3.3 Quizz 3

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3.3.1 Problem 1

Determine whether or not the given matrix $A$ is diagonalizable. If it is find a diagonalizing matrix $P$ and a diagonal matrix $D$ such that $P^{-1}AP=D$$$A=\left[\begin{array}{cc}3& 0\\ 8& -1\end{array}\right]$$ Solution

The first step is to find the eigenvalues. This is found by solving$$\begin{array}{rl}det(A-\lambda I)& =0\\ \left|\begin{array}{cc}3-\lambda & 0\\ 8& -1-\lambda \end{array}\right|& =0\\ (3-\lambda )(-1-\lambda )& =0\end{array}$$

Hence the eigenvalues are ${\lambda }_1=3,{\lambda }_2=-1$. Since the eigenvalues are unique, then the matrix is diagonalizable. We need to determine the corresponding eigenvector in order to find $P$

${\lambda }_1=3$

Solving$$\begin{array}{rl}\left[\begin{array}{cc}3-{\lambda }_1& 0\\ 8& -1-{\lambda }_1\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]& =\left[\begin{array}{c}0\\ 0\end{array}\right]\\ \left[\begin{array}{cc}3-3& 0\\ 8& -1-3\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]& =\left[\begin{array}{c}0\\ 0\end{array}\right]\\ \left[\begin{array}{cc}0& 0\\ 8& -4\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]& =\left[\begin{array}{c}0\\ 0\end{array}\right]\end{array}$$

Hence $v_1=t$ is free variable and $v_2$ is base variable. Second row gives  $8t-4v_2=0$ or $v_2=2t$. Therefore the eigenvector is $$\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]=\left[\begin{array}{c}t\\ 2t\end{array}\right]=t\left[\begin{array}{c}1\\ 2\end{array}\right]$$ Let $t=1$, then $${\overrightarrow{v}}_1=\left[\begin{array}{c}1\\ 2\end{array}\right]$$ ${\lambda }_2=-1$

Solving$$\begin{array}{rl}\left[\begin{array}{cc}3-{\lambda }_2& 0\\ 8& -1-{\lambda }_2\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]& =\left[\begin{array}{c}0\\ 0\end{array}\right]\\ \left[\begin{array}{cc}3+1& 0\\ 8& -1+1\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]& =\left[\begin{array}{c}0\\ 0\end{array}\right]\\ \left[\begin{array}{cc}4& 0\\ 8& 0\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]& =\left[\begin{array}{c}0\\ 0\end{array}\right]\end{array}$$

$R_2=R_2-2R_1$ gives$$\left[\begin{array}{cc}4& 0\\ 0& 0\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$$ Hence $v_2=t$ is free variable and $v_1$ is base variable. First row gives  $4v_1=0$ or $v_1=0$. Therefore the eigenvector is $$\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]=\left[\begin{array}{c}0\\ t\end{array}\right]=t\left[\begin{array}{c}0\\ 1\end{array}\right]$$ Let $t=1$, then $${\overrightarrow{v}}_2=\left[\begin{array}{c}0\\ 1\end{array}\right]$$ Now that both eigenvectors are found, then $$\begin{array}{rl}P& =\left[\begin{array}{cc}{\overrightarrow{v}}_1& {\overrightarrow{v}}_2\end{array}\right]\\ & =\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]\end{array}$$

And $D$ is the diagonal matrix of the eigenvalues arranged in same order as the corresponding eigenvectors. (Will verify below). Hence$$D=\left[\begin{array}{cc}3& 0\\ 0& -1\end{array}\right]$$ Therefore $$\begin{array}{rl}{P}^{-1}AP& =D\\ \text{(1)}& {\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]}^{-1}\left[\begin{array}{cc}3& 0\\ 8& -1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]& =\left[\begin{array}{cc}3& 0\\ 0& -1\end{array}\right]\end{array}$$

To verify the above, the LHS of (1) is evaluated directly, to confirm that $D$ is indeed the result and it is diagonal of the eigenvalues. The first step is to find ${\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]}^{-1}$. Since this is $2\times 2$ then$${\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]}^{-1}=\frac{1}{det(P)}\left[\begin{array}{cc}1& 0\\ -2& 1\end{array}\right]$$ But $det(P)=1$. The above becomes$${\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]}^{-1}=\left[\begin{array}{cc}1& 0\\ -2& 1\end{array}\right]$$ Therefore the LHS of (1) becomes$$\begin{array}{rl}{\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]}^{-1}\left[\begin{array}{cc}3& 0\\ 8& -1\end{array}\right]& =\left[\begin{array}{cc}1& 0\\ -2& 1\end{array}\right]\left[\begin{array}{cc}3& 0\\ 8& -1\end{array}\right]\\ & =\left[\begin{array}{cc}1\times 3& 0\\ -2\times 3+1\times 8& -1\end{array}\right]\\ & =\left[\begin{array}{cc}3& 0\\ -6+8& -1\end{array}\right]\\ & =\left[\begin{array}{cc}3& 0\\ 2& -1\end{array}\right]\end{array}$$

And now LHS of (1) becomes$$\begin{array}{rl}\left[\begin{array}{cc}3& 0\\ 2& -1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]& =\left[\begin{array}{cc}3& 0\\ 2\times 1-1\times 2& -1\end{array}\right]\\ & =\left[\begin{array}{cc}3& 0\\ 0& -1\end{array}\right]\end{array}$$

Hence$$D=\left[\begin{array}{cc}3& 0\\ 0& -1\end{array}\right]$$ Which confirms $D$ is the matrix whose diagonal elements are the eigenvalues of $A$.

