MATH383: A first course in differential equationsMarch 3, 2020

Solve the following problems (3 course points each). Present a brief motivation of your method of solution. Explicitly state any conditions that must be met for solution procedure to be valid. Sketch out a solution for yourself on scratch paper, and then neatly transcribe so the solution you present is readily legible.

No credit is awarded for statement of the final answer to a problem without presentation of the solution procedure.

1. At $t=0$ an object is placed in a room with temperature of ${20}^{\circ }$ C. The temperature of the object drops by $5^{\circ }$ C in 4 minutes and by $7^{\circ }$ C in 8 minutes. What was the temperature of the object at $t=0$?

   Solution. The initial temperature is $T\left(0\right)=T_0$, and the rate of change of the object temperature $T^{\text{'}}$ is proportional to the difference with respect to ambient temperature $T_m={20}^{\circ }$ C,

   ${\displaystyle T^{\text{'}}=-k(T-T_m),T\left(0\right)=T_0.}$

   The solution to the above initial value problem (IVP) is $T\left(t\right)=T_m+(T_0-T_m)e^{-kt}$, which evaluated at $t=4$ and $t=8$ gives

   ${\displaystyle T\left(4\right)=20+(T_0-20)e^{-4k},T\left(8\right)=20+(T_0-20)e^{-8k}.}$

   From problem information $T_0-T\left(4\right)=5,T_0-T\left(8\right)=7$, or $T\left(4\right)=T_0-5$, $T\left(8\right)=T_0-7$, which leads to

   ${\displaystyle \{\begin{array}{l}{T}_0-5=20+(T_0-20)e^{-4k}\\ T_0-7=20+(T_0-20)e^{-8k}\end{array}.\Rightarrow \{\begin{array}{l}{e}^{-4k}=\frac{T_0-25}{T_0-20}\\ e^{-8k}=\frac{T_0-27}{T_0-20}\end{array}..}$

   Recall that ${\left(e^{-4k}\right)}^2=e^{-4k}\cdot e^{-4k}=e^{-8k}$ to obtain

   ${\displaystyle {\left(\frac{T_0-25}{T_0-20}\right)}^2=\frac{T_0-27}{T_0-20}\Rightarrow {(25-T_0)}^2=(27-T_0)(20-T_0)\Rightarrow 625-50T_0+T_0^2=540-47T_0+T_0^2\Rightarrow 3T_0=85,}$

   and the initial temperature is $T_0=85/3\cong 28.3^{\circ }$ C.

2. Determine whether $𝑩$ is a basis set for real-valued 2 by 2 matrices

   ${\displaystyle 𝑩=\left\{\begin{array}{lll}{𝑩}_1,& {𝑩}_2,& {𝑩}_3\end{array}\right\}=\left\{\begin{array}{lll}\left(\begin{array}{ll}1& 3\\ 2& 1\end{array}\right),& \left(\begin{array}{ll}-1& 2\\ 1& 0\end{array}\right),& \left(\begin{array}{ll}0& 1\\ 0& -4\end{array}\right)\end{array}\right\}.}$

   Solution. Note that a 2 by 2 matrix can also be represented as a 4 component vector (HW06 2.Ex2 solution)

   ${\displaystyle 𝑨=\left(\begin{array}{ll}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right)\to 𝒂=\left(\begin{array}{llll}{a}_{11}& a_{12}& a_{21}& a_{22}\end{array}\right),𝒂\in {ℝ}^4.}$

   Recall that the dimension of a vector space is the number of vectors in a basis, and the fact that the dimension of ${ℝ}^4$ is 4, while $𝑩$ only contains 3 elements implies that $𝑩$ is not a basis set (this observation is sufficient to solve the problem). To obtain a counterexample (optional), try to obtain some general $𝑨$ by a linear combination of elements of $𝑩$

   ${\displaystyle c_1{𝑩}_1+c_2{𝑩}_2+c_3{𝑩}_3=\left(\begin{array}{ll}{c}_1-c_2& 3c_1+2c_2+c_3\\ 2c_1+c_2& c_1-4c_3\end{array}\right)=\left(\begin{array}{ll}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right),}$

   leading to a system of 4 equations with 3 unknowns

   ${\displaystyle \{\begin{array}{lllll}{c}_1& -c_2& & =& a_{11}\\ 3c_1& +2c_2& +c_3& =& a_{12}\\ 2c_1& +c_2& & =& a_{21}\\ c_1& & -4c_3& =& a_{22}\end{array}..}$

   Try to find a simple counterexample: set $a_{11}=0,a_{21}=3$ to obtain $c_1=c_2=1$, leading to the equations

   ${\displaystyle \{\begin{array}{lll}{c}_3& =& a_{12}-5\\ -4c_3& =& a_{22}-1\end{array}.,}$

   that cannot be both true, for example for $a_{22}=1$, and $a_{12}=6$ the contradictory conditions $c_3=1$ (first equation), $c_3=0$ (second equation) are obtained, and $𝑩$ is not a basis since it does not span the set of 2 by 2 matrices, in particular the matrix

   ${\displaystyle 𝑨=\left(\begin{array}{ll}0& 6\\ 3& 1\end{array}\right),}$

   cannot be represented by a linear combination of elements of $𝑩$.

