Problem 2.
If the temperature of the bread is $120^oC$, of the air is $30^oC$ and $K_1=0.0366$, $K_2=-0.0002$ determine the temperature of the bread and air $60$ minutes later.
(By linear algebraic method.)
We use the Newton's law of cooling,
\begin{align*}
M'(t) &=K_2(M(t)-T(t)),\\
T'(t) &=K_1(M(t)-T(t)).
\end{align*}
where $T(t)$ is the temperature of the bread, $M(t)$ is the temperature of the air and $K_1\gt 0$, $K_2\lt 0$ are constants. Denote $T_0:=T(0)$, $M_0:=M(0)$, $M_0\lt T_0$.
Introduce the vector \[ \mathbf{x}(t):= \left[ \begin{array}{cc} M(t)\\ T(t) \end{array} \right], \] and the matrix \[ \mathbf{A}:= \left[ \begin{array}{rr} K_2 & -K_2\\ K_1 & -K_1 \end{array} \right]. \] Then we can write \[ \mathbf{x}'(t)=\mathbf{A}\mathbf{x}(t). \] The general solution of this equation is \[ \mathbf{x}(t)=e^{\mathbf{A}t}\mathbf{c}, \] where $\mathbf{c}\in\mathbf{R}^2$. The eigenvalues and corresponding eigenvectors of $\mathbf{A}$ are \begin{align*} \lambda_1 &=-K_1+K_2,\quad \mathbf{s}_1 = \left[ \begin{array}{c} \frac{K_2}{K_1}\\ 1 \end{array} \right], \\ \lambda_2 &=0, \qquad \mathbf{s}_2 = \left[ \begin{array}{c} 1\\ 1 \end{array} \right]. \end{align*} Since $\lambda_1\neq\lambda_2$ are real numbers, \begin{align*} \mathbf{x}(t) &=[\mathbf{s}_1 \quad \mathbf{s}_2] \left[ \begin{array}{rr} e^{-(K_1-K_2)t} & 0\\ 0 & e^{0t} \end{array} \right] \left[ \begin{array}{c} c_1\\ c_2 \end{array} \right] \\ \\ &= \left[ \begin{array}{c} \dfrac{K_2\, e^{-(K_1-K_2)t}c_1}{K_1}+c_2\\[1em] e^{-(K_1-K_2)t}c_1+c_2 \end{array} \right]. \end{align*} Using the initial conditions we have \[ \left[ \begin{array}{c} M_0\\ T_0 \end{array} \right] =\mathbf{x}(0)=\left[ \begin{array}{c} \dfrac{K_2\, c_1}{K_1}+c_2\\[0.8em] c_1+c_2 \end{array} \right]. \] Solving this system we obtain \[ c_1=-\frac{K_1(M_0-T_0)}{K_1-K_2},\quad c_2=\frac{K_1M_0-K_2T_0}{K_1-K_2}. \] Substituting back \[ \mathbf{x}(t)= \left[ \begin{array}{c} \dfrac{-K_2 e^{-(K_1-K_2)t} (M_0-T_0)+K_1M_0-K_2T_0}{K_1-K_2}\\[1em] \dfrac{-K_1 e^{-(K_1-K_2)t} (M_0-T_0)+K_1M_0-K_2T_0}{K_1-K_2} \end{array} \right]. \] We note that \[ \lim_{t\to\infty}\mathbf{x}(t)= \left[ \begin{array}{c} \dfrac{K_1M_0-K_2T_0}{K_1-K_2}\\[0.8em] \dfrac{K_1M_0-K_2T_0}{K_1-K_2} \end{array} \right]. \] For the given data we obtain \[ \mathbf{x}(t)=\left[ \begin{array}{c} 30.489-0.489e^{-0.0368t}\\ 89.511e^{-0.0368t}+30.489 \end{array} \right] \] and \[ \mathbf{x}(60)=\left[ \begin{array}{c} 30.435\\ 40.328 \end{array} \right].\quad\blacksquare \]
