MATH383: A first course in differential equationsMarch 26, 2020

Due date: April 2, 2020, 11:55PM.

Bibliography: Trench, 10.1-6.4

1. Exercise 5a-d, p. 516

   1. $x^{\text{'}\text{'}\text{'}}=f(t,x,y,y^{\text{'}})$, $y^{\text{'}\text{'}}=g(t,y,y^{\text{'}})$. Introduce notations

      ${\displaystyle z_1=x,z_2=x^{\text{'}},z_3=x^{\text{'}\text{'}},z_4=y,z_5=y^{\text{'}}}$

      and obtain system of 5 first-order equations

      ${\displaystyle \{\begin{array}{l}{z}_1^{\text{'}}=z_2\\ z_2^{\text{'}}=z_3\\ z_3^{\text{'}}=f(t,z_1,z_4,z_5)\\ z_4^{\text{'}}=z_5\\ z_5^{\text{'}}=g(t,z_4,z_5)\end{array}..}$

   2. $u^{\text{'}}=f(t,u,v,v^{\text{'}},w^{\text{'}}),v^{\text{'}\text{'}}=g(t,u,v,v^{\text{'}},w),w^{\text{'}\text{'}}=h(t,u,v,v^{\text{'}},w,w^{\text{'}})$. Introduce notations

      ${\displaystyle z_1=u,z_2=v,z_3=v^{\text{'}},z_4=w,z_5=w^{\text{'}}}$

      and obtain system of 5 first-order equations

      ${\displaystyle \{\begin{array}{l}{z}_1^{\text{'}}=f(t,z_1,z_2,z_5)\\ z_2^{\text{'}}=z_3\\ z_3^{\text{'}}=g(t,z_1,z_2,z_3,z_4)\\ z_4^{\text{'}}=z_5\\ z_5^{\text{'}}=h(t,z_1,z_2,z_3,z_4,z_5)\end{array}..}$

   3. $y^{\text{'}\text{'}\text{'}}=f(t,y,y^{\text{'}},y^{\text{'}\text{'}})$. Introduce notations

      ${\displaystyle z_1=y,z_2=y^{\text{'}},z_3=y^{\text{'}}{}^{\text{'}}}$

      and obtain system of 3 first-order equations

      ${\displaystyle \{\begin{array}{l}{z}_1^{\text{'}}=z_2\\ z_2^{\text{'}}=z_3\\ z_3^{\text{'}}=f(t,z_1,z_2)\end{array}..}$

   4. $y^{\left(4\right)}=f(t,y)$. Introduce notations

      ${\displaystyle z_1=y,z_2=y^{\text{'}},z_3=y^{\text{'}\text{'}},z_4=y^{\text{'}\text{'}\text{'}}}$

      and obtain system of 4 first-order equations

      ${\displaystyle \{\begin{array}{l}{z}_1^{\text{'}}=z_2\\ z_2^{\text{'}}=z_3\\ z_3^{\text{'}}=z_4\\ z_4^{\text{'}}=f(t,z_1)\end{array}..}$

2. Exercises 8a,b,e,f, p. 521

   a) From

   ${\displaystyle 𝒀=\left(\begin{array}{ll}{e}^{6t}& e^{-2t}\\ e^{6t}& -e^{-2t}\end{array}\right),𝑨=\left(\begin{array}{ll}2& 4\\ 4& 2\end{array}\right)}$

   compute

   ${\displaystyle {𝒀}^{\text{'}}=\left(\begin{array}{ll}6e^{6t}& -2e^{-2t}\\ 6e^{6t}& 2e^{-2t}\end{array}\right)=\left(\begin{array}{ll}2& 4\\ 4& 2\end{array}\right)\left(\begin{array}{ll}{e}^{6t}& e^{-2t}\\ e^{6t}& -e^{-2t}\end{array}\right)=\left(\begin{array}{ll}(2+4)e^{6t}& (2-4)e^{-2t}\\ (4+2)e^{6t}& (4-2)e^{-2t}\end{array}\right)?.}$

   b) From

   ${\displaystyle 𝒀=\left(\begin{array}{ll}{e}^{-4t}& -2e^{3t}\\ e^{-4t}& 5e^{3t}\end{array}\right),𝑨=\left(\begin{array}{ll}-2& -2\\ -5& 1\end{array}\right)}$

