Solving coupled linear systems exactly Problem set B One question at a time

Write each system in matrix form.

1. (a) \(\dfrac{dx}{dt}=3x-5y\) and \(\dfrac{dy}{dt}=6x+2y\) 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.
   • \(\mathbf M=\begin{pmatrix}3&-5\\6&2\end{pmatrix}\)

2. (b) \(\dfrac{dx}{dt}=y-4x\) and \(\dfrac{dy}{dt}=2x\) 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.
   • \(\mathbf M=\begin{pmatrix}-4&1\\2&0\end{pmatrix}\)

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 \(\mathbf M=\begin{pmatrix}4&1\\2&3\end{pmatrix}\).

   • \(\det(\mathbf M-\lambda\mathbf I)=\lambda^2-7\lambda+10=(\lambda-2)(\lambda-5)=0\)
   • \(\lambda=2:\ 2x+y=0\Rightarrow\mathbf p_1=\begin{pmatrix}1\\-2\end{pmatrix}\)
   • \(\lambda=5:\ -x+y=0\Rightarrow\mathbf p_2=\begin{pmatrix}1\\1\end{pmatrix}\)

2. (b) Find the eigenvalues and eigenvectors of \(\mathbf M=\begin{pmatrix}1&1\\4&-2\end{pmatrix}\).

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

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

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

Consider the system \(\dfrac{dx}{dt}=y\) and \(\dfrac{dy}{dt}=2x-y\). The eigenvalues of the matrix are \(-2\) and \(1\). Given that \(x(0)=6\) and \(y(0)=3\), find \(x(t)\) and \(y(t)\).

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

Consider the system \(\dfrac{dx}{dt}=x+2y\) and \(\dfrac{dy}{dt}=3x\). The eigenvalues of the matrix are \(-2\) and \(3\). Given that \(x(0)=4\) and \(x'(0)=2\), find \(x(t)\).

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

A drug is injected into the blood. After \(t\) hours, the blood holds \(x\) mg of the drug and body tissue holds \(y\) mg. The amounts are modelled by \(\dfrac{dx}{dt}=-3x+y\) and \(\dfrac{dy}{dt}=4x-3y\). At the start the blood holds 8 mg and the tissue 4 mg.

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

   • \(\mathbf M=\begin{pmatrix}-3&1\\4&-3\end{pmatrix}\)

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

   • \(\det(\mathbf M-\lambda\mathbf I)=\lambda^2+6\lambda+5=(\lambda+5)(\lambda+1)=0\)
   • \(\lambda=-5:\ 2x+y=0\Rightarrow\mathbf p_1=\begin{pmatrix}1\\-2\end{pmatrix}\)
   • \(\lambda=-1:\ -2x+y=0\Rightarrow\mathbf p_2=\begin{pmatrix}1\\2\end{pmatrix}\)

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

   • General solution: \(\begin{pmatrix}x\\y\end{pmatrix}=Ae^{-5t}\begin{pmatrix}1\\-2\end{pmatrix}+Be^{-t}\begin{pmatrix}1\\2\end{pmatrix}\)
   • At t = 0: \(A+B=8,\ -2A+2B=4\)
   • \(A=3,\ B=5\)
   • \(x=3e^{-5t}+5e^{-t},\ y=-6e^{-5t}+10e^{-t}\)

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

   • \(x(1)=3e^{-5(1)}+5e^{-(1)}=1.86\)
   • Both exponents are negative, so both terms tend to 0: the drug is eventually cleared

Consider the system \(\dfrac{dx}{dt}=4x+y\) and \(\dfrac{dy}{dt}=3x+2y\) with \(x(0)=6\) and \(y(0)=2\).

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

   • \(\det(\mathbf M-\lambda\mathbf I)=\lambda^2-6\lambda+5=(\lambda-1)(\lambda-5)=0\)
   • \(\lambda=1:\ 3x+y=0\Rightarrow\mathbf p_1=\begin{pmatrix}1\\-3\end{pmatrix}\)
   • \(\lambda=5:\ -x+y=0\Rightarrow\mathbf p_2=\begin{pmatrix}1\\1\end{pmatrix}\)

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

   • General solution: \(\begin{pmatrix}x\\y\end{pmatrix}=Ae^{t}\begin{pmatrix}1\\-3\end{pmatrix}+Be^{5t}\begin{pmatrix}1\\1\end{pmatrix}\)
   • At t = 0: \(A+B=6,\ -3A+B=2\)
   • \(A=1,\ B=5\)
   • \(x=e^{t}+5e^{5t},\ y=-3e^{t}+5e^{5t}\)

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

   • \(x(0.5)=62.56,\ y(0.5)=55.97\)
   • \(x(0.5)=62.6,\ y(0.5)=56.0\)