3. Let

   ${\displaystyle 𝒮=\{\left(\begin{array}{l}2s-t\\ s\\ t\\ -s\end{array}\right),s,t\in ℝ\}.}$

   1. Prove that $(𝒮,+,ℝ,\cdot )$ a subspace of $({ℝ}^4,+,ℝ,\cdot )$.

      Solution. Note that for $s=t=0$, the vector $𝟎=(0,0,0,0)$ is verified to be an element of $𝒮$. Verify closure, $c_1,c_2\in ℝ$, ${𝒖}_1,{𝒖}_2\in 𝒮$

      ${\displaystyle {𝒖}_1=\left(\begin{array}{l}2s_1-t_1\\ s_1\\ t_1\\ -s_1\end{array}\right),{𝒖}_2=\left(\begin{array}{l}2s_2-t_2\\ s_2\\ t_2\\ -s_2\end{array}\right),c_1{𝒖}_1+c_2{𝒖}_2=\left(\begin{array}{l}2(c_1{s}_1+c_2{s}_2)-(c_1{t}_1+c_2{t}_2)\\ c_1{s}_1+c_2{s}_2\\ c_1{t}_1+c_2{t}_2\\ -c_1{s}_1-c_2{s}_2\end{array}\right)=\left(\begin{array}{l}2s-t\\ s\\ t\\ -s\end{array}\right)\in 𝒮}$

      with $s=$$c_1{s}_1+c_2{s}_2$, $t=c_1{t}_1+c_2{t}_2$, so $(𝒮,+,ℝ,\cdot )$ is indeed a subspace of $({ℝ}^4,+,ℝ,\cdot )$.

   2. Find two vectors that span $𝒮$.

      Solution. Write

      ${\displaystyle \left(\begin{array}{l}2s-t\\ s\\ t\\ -s\end{array}\right)=s\left(\begin{array}{l}2\\ 1\\ 0\\ -1\end{array}\right)+t\left(\begin{array}{l}-1\\ 0\\ 1\\ 0\end{array}\right)}$

      and obtain that the vectors

      ${\displaystyle \{\left(\begin{array}{l}2\\ 1\\ 0\\ -1\end{array}\right),\left(\begin{array}{l}-1\\ 0\\ 1\\ 0\end{array}\right)\}}$

      span $𝒮$.

4. Find and subsequently sketch the solution to the initial value problem

   ${\displaystyle y^{\text{'}\text{'}}-14y^{\text{'}}+49y=0,y\left(1\right)=2,y^{\text{'}}\left(1\right)=11.}$

   Solution. Try $y=e^{rt}$ in the above homogeneous second-order differential equation to obtain

   ${\displaystyle r^2-14r+49={(r-7)}^2=0}$

   with double root $r=7$. Two independent solutions are $y_1=e^{7t}$, $y_2=t{e}^{7t}$, with derivatives $y_1^{\text{'}}=7e^{7t}$, $y_2^{\text{'}}=(1+7t)e^{7t}$. Impose initial conditions on linear combination $y=c_1{y}_1+c_2{y}_2$

   ${\displaystyle \{\begin{array}{l}{c}_1{y}_1\left(1\right)+c_2{y}_2\left(1\right)=y\left(1\right)\\ c_1{y}_1^{\text{'}}\left(1\right)+c_2{y}_2^{\text{'}}\left(1\right)=y^{\text{'}}\left(1\right)\end{array}.\Rightarrow \{\begin{array}{l}{c}_1{e}^7+c_2{e}^7=2\\ 7c_1{e}^7+8c_2{e}^7=11\end{array}.\Rightarrow \{\begin{array}{l}{c}_1+c_2=2e^{-7}\\ 7c_1+8c_2=11e^{-7}\end{array}.\Rightarrow \{\begin{array}{l}{c}_1=5e^{-7}\\ c_2=-3e^{-7}\end{array}.,}$

   and the solution is

   ${\displaystyle y\left(x\right)=(5-3t)e^{7(t-1)}.}$