3.3.2 Problem 2

Find the general solution of the homogeneous differential equation $y^{\prime \prime \prime }+y^{\prime }-10y=0$

Solution

This is a linear 3rd order constant coefficient ODE. Hence the method of characteristic equation will be used. Let the solution be $y=A{e}^{\lambda x}$. Substituting this into the ODE gives$$\begin{array}{rl}A{\lambda }^3{e}^{\lambda x}+A\lambda e^{\lambda x}-10A{e}^{\lambda x}& =0\\ A{e}^{\lambda x}({\lambda }^3+\lambda -10)& =0\end{array}$$

Which simplifies (for non-trivial $y$) to the characteristic equation which is a polynomial in $\lambda$ $${\lambda }^3+\lambda -10=0$$ By inspection, we see that $\lambda =2$ is a root. Therefore a factor of the equation is $(\lambda -2)$. Now doing long division $\frac{{\lambda }^3+\lambda -10}{(\lambda -2)}$ gives ${\lambda }^2+2\lambda +5$.

Figure 3.3:Polynmial long division to find remainder

Hence the above polynomial can be written as$$\begin{array}{}\text{(1)}& (\lambda -2)({\lambda }^2+2\lambda +5)=0\end{array}$$ Now the roots for $({\lambda }^2+2\lambda +5)=0$ are found using the quadratic formula. $$\begin{array}{rl}\lambda & =-\frac{b}{2a}\pm \frac{1}{2a}\sqrt{b^2-4ac}\\ & =-\frac{2}{2}\pm \frac{1}{2}\sqrt{4-4\times 5}\\ & =-1\pm \frac{1}{2}\sqrt{-16}\\ & =-1\pm \frac{1}{2}\left(4i\right)\\ & =-1\pm 2i\end{array}$$

Hence the roots of the characteristic equation are$$\begin{array}{rl}{\lambda }_1& =2\\ {\lambda }_2& =-1+2i\\ {\lambda }_3& =-1-2i\end{array}$$

Therefore the basis solution are $\{e^{2x},e^{(-1+2i)x},e^{(-1-2i)x}\}$ and the general solution is a linear combination of these basis solutions which gives$$y=A{e}^{2x}+B{e}^{(-1+2i)x}+C{e}^{(-1-2i)x}$$ Which can be simplified to $$\begin{array}{}\text{(2)}& y=A{e}^{2x}+e^{-x}(B{e}^{2ix}+C{e}^{-2ix})\end{array}$$ By using Euler formula, the above can be simplified further as follows$$\begin{array}{rl}B{e}^{2ix}+C{e}^{-2ix}& =B(\cos \left(2x\right)+i\sin \left(2x\right))+C(\cos \left(2x\right)-i\sin \left(2x\right))\\ & =(B+C)\cos \left(2x\right)+\sin \left(2x\right)\left(i(B-C)\right)\end{array}$$

Let $(B+C)=B_0$ a new constant and let $i(B-C)=C_0$ a new constant, the above becomes$$B{e}^{2ix}+C{e}^{-2ix}=B_0\cos \left(2x\right)+C_0\sin \left(2x\right)$$ Substituting the above back in (2) gives the general solution as$$y=A{e}^{2x}+e^{-x}(B_0\cos \left(2x\right)+C_0\sin \left(2x\right))$$ The constants $A,B_0,C_0$ can be found from initial conditions if given.

3.3.3 Problem 3

Using the method of undermined coefficients, compute the general solution of the given equation $y^{\prime \prime }+3y^{\prime }+2y=2\sin (x)$

Solution

The solution is  $$y=y_h+y_p$$ Where $y_h$ is the solution to the homogenous ode $y^{\prime \prime }+3y^{\prime }+2y=0$ and $y_p$ is a particular solution to the given ODE. The ode $y^{\prime \prime }+3y^{\prime }+2y=0$ is linear second order constant coefficient ODE. Hence the method of characteristic equation will be used. Let the solution be $y_h=A{e}^{\lambda x}$. Substituting this into $y^{\prime \prime }+3y^{\prime }+2y=0$$$\begin{array}{rl}A{\lambda }^2{e}^{\lambda x}+A\lambda 3e^{\lambda x}+2A{e}^{\lambda x}& =0\\ A{e}^{\lambda x}({\lambda }^2+3\lambda +2)& =0\end{array}$$

And for non trivial solution the above simplifies to$$\begin{array}{rl}{\lambda }^2+3\lambda +2& =0\\ (\lambda +1)(\lambda +2)& =0\end{array}$$