   compute

   ${\displaystyle {𝒀}^{\text{'}}=\left(\begin{array}{ll}-4e^{-4t}& -6e^{3t}\\ -4e^{-4t}& 15e^{3t}\end{array}\right)=\left(\begin{array}{ll}-2& -2\\ -5& 1\end{array}\right)\left(\begin{array}{ll}{e}^{-4t}& -2e^{3t}\\ e^{-4t}& 5e^{3t}\end{array}\right)=\left(\begin{array}{ll}\begin{array}{ll}(-2-2)e^{-4t}& (4-10)e^{3t}\\ (-5+1)e^{-4t}& (10+5)e^{3t}\end{array}& \end{array}\right)?.}$

   e) From

   ${\displaystyle 𝒀=\left(\begin{array}{lll}{e}^t& e^{-t}& e^{-2t}\\ e^t& 0& -2e^{-2t}\\ 0& 0& e^{-2t}\end{array}\right),𝑨=\left(\begin{array}{lll}-1& 2& 3\\ 0& 1& 6\\ 0& 0& -2\end{array}\right)}$

   compute ${𝒀}^{\text{'}}$

   ${\displaystyle \left(\begin{array}{lll}{e}^t& -e^{-t}& -2e^{-2t}\\ e^t& 0& 4e^{-2t}\\ 0& 0& -2e^{-2t}\end{array}\right)=\left(\begin{array}{lll}-1& 2& 3\\ 0& 1& 6\\ 0& 0& -2\end{array}\right)\left(\begin{array}{lll}{e}^t& e^{-t}& e^{-2t}\\ e^t& 0& -2e^{-2t}\\ 0& 0& e^{-2t}\end{array}\right)=\left(\begin{array}{lll}(-1+2)e^t& (-1)e^{-t}& (-1-4+3)e^{-2t}\\ \left(1\right)e^t& 0& (-2+6)e^{-2t}\\ 0& 0& (-2)e^{-2t}\end{array}\right)?.}$

   f) (same procedure as above)

3. Exercises 7,8, p. 528-9

   Ex 7) (a) is verified in (2a) above.

   (b) In the general solution

   ${\displaystyle 𝒚\left(t\right)=c_1\left(\begin{array}{l}{e}^{6t}\\ e^{6t}\end{array}\right)+c_2\left(\begin{array}{l}{e}^{-2t}\\ -e^{-2t}\end{array}\right)}$

   impose initial condition

   ${\displaystyle 𝒚\left(0\right)=c_1\left(\begin{array}{l}1\\ 1\end{array}\right)+c_2\left(\begin{array}{l}1\\ -1\end{array}\right)=\left(\begin{array}{l}-3\\ 9\end{array}\right)}$

   a linear system for $(c_1,c_2)$, with solution $c_1=3$, $c_2=-6$, hence

   c) Replace $t=0$ in

   ${\displaystyle 𝒀\left(t\right)=\left(\begin{array}{ll}{e}^{6t}& e^{-2t}\\ e^{6t}& -e^{-2t}\end{array}\right)}$

   to obtain

   ${\displaystyle 𝒀\left(0\right)=\left(\begin{array}{ll}1& 1\\ 1& -1\end{array}\right)}$

   with inverse

   ${\displaystyle {\left[𝒀\left(0\right)\right]}^{-1}=\frac{1}{2}\left(\begin{array}{ll}1& 1\\ 1& -1\end{array}\right)}$

   and compute

   ${\displaystyle 𝒀\left(t\right){\left[𝒀\left(0\right)\right]}^{-1}𝒌=\left(\begin{array}{ll}{e}^{6t}& e^{-2t}\\ e^{6t}& -e^{-2t}\end{array}\right)\frac{1}{2}\left(\begin{array}{ll}1& 1\\ 1& -1\end{array}\right)\left(\begin{array}{l}-3\\ 9\end{array}\right)=\left(\begin{array}{ll}{e}^{6t}& e^{-2t}\\ e^{6t}& -e^{-2t}\end{array}\right)\left(\begin{array}{l}3\\ -6\end{array}\right)?}$

4. Exercises 16,17,20,21, p. 541

16.