Solving coupled linear systems exactly Problem set A Worked solutions

Write each system in matrix form.

1. (a) \(\dfrac{dx}{dt}=4x+3y\) and \(\dfrac{dy}{dt}=2x-y\) are written as \(\begin{pmatrix}x'\\y'\end{pmatrix}=\mathbf M\begin{pmatrix}x\\y\end{pmatrix}\). Write down \(\mathbf M\).

   • Row 1: the coefficients of x and y in dx/dt. Row 2: those in dy/dt.
   • \(\mathbf M=\begin{pmatrix}4&3\\2&-1\end{pmatrix}\)

2. (b) \(\dfrac{dx}{dt}=2y-3x\) and \(\dfrac{dy}{dt}=x\) are written in the same way. Write down \(\mathbf M\).

   • A term that is missing has coefficient 0.
   • The terms may be written in any order: match each to x or y.
   • \(\mathbf M=\begin{pmatrix}-3&2\\1&0\end{pmatrix}\)

Use the characteristic equation, then solve for each eigenvector. Any non-zero multiple of an eigenvector is correct.

1. (a) Find the eigenvalues and eigenvectors of \(\mathbf M=\begin{pmatrix}3&1\\2&2\end{pmatrix}\).

   • \(\det(\mathbf M-\lambda\mathbf I)=\lambda^2-5\lambda+4=(\lambda-1)(\lambda-4)=0\)
   • \(\lambda=1:\ 2x+y=0\Rightarrow\mathbf p_1=\begin{pmatrix}1\\-2\end{pmatrix}\)
   • \(\lambda=4:\ -x+y=0\Rightarrow\mathbf p_2=\begin{pmatrix}1\\1\end{pmatrix}\)

2. (b) Find the eigenvalues and eigenvectors of \(\mathbf M=\begin{pmatrix}0&1\\6&1\end{pmatrix}\).

   • \(\det(\mathbf M-\lambda\mathbf I)=\lambda^2-\lambda-6=(\lambda+2)(\lambda-3)=0\)
   • \(\lambda=-2:\ 2x+y=0\Rightarrow\mathbf p_1=\begin{pmatrix}1\\-2\end{pmatrix}\)
   • \(\lambda=3:\ -3x+y=0\Rightarrow\mathbf p_2=\begin{pmatrix}1\\3\end{pmatrix}\)

Consider the system \(\dfrac{dx}{dt}=2x+2y\) and \(\dfrac{dy}{dt}=x+3y\). Find the eigenvalues and eigenvectors of the matrix of the system, and write down the general solution.

• Matrix: \(\mathbf M=\begin{pmatrix}2&2\\1&3\end{pmatrix}\)
• Eigenvalues 1 and 4; eigenvectors \(\begin{pmatrix}2\\-1\end{pmatrix}\) and \(\begin{pmatrix}1\\1\end{pmatrix}\) (calculator, or the characteristic equation)
• General solution: \(\begin{pmatrix}x\\y\end{pmatrix}=Ae^{t}\begin{pmatrix}2\\-1\end{pmatrix}+Be^{4t}\begin{pmatrix}1\\1\end{pmatrix}\)

Consider the system \(\dfrac{dx}{dt}=y\) and \(\dfrac{dy}{dt}=2x+y\). The eigenvalues of the matrix are \(-1\) and \(2\). Given that \(x(0)=5\) and \(y(0)=4\), find \(x(t)\) and \(y(t)\).

• Eigenvectors, from \((\mathbf M-\lambda\mathbf I)\mathbf v=\mathbf 0\): \(\begin{pmatrix}1\\-1\end{pmatrix}\) for \(\lambda=-1\) and \(\begin{pmatrix}1\\2\end{pmatrix}\) for \(\lambda=2\)
• General solution: \(\begin{pmatrix}x\\y\end{pmatrix}=Ae^{-t}\begin{pmatrix}1\\-1\end{pmatrix}+Be^{2t}\begin{pmatrix}1\\2\end{pmatrix}\)
• At t = 0: \(A+B=5,\ -A+2B=4\)
• \(A=2,\ B=3\)
• \(x=2e^{-t}+3e^{2t},\ y=-2e^{-t}+6e^{2t}\)

Consider the system \(\dfrac{dx}{dt}=x+y\) and \(\dfrac{dy}{dt}=2x\). The eigenvalues of the matrix are \(-1\) and \(2\). Given that \(x(0)=4\) and \(x'(0)=5\), find \(x(t)\).