Hence the roots are $\lambda =-1,\lambda =-2$. Therefore the basis solutions for $y_h$ are $\{e^{-x},e^{-2x}\}$ and $y_h$ is linear combination of these basis. Therefore $$\begin{array}{}\text{(1)}& y_h=c_1{e}^{-x}+c_2{e}^{-2x}\end{array}$$ Now $y_p$ is found. Since the RHS is $\sin (x)$ then the trial solution is $$\begin{array}{}\text{(2)}& y_p=A\cos (x)+B\sin (x)\end{array}$$ This shows that the basis for $y_p$ are $\{\sin x,\cos x\}$. There are no duplication between these basis and the basis for $y_h$, so no need to multiply by an extra $x$. Using (2) gives$$\begin{array}{}\text{(3)}& y_p^{\prime }& =-A\sin (x)+B\cos (x)\text{(4)}& y_p^{\prime \prime }& =-A\cos (x)-B\sin (x)\end{array}$$

Substituting (2,3,4) back into the given ODE gives$$\begin{array}{rl}{y}_p^{\prime \prime }+3y_p^{\prime }+2y_p& =2\sin (x)\\ (-A\cos (x)-B\sin (x))+3(-A\sin (x)+B\cos (x))+2(A\cos (x)+B\sin (x))& =2\sin \left(x\right)\\ \cos (x)(-A+3B+2A)+\sin (x)(-B-3A+2B)& =2\sin (x)\\ \cos (x)(3B+A)+\sin (x)(-3A+B)& =2\sin (x)\end{array}$$

Comparing coefficients on both sides gives two equations to solve for $A,B$ $$\begin{array}{rl}3B+A& =0\\ -3A+B& =2\end{array}$$

Multiplying the second equation by $-3$ gives $$\begin{array}{rl}3B+A& =0\\ 9A-3B& =-6\end{array}$$

Adding the above two equations gives $10A=-6$. Hence $A=-\frac{3}{5}$ and therefore $3B=\frac{3}{5}$ or $B=\frac{1}{5}$. Substituting these values of $A,B$into (2) gives$$y_p=-\frac{3}{5}\cos (x)+\frac{1}{5}\sin (x)$$ Hence the solution becomes$$\begin{array}{rl}y& =y_h+y_p\\ & =(c_1{e}^{-x}+c_2{e}^{-2x})+(-\frac{3}{5}\cos \left(x\right)+\frac{1}{5}\sin (x))\\ & =c_1{e}^{-x}+c_2{e}^{-2x}-\frac{3}{5}\cos (x)+\frac{1}{5}\sin (x)\end{array}$$

3.3.4 Problem 4

Show that the given vector functions are linearly independent $${\overrightarrow{x}}_1(t)=\left[\begin{array}{c}{e}^t\\ 2e^t\end{array}\right]{\overrightarrow{x}}_2(t)=\left[\begin{array}{c}\sin t\\ \cos t\end{array}\right]$$ Solution

These functions are defined for all $t$. Hence domain is $t\in (-\mathrm{\infty },\mathrm{\infty })$. The Wronskian of these vectors is$$\begin{array}{rl}W(t)& =\left|\begin{array}{cc}{e}^t& \sin t\\ 2e^t& \cos t\end{array}\right|\\ & =e^t\cos t-2e^t\sin t\\ & =e^t(\cos t-2\sin t)\end{array}$$

We just need find one value $t_0$ where $W\left(t_0\right)\ne 0$ to show linearly independence. At $t=0$ the above becomes$$W(t=0)=1$$ Therefore the given vector functions are linearly independent. An alternative method is to write $$\begin{array}{rl}{c}_1{\overrightarrow{x}}_1(t)+c_2{\overrightarrow{x}}_2(t)& =\overrightarrow{0}\\ c_1\left[\begin{array}{c}{e}^t\\ 2e^t\end{array}\right]+c_2\left[\begin{array}{c}\sin t\\ \cos t\end{array}\right]& =\left[\begin{array}{c}0\\ 0\end{array}\right]\end{array}$$

If the above is true only for $c_1=0,c_2=0$ then ${\overrightarrow{x}}_1\left(t\right),{\overrightarrow{x}}_2(t)$ are linearly independent. The above can be written as$$\left[\begin{array}{cc}{e}^t& \sin t\\ 2e^t& \cos t\end{array}\right]\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$$ $R_2=R_2-2R_1$$$\left[\begin{array}{cc}{e}^t& \sin t\\ 0& \cos t-2\sin t\end{array}\right]\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$$ Row two gives $$c_2(\cos t-2\sin t)=0$$ For this to be true for any $t$ in the interval $t\in (-\mathrm{\infty },\mathrm{\infty })$, then only solution is $c_2=0$. First row now gives $$c_1{e}^t=0$$ But $e^t$ is never zero which means $c_1=0$.

Since the only solution to $c_1{\overrightarrow{x}}_1+c_2{\overrightarrow{x}}_2=\overrightarrow{0}$ is $c_1=c_2=0$, then this shows that ${\overrightarrow{x}}_1,{\overrightarrow{x}}_2$ are linearly independent.