Introduce the vector \[ \mathbf{x}(t):= \left[ \begin{array}{cc} M(t)\\ T(t) \end{array} \right], \] and the matrix \[ \mathbf{A}:= \left[ \begin{array}{rr} K_2 & -K_2\\ K_1 & -K_1 \end{array} \right]. \] Then we can write \[ \mathbf{x}'(t)=\mathbf{A}\mathbf{x}(t). \] The general solution of this equation is \[ \mathbf{x}(t)=e^{\mathbf{A}t}\mathbf{c}, \] where $\mathbf{c}\in\mathbf{R}^2$. The eigenvalues and corresponding eigenvectors of $\mathbf{A}$ are \begin{align*} \lambda_1 &=-K_1+K_2,\quad \mathbf{s}_1 = \left[ \begin{array}{c} \frac{K_2}{K_1}\\ 1 \end{array} \right], \\ \lambda_2 &=0, \qquad \mathbf{s}_2 = \left[ \begin{array}{c} 1\\ 1 \end{array} \right]. \end{align*} Since $\lambda_1\neq\lambda_2$ are real numbers, \begin{align*} \mathbf{x}(t) &=[\mathbf{s}_1 \quad \mathbf{s}_2] \left[ \begin{array}{rr} e^{-(K_1-K_2)t} & 0\\ 0 & e^{0t} \end{array} \right] \left[ \begin{array}{c} c_1\\ c_2 \end{array} \right] \\ \\ &= \left[ \begin{array}{c} \dfrac{K_2\, e^{-(K_1-K_2)t}c_1}{K_1}+c_2\\[1em] e^{-(K_1-K_2)t}c_1+c_2 \end{array} \right]. \end{align*} Using the initial conditions we have \[ \left[ \begin{array}{c} M_0\\ T_0 \end{array} \right] =\mathbf{x}(0)=\left[ \begin{array}{c} \dfrac{K_2\, c_1}{K_1}+c_2\\[0.8em] c_1+c_2 \end{array} \right]. \] Solving this system we obtain \[ c_1=-\frac{K_1(M_0-T_0)}{K_1-K_2},\quad c_2=\frac{K_1M_0-K_2T_0}{K_1-K_2}. \] Substituting back \[ \mathbf{x}(t)= \left[ \begin{array}{c} \dfrac{-K_2 e^{-(K_1-K_2)t} (M_0-T_0)+K_1M_0-K_2T_0}{K_1-K_2}\\[1em] \dfrac{-K_1 e^{-(K_1-K_2)t} (M_0-T_0)+K_1M_0-K_2T_0}{K_1-K_2} \end{array} \right]. \] We note that \[ \lim_{t\to\infty}\mathbf{x}(t)= \left[ \begin{array}{c} \dfrac{K_1M_0-K_2T_0}{K_1-K_2}\\[0.8em] \dfrac{K_1M_0-K_2T_0}{K_1-K_2} \end{array} \right]. \] For the given data we obtain \[ \mathbf{x}(t)=\left[ \begin{array}{c} 30.489-0.489e^{-0.0368t}\\ 89.511e^{-0.0368t}+30.489 \end{array} \right] \] and \[ \mathbf{x}(60)=\left[ \begin{array}{c} 30.435\\ 40.328 \end{array} \right].\quad\blacksquare \]
(By Laplace transformation.)
We use the Newton's law of cooling,
\begin{align*}
M'(t) &=K_2(M(t)-T(t)),\\
T'(t) &=K_1(M(t)-T(t)).
\end{align*}
where $T(t)$ is the temperature of the bread, $M(t)$ is the temperature of the air and $K_1\gt 0$, $K_2\lt 0$ are constants. Denote $T_0:=T(0)$, $M_0:=M(0)$, $M_0\lt T_0$.