• Put t = 0 in \(x'=x+y\): \(5=1(4)+1(y(0))\), so \(y(0)=1\)
• Eigenvectors \(\begin{pmatrix}1\\-2\end{pmatrix}\) and \(\begin{pmatrix}1\\1\end{pmatrix}\); general solution \(\begin{pmatrix}x\\y\end{pmatrix}=Ae^{-t}\begin{pmatrix}1\\-2\end{pmatrix}+Be^{2t}\begin{pmatrix}1\\1\end{pmatrix}\)
• At t = 0: \(A+B=4,\ -2A+B=1\), so \(A=1,\ B=3\)
• The x row: \(x=e^{-t}+3e^{2t}\)

Two tanks are connected by pipes. After \(t\) hours, tank 1 holds \(x\) kg of salt and tank 2 holds \(y\) kg. The amounts are modelled by \(\dfrac{dx}{dt}=-3x+y\) and \(\dfrac{dy}{dt}=2x-2y\). At the start tank 1 holds 9 kg and tank 2 holds none.

1. (a) Write the system in the form \(\begin{pmatrix}x'\\y'\end{pmatrix}=\mathbf M\begin{pmatrix}x\\y\end{pmatrix}\).

   • \(\mathbf M=\begin{pmatrix}-3&1\\2&-2\end{pmatrix}\)

2. (b) Find the eigenvalues and eigenvectors of \(\mathbf M\).

   • \(\det(\mathbf M-\lambda\mathbf I)=\lambda^2+5\lambda+4=(\lambda+4)(\lambda+1)=0\)
   • \(\lambda=-4:\ x+y=0\Rightarrow\mathbf p_1=\begin{pmatrix}1\\-1\end{pmatrix}\)
   • \(\lambda=-1:\ -2x+y=0\Rightarrow\mathbf p_2=\begin{pmatrix}1\\2\end{pmatrix}\)

3. (c) Find \(x(t)\) and \(y(t)\).

   • General solution: \(\begin{pmatrix}x\\y\end{pmatrix}=Ae^{-4t}\begin{pmatrix}1\\-1\end{pmatrix}+Be^{-t}\begin{pmatrix}1\\2\end{pmatrix}\)
   • At t = 0: \(A+B=9,\ -A+2B=0\)
   • \(A=6,\ B=3\)
   • \(x=6e^{-4t}+3e^{-t},\ y=-6e^{-4t}+6e^{-t}\)

4. (d) Find the amount of salt in tank 1 after 1 hour, and say what happens to both amounts as \(t\to\infty\).

   • \(x(1)=6e^{-4(1)}+3e^{-(1)}=1.214\)
   • Both exponents are negative, so both terms tend to 0: the tanks end up with no salt

Consider the system \(\dfrac{dx}{dt}=5x-y\) and \(\dfrac{dy}{dt}=3x+y\) with \(x(0)=6\) and \(y(0)=10\).

1. (a) Find the eigenvalues and eigenvectors of the matrix of the system.

   • \(\det(\mathbf M-\lambda\mathbf I)=\lambda^2-6\lambda+8=(\lambda-2)(\lambda-4)=0\)
   • \(\lambda=2:\ 3x-y=0\Rightarrow\mathbf p_1=\begin{pmatrix}1\\3\end{pmatrix}\)
   • \(\lambda=4:\ x-y=0\Rightarrow\mathbf p_2=\begin{pmatrix}1\\1\end{pmatrix}\)

2. (b) Find \(x(t)\) and \(y(t)\).

   • General solution: \(\begin{pmatrix}x\\y\end{pmatrix}=Ae^{2t}\begin{pmatrix}1\\3\end{pmatrix}+Be^{4t}\begin{pmatrix}1\\1\end{pmatrix}\)
   • At t = 0: \(A+B=6,\ 3A+B=10\)
   • \(A=2,\ B=4\)
   • \(x=2e^{2t}+4e^{4t},\ y=6e^{2t}+4e^{4t}\)

3. (c) Find \(x\) and \(y\) when \(t=0.5\), to 3 significant figures.

   • \(x(0.5)=34.99,\ y(0.5)=45.87\)
   • \(x(0.5)=35.0,\ y(0.5)=45.9\)