MATH383: A first course in differential equationsApril 7, 2020

1. Transform the following system of first-order differential equations

into a system of two third-order equations for dependent variables $u\left(t\right),v\left(t\right)$.

Solution. Take eq 5 to be $y_5^{\text{'}}=y_6$, $y_1=u$, $u^{\text{'}}=y_1^{\text{'}}=y_2$, $u^{\text{'}\text{'}}=y_2^{\text{'}}=y_3$, $u^{\text{'}\text{'}\text{'}}=y_3^{\text{'}}=f(t,u,v,v^{\text{'}},v^{\text{'}\text{'}})$. Take $y_4=v$, $v^{\text{'}}=y_4^{\text{'}}=y_5$, $y_5^{\text{'}}=y_6=v^{\text{'}\text{'}}$, $y_6^{\text{'}}=v^{\text{'}\text{'}\text{'}}=g(t,v,v^{\text{'}},v^{\text{'}\text{'}})$. Assuming correction of 5th equation

If $y_4^{\text{'}}=y_5$. $y_1=u$, $u^{\text{'}}=y_1^{\text{'}}=y_2$, $u^{\text{'}\text{'}}=y_2^{\text{'}}=y_3$, $u^{\text{'}\text{'}\text{'}}=y_3^{\text{'}}=f(t,u,)$. Take $y_4=v$, $v^{\text{'}}=y_4^{\text{'}}=y_5.$ Would need to introduce additional variable, w, cannot be put into form of two third order equations.

2. Consider the system of differential equations ${𝒚}^{\text{'}}=𝑨𝒚$,

Determine $a,b$ for the fundamental set of solutions to be:

1. $\{e^{2t},t{e}^{2t}\}$

2. $\{e^{-6t}\cos 4t,e^{-6t}\sin 4t\}$

3. $\{e^{6t},e^{14t}\}$

Solution. Determine roots of characteristic polynomial

Roots are

(a) The set $\{e^{2t},t{e}^{2t}\}$ indicates a double root ${\lambda }_1={\lambda }_2$, $a=\pm b$, $a=1$

(b) The set $\{e^{-6t}\cos 4t,e^{-6t}\sin 4t\}$ indicates complex conjugate pair ${\lambda }_{1,2}=\alpha \pm i\beta$, with $\alpha =-6$, $\beta =4$. Complex conjugate roots arise if $a^2-b^2<0$, $2a=-6\Rightarrow a=-3$, $\sqrt{a^2-b^2}=\sqrt{(-1)(b^2-a^2)}=i\sqrt{b^2-a^2}$, $\sqrt{b^2-a^2}=4\Rightarrow b^2-9=16\Rightarrow b=\pm 5$.

(c) The set $\{e^{6t},e^{14t}\}$ indicates distinct roots

3. Complete the set $\left\{e^{(p+q)t}\right\}$ to form a fundamental set of solutions for the system of differential equations ${𝒚}^{\text{'}}=𝑨𝒚$

Solution. The system ${𝒚}^{\text{'}}=𝑨𝒚$ has 3 fundamental solutions, of which one is $e^{(p+q)t}$. The characteristic polynomial is

and is of degree 3. We must determine two more eigenvalues. Since $p+q$ is a root, the polynomial can be written as

with $Q\left(\lambda \right)$ a polynomial of degree 2, $Q\left(\lambda \right)={\lambda }^2+a\lambda +b$,

Take $\lambda =0$

Assume $p+q\ne 0$ (otherwise $e^{(p+q)t}$ is not a non-trivial solution)

Take $\lambda =p$

Assume $q\ne 0$,

Assume $p\ne 0$,

Roots of $Q\left(\lambda \right)$ are

and the completion of the fundamental set is $e^{{\lambda }_{1}t},e^{{\lambda }_{2}t}.$

4. Find a fundamental set of solutions for the system of differential equations${𝒚}^{\text{'}}=𝑨𝒚$,

knowing that as $t\to \infty$, $e^{-t}𝒚\left(t\right)$ remains finite and non-zero.

Solution. The system has 3 fundamental solutions from which the general solution is

Multiply by $e^{-t}$

and ${\lambda }_1=1$, otherwise if ${\lambda }_1>1$, $e^{-t}𝒚\left(t\right)\to \infty$, or if ${\lambda }_1<1$ $e^{-t}𝒚\left(t\right)\to 0.$ Characteristic polynomial is

Indeed

Define $Q\left(\lambda \right)={\lambda }^2+a\lambda +b$, through

At $\lambda =0$

At $\lambda =2$

Roots of $Q\left(\lambda \right)$ are

and the completion of the fundamental set is $e^{{\lambda }_{1}t},e^{{\lambda }_{2}t}.$