3.5E: An Application to Systems of Differential Equations Exercises

Exercises for 1

solutions

2

Use Theorem [thm:010427] to find the general solution to each of the following systems. Then find a specific solution satisfying the given boundary condition.

1. \( \begin{array}[t]{rrrl} f_1^{\prime} & = 2f_1 + 4f_2, && f_1(0)=0 \\ f_2^{\prime} & = 3f_1 + 3f_2, && f_2(0)=1 \end{array}\)
2. \( \begin{array}[t]{rrrlr} f_1^{\prime} & = -f_1 + 5f_2,& & f_1(0)=&1 \\ f_2^{\prime} & = f_1 + 3f_2, && f_2(0)=&-1 \end{array}\)
3. \( \begin{array}[t]{rr} f_1^{\prime} = & 4f_2 + 4f_3 \\ f_2^{\prime} = & f_1 + f_2 -2f_3 \\ f_3^{\prime} = & -f_1 + f_2 +4f_3 \\ \end{array}\)

   \(f_1(0) = f_2(0) = f_3(0)=1\)

4. \( \begin{array}[t]{rlrr} f_1^{\prime} = & 2f_1 + &f_2 + &2f_3 \\ f_2^{\prime} = & 2f_1 + &2f_2 - &2f_3 \\ f_3^{\prime} = & 3f_1 + &f_2 + &f_3 \\ \end{array}\)

   \(f_1(0) = f_2(0) = f_3(0)=1\)

2. \(c_1 \left[ \begin{array}{r} 1 \\ 1 \end{array} \right] e^{4x} + c_2 \left[ \begin{array}{r} 5 \\ -1 \end{array} \right] e^{-2x}; c_1 = -\frac{2}{3}, c_2 = \frac{1}{3}\)

3. \(c_1 \left[ \begin{array}{r} -8 \\ 10 \\ 7 \end{array} \right] e^{-x} + c_2 \left[ \begin{array}{r} 1 \\ -2 \\ 1 \end{array} \right] e^{2x} + c_3 \left[ \begin{array}{r} 1 \\ 0 \\ 1 \end{array} \right] e^{4x}\);
   \(c_1 = 0, c_2 = -\frac{1}{2}, c_3=\frac{3}{2}\)

Show that the solution to \(f^{\prime}= af\) satisfying \(f(x_{0}) = k\) is \(f(x) = ke^{a(x-x_0)}\).

A radioactive element decays at a rate proportional to the amount present. Suppose an initial mass of 10 g decays to 8 g in 3 hours.

1. Find the mass \(t\) hours later.
2. Find the half-life of the element—the time taken to decay to half its mass.

2. The solution to (a) is \(m(t) = 10 \left( \frac{4}{5} \right)^{t/3}\). Hence we want \(t\) such that \(10 \left( \frac{4}{5} \right)^{t/3}=5\). We solve for \(t\) by taking natural logarithms:

   \[t = \frac{3 \ln (\frac{1}{2})}{\ln (\frac{4}{5})} = 9.32 \mbox{ hours}. \nonumber \]

The population \(N(t)\) of a region at time \(t\) increases at a rate proportional to the population. If the population doubles every 5 years and is 3 million initially, find \(N(t)\).

Let \(A\) be an invertible diagonalizable \(n \times n\) matrix and let \(\mathbf{b}\) be an \(n\)-column of constant functions. We can solve the system \(\mathbf{f}^{\prime} = A\mathbf{f} + \mathbf{b}\) as follows:

1. If \(\mathbf{g}\) satisfies \(\mathbf{g}^{\prime} = A\mathbf{g}\) (using Theorem [thm:010514]), show that \(\mathbf{f} = \mathbf{g} - A^{-1}\mathbf{b}\) is a solution to \(\mathbf{f}^{\prime} = A\mathbf{f} + \mathbf{b}\).
2. Show that every solution to \(\mathbf{f}^{\prime} = A\mathbf{f} + \mathbf{b}\) arises as in (a) for some solution \(\mathbf{g}\) to \(\mathbf{g}^{\prime} = A\mathbf{g}\).

