transform:

$$ \begin{cases} y'_1(t)=y_1(t)+y_2(t)+1\\ y'_2(t)=y_2(t)-y_1(t)+1 \end{cases}\Longleftrightarrow $$ $$ \begin{cases} \mathcal{L}_t\left[y'_1(t)\right]_{(s)}=\mathcal{L}_t\left[y_1(t)+y_2(t)+1\right]_{(s)}\\ \mathcal{L}_t\left[y'_2(t)\right]_{(s)}=\mathcal{L}_t\left[y_2(t)-y_1(t)+1\right]_{(s)} \end{cases}\Longleftrightarrow $$

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Use:

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

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$$ \begin{cases} s\text{Y}_1(s)-y_1(0)=\text{Y}_1(s)+\text{Y}_2(s)+\frac{1}{s}\\ s\text{Y}_2(s)-y_2(0)=\text{Y}_2(s)-\text{Y}_1(s)+\frac{1}{s} \end{cases}\Longleftrightarrow $$

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Use the initial conditions $y_1(0)=1$ and $y_2(0)=-1$:

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$$ \begin{cases} s\text{Y}_1(s)-1=\text{Y}_1(s)+\text{Y}_2(s)+\frac{1}{s}\\ s\text{Y}_2(s)+1=\text{Y}_2(s)-\text{Y}_1(s)+\frac{1}{s} \end{cases}\Longleftrightarrow $$ $$ \begin{cases} s\text{Y}_1(s)-\text{Y}_1(s)=\text{Y}_2(s)+\frac{1}{s}+1\\ s\text{Y}_2(s)-\text{Y}_2(s)=\frac{1}{s}-1-\text{Y}_1(s) \end{cases}\Longleftrightarrow $$ $$ \begin{cases} \text{Y}_1(s)\left[s-1\right]=\text{Y}_2(s)+\frac{1}{s}+1\\ \text{Y}_2(s)\left[s-1\right]=\frac{1}{s}-1-\text{Y}_1(s) \end{cases}\Longleftrightarrow $$ $$ \begin{cases} \text{Y}_1(s)=\frac{\text{Y}_2(s)+\frac{1}{s}+1}{s-1}\\ \text{Y}_2(s)=\frac{\frac{1}{s}-1-\text{Y}_1(s)}{s-1} \end{cases} $$

Now, using substitution:

• $$\text{Y}_1(s)=\frac{s-1}{2+s(s-2)}$$
• $$\text{Y}_2(s)=-\frac{1}{s}-\frac{1}{2+s(s-2)}$$

With inverse Laplace transform:

• $$y_1(t)=e^t\cos(t)$$
• $$y_2(t)=-1-e^t\sin(t)$$

So, for $y_1(\pi)$:

$$y_1(t)=e^t\cos(t)\to y_1(\pi)=e^{\pi}\cos(\pi)=e^{\pi}\cdot-1=-e^{\pi}\approx-23.1407$$

Answer 2

Another solution, using linear algebra.

Consider the matrix $A = \pmatrix{ 1 & 1 \\ -1 & 1}$ and $b(t) = \pmatrix{1 \\ 1}$. Let $Y(t) = \pmatrix{y_1(t) \\ y_2(t)}$.

The problem is equivalent to solving

$$ Y'(t) = AY(t) + b(t).$$

Remark that the characteristic polynomial of $A$ is $X^2 - 2X + 2 = (X-1+i)(X-1-i)$, meaning $A$ is diagonalizable. Two associated eigenvectors are $v_1 = \pmatrix{1 \\ -i}$ and $v_2 = \pmatrix{1\\i}$.

Let $P = \pmatrix{1 & 1 \\ -i & i}$ such that $P^{-1}AP = D = \pmatrix{1-i & 0 \\ 0 & 1+i }$, and $Y = PX$. One has

$$ Y' = AY + b \Longleftrightarrow PX' = APX + b$$

hence $$X' = P^{-1}APYX + P^{-1}b = DX + P^{-1}b$$

One has $P^{-1} = \pmatrix{\frac{1}{2} & \frac{i}{2} \\ \frac{1}{2} & -\frac{i}{2}}$ hence $P^{-1}b(t) = \frac{1}{2}\pmatrix{1+i \\ 1-i}$. The system is then equivalent to

$$ \cases{x_1' = (1-i)x_1 + \frac{1}{2}(1+i) \\ x_2' = (1+i)x_2 + \frac{1}{2}(1-i)} $$

which solutions are

$$ \cases{x_1(t) = c_1 e^{(1-i)t}-\frac{i}{2} \\ x_2(t) = c_2 e^{(1+i)t}+\frac{i}{2} }, \qquad (c_1, c_2) \in \mathbb{R}^2.$$

Initial conditions lead to

$$ c_1 = c_2 = \frac{1}{2}. $$

Now, using $Y=PX$, one finally has

$$ \bbox[lightgreen,5px,border:2px solid green]{\cases{y_1(t) = e^t\left(\dfrac{e^{it} + e^{-it}}{2}\right) = e^t \cos(t) \\ y_2(t) = ie^t\left(\dfrac{e^{it} - e^{-it}}{2}\right) - 1 = -e^t \sin(t) - 1}}. $$

One can directly deduce that $\bbox[lightgreen,5px,border:2px solid green]{y_1(\pi) = -e^{\pi}}$.

Answer 3

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

Lets $\ds{% \,\mathsf{A} \equiv \pars{\begin{array}{rr} \ds{0} & \ds{-\ic} \\ \ds{\ic} & \ds{0} \end{array}}\ \mbox{and}\ \,\mathsf{A}_{0} \equiv \pars{\begin{array}{rr} \ds{1} & \ds{0} \\ \ds{0} & \ds{1} \end{array}}}$. Note that $\ds{\,\mathsf{A}^{2} = \,\mathsf{A}_{0}}$.

\begin{align} &\pars{\totald{}{x} - \,\mathsf{A}_{0} - \ic\,\mathsf{A}} {\,\mathrm{y}_{1}\pars{x} \choose \,\mathrm{y}_{2}\pars{x} + 1} = {0 \choose 0}\ \imp\ {\,\mathrm{y}_{1}\pars{x} \choose \,\mathrm{y}_{2}\pars{x} + 1} = \exp\pars{\bracks{\,\mathsf{A}_{0} + \ic\,\mathsf{A}}x}{1 \choose 0} \end{align}

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\begin{align} {\,\mathrm{y}_{1}\pars{x} \choose \,\mathrm{y}_{2}\pars{x} + 1} & = \exp\pars{x}\exp\pars{\ic\,\mathsf{A}x}{1 \choose 0} = \exp\pars{x}\bracks{\cos\pars{x} + \ic\,\mathsf{A}\sin\pars{x}}{1 \choose 0} \\[4mm] & = \exp\pars{x} \pars{\begin{array}{rr} \ds{\cos\pars{x}} & \ds{\sin\pars{x}} \\[2mm] \ds{-\sin\pars{x}} & \ds{\cos\pars{x}} \end{array}}{1 \choose 0} \end{align}

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$$ \,\mathrm{y}_{1}\pars{x} = \expo{x}\cos\pars{x}\quad\imp\quad\, \color{#f00}{\mathrm{y}_{1}\pars{\pi}} = \color{#f00}{-\expo{\pi}} $$