Introduce the vector \[ \mathbf{x}(t):= \left[ \begin{array}{cc} M(t)\\ T(t) \end{array} \right], \] and the matrix \[ \mathbf{A}:= \left[ \begin{array}{rr} K_2 & -K_2\\ K_1 & -K_1 \end{array} \right]. \] Then we can write \[ \mathbf{x}'(t)=\mathbf{A}\mathbf{x}(t). \] Let $\widehat{\mathbf{x}}(s)$ denote the Laplace transform of $\mathbf{x}(t)$. Then taking the Laplace transform of the system we get \[ s\widehat{\mathbf{x}}(s)-\mathbf{x}_0=A\widehat{\mathbf{x}}(s), \] or equivalently \[ (sE-A)\widehat{\mathbf{x}}(s)=\mathbf{x}_0, \] where $E$ is the unit matrix. Since \[ (sE-A)^{-1}= \left[ \begin{array}{rr} \dfrac{s+K_1}{s(s+K_1-K_2)} & -\dfrac{K_2}{s(s+K_1-K_2)}\\[0.8em] \dfrac{K_1}{s(s+K_1-K_2)} & \dfrac{s-K_2}{s(s+K_1-K_2)} \end{array} \right], \] we obtain \begin{align*} \widehat{\mathbf{x}}(s) &=(sE-A)^{-1}\mathbf{x}_0\\[1em] &= \left[ \begin{array}{c} \dfrac{(s+K_1)M_0-K_2T_0}{s(s+K_1-K_2)} \\[0.8em] \dfrac{K_1 M_0+(s-K_2)T_0}{s(s+K_1-K_2)} \end{array} \right] \\[1em] &= \left[ \begin{array}{c} \dfrac{K_1 M_0-K_2 T_0}{s(K_1-K_2)}-\dfrac{K_2(M_0-T_0)}{(s+K_1-K_2)(K_1-K_2)} \\[0.8em] \dfrac{K_1 M_0-K_2 T_0}{s(K_1-K_2)}-\dfrac{K_1(M_0-T_0)}{(s+K_1-K_2)(K_1-K_2)} \end{array} \right]. \end{align*}
Introduce the vector \[ \mathbf{x}(t):= \left[ \begin{array}{cc} M(t)\\ T(t) \end{array} \right], \] and the matrix \[ \mathbf{A}:= \left[ \begin{array}{rr} K_2 & -K_2\\ K_1 & -K_1 \end{array} \right]. \] Then we can write \[ \mathbf{x}'(t)=\mathbf{A}\mathbf{x}(t). \] Let $\widehat{\mathbf{x}}(s)$ denote the Laplace transform of $\mathbf{x}(t)$. Then taking the Laplace transform of the system we get \[ s\widehat{\mathbf{x}}(s)-\mathbf{x}_0=A\widehat{\mathbf{x}}(s), \] or equivalently \[ (sE-A)\widehat{\mathbf{x}}(s)=\mathbf{x}_0, \] where $E$ is the unit matrix. Since \[ (sE-A)^{-1}= \left[ \begin{array}{rr} \dfrac{s+K_1}{s(s+K_1-K_2)} & -\dfrac{K_2}{s(s+K_1-K_2)}\\[0.8em] \dfrac{K_1}{s(s+K_1-K_2)} & \dfrac{s-K_2}{s(s+K_1-K_2)} \end{array} \right], \] we obtain \begin{align*} \widehat{\mathbf{x}}(s) &=(sE-A)^{-1}\mathbf{x}_0\\[1em] &= \left[ \begin{array}{c} \dfrac{(s+K_1)M_0-K_2T_0}{s(s+K_1-K_2)} \\[0.8em] \dfrac{K_1 M_0+(s-K_2)T_0}{s(s+K_1-K_2)} \end{array} \right] \\[1em] &= \left[ \begin{array}{c} \dfrac{K_1 M_0-K_2 T_0}{s(K_1-K_2)}-\dfrac{K_2(M_0-T_0)}{(s+K_1-K_2)(K_1-K_2)} \\[0.8em] \dfrac{K_1 M_0-K_2 T_0}{s(K_1-K_2)}-\dfrac{K_1(M_0-T_0)}{(s+K_1-K_2)(K_1-K_2)} \end{array} \right]. \end{align*}
Applying the inverse Laplace transformation
we obtain
\[
\mathbf{x}(t)=
\left[
\begin{array}{c}
\dfrac{-K_2 e^{-(K_1-K_2)t} (M_0-T_0)+K_1M_0-K_2T_0}{K_1-K_2}\\[1em]
\dfrac{-K_1 e^{-(K_1-K_2)t} (M_0-T_0)+K_1M_0-K_2T_0}{K_1-K_2}
\end{array}
\right].
\]
We note that
\[
\lim_{t\to\infty}\mathbf{x}(t)=
\left[
\begin{array}{c}
\dfrac{K_1M_0-K_2T_0}{K_1-K_2}\\[0.8em]
\dfrac{K_1M_0-K_2T_0}{K_1-K_2}
\end{array}
\right].
\]
For the given data we obtain
\[
\mathbf{x}(t)=\left[
\begin{array}{c}
30.489-0.489e^{-0.0368t}\\
89.511e^{-0.0368t}+30.489
\end{array}
\right]
\]
and
\[
\mathbf{x}(60)=\left[
\begin{array}{c}
30.435\\
40.328
\end{array}
\right].\quad\blacksquare
\]