1. If \(\mathbf{g}^{\prime} = A\mathbf{g}\), put \(\mathbf{f} = \mathbf{g} - A^{-1}\mathbf{b}\). Then \(\mathbf{f}^{\prime} = \mathbf{g}^{\prime}\) and \(A\mathbf{f} = A\mathbf{g} - \mathbf{b}\), so \(\mathbf{f}^{\prime} = \mathbf{g}^{\prime} = A\mathbf{g} = A\mathbf{f} + \mathbf{b}\), as required.

[ex:3_5_6] Denote the second derivative of \(f\) by \(f^{\prime \prime} = (f^{\prime})^{\prime}\). Consider the second order differential equation

\[\label{eq:secondorderdiff} f^{\prime \prime} - a_1 f^{\prime}-a_2 f = 0, \quad a_1 \mbox{ and } a_2 \mbox{ real numbers} \]

1. \(\left\lbrace \begin{array}{l} f_1^{\prime} = a_1f_1 + f_2 \\ f_2^{\prime} = a_2f_1 \end{array}, \right.\)
   that is \(\left. \left[ \arraycolsep=5pt \begin{array}{c} f_1^{\prime} \\ f_2^{\prime} \end{array}\right] = \left[ \arraycolsep=5pt \begin{array}{cc} a_1 & 1 \\ a_2 & 0 \end{array}\right] \left[ \arraycolsep=5pt \begin{array}{c} f_1 \\ f_2 \end{array}\right] \right.\)
2. Conversely, if \(\left[ \begin{array}{c} f_1 \\ f_2 \end{array}\right]\) is a solution to the system in (a), show that \(f_{1}\) is a solution to Equation [eq:secondorderdiff].

2. Assume that \(f_{1}^{\prime} = a_{1}f_{1} + f_{2}\) and \(f_{2}^{\prime} = a_{2}f_{1}\). Differentiating gives \(f_{1}^{\prime \prime} = a_{1}f_{1}^{\prime} + f_{2}{}^{\prime} = a_{1}f_{1}^{\prime} + a_{2}f_{1}\), proving that \(f_{1}\) satisfies Equation [eq:secondorderdiff].

[ex:3_5_7] Writing \(f^{\prime \prime\prime} = (f^\prime \prime)^{\prime}\), consider the third order differential equation

\[f^{\prime \prime\prime} - a_1 f^{\prime \prime} - a_2 f^{\prime} - a_3 f = 0 \nonumber \]

where \(a_{1}\), \(a_{2}\), and \(a_{3}\) are real numbers. Let
\(f_{1} = f\), \(f_{2} = f^{\prime}- a_{1}f\) and \(f_{3} = f^\prime \prime - a_{1}f{}^{\prime} - a_{2}f^\prime \prime\).

1. Show that \(\left[ \begin{array}{c} f_1 \\ f_2 \\ f_3 \end{array}\right]\) is a solution to the system

   \(\left\lbrace \begin{array}{l} f_1^{\prime} = a_1f_1 + f_2 \\ f_2^{\prime} = a_2f_1 + f_3 \\ f_3^{\prime} = a_3f_1 \end{array}, \right.\)
   that is \(\left. \left[ \arraycolsep=5pt \begin{array}{c} f_1^{\prime} \\ f_2^{\prime} \\ f_3^{\prime} \end{array}\right] = \left[ \arraycolsep=5pt \begin{array}{ccc} a_1 & 1 & 0 \\ a_2 & 0 & 1 \\ a_3 & 0 & 0 \end{array}\right] \left[ \arraycolsep=5pt \begin{array}{c} f_1 \\ f_2 \\ f_3 \end{array}\right]\right.\)

2. Show further that if \(\left[ \begin{array}{c} f_1 \\ f_2 \\ f_3 \end{array}\right]\) is any solution to this system, then \(f = f_{1}\) is a solution to Equation [eq:secondorderdiff].

Remark. A similar construction casts every linear differential equation of order \(n\) (with constant coefficients) as an \(n \times n\) linear system of first order equations. However, the matrix need not be diagonalizable, so other methods have been developed.