Characteristic Equations with Complex Roots

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 4.3 – Auxiliary Equations with Complex Roots

Expected Educational Results

• Objective 11–1: I can identify complex solutions to the characteristic equation for $n^{\text{th}}$$n^{\text{th}}$-degree homogeneous linear ODEs.

• Objective 11–2: I understand the form of the solution to an $n^{\text{th}}$$n^{\text{th}}$-degree homogeneous linear ODE for complex roots to the characteristic equation.

• Objective 11–3: I can find the most general solution to $n^{\text{th}}$$n^{\text{th}}$-degree homogeneous linear ODEs.

• Objective 11–4: I can find the solution to $n^{\text{th}}$$n^{\text{th}}$-degree homogeneous linear IVPs.

Complex Solutions to Characteristic Equations

Let $z(t)=u(t)+iv(t)$$z(t)=u(t)+iv(t)$ be a solution to ${\displaystyle a{y}^{\prime \prime }+b{y}^{\prime }+cy=0}$$\displaystyle ay^{\,\prime\prime}+by^{\,\prime}+cy=0$, where $a,b,c\in \mathbb{R}$$a, b, c\in\mathbb{R}$. Then the real part $u(t)$$u(t)$ and the imaginary part $v(t)$$v(t)$ are real-valued solutions.

If $r=\alpha \pm i\beta$$r=\alpha\pm i\beta$ are complex conjugate solutions to ${\displaystyle a{y}^{\prime \prime }+b{y}^{\prime }+cy=0}$$\displaystyle ay^{\,\prime\prime}+by^{\,\prime}+cy=0$, then two linearly independent solutions are $e^{\alpha t}\cos (\beta t)$$e^{\alpha t}\cos{(\beta t)}$ and $e^{\alpha t}\sin (\beta t)$$e^{\alpha t}\sin{(\beta t)}$, and the general solution is\newline $y(t)=c_1{e}^{\alpha t}\cos (\beta t)+c_2{e}^{\alpha t}\sin (\beta t)$$y(t)=c_1e^{\alpha t}\cos{(\beta t)}+c_2e^{\alpha t}\sin{(\beta t)}$, where $c_1$$c_1$ and $c_2$$c_2$ are arbitrary constants.

Investigation 09

Let $y=e^{rt}$$y=e^{rt}$, where $r=\alpha \pm i\beta$$r=\alpha\pm i\beta$, be solutions to $a{y}^{\prime \prime }+b{y}^{\prime }+cy=0$$ay^{\,\prime\prime}+by^{\,\prime}+cy=0$.

1. Use Euler's formula to rewrite $y_a=e^{(\alpha +i\beta )t}$$y_a=e^{(\alpha + i\beta)t}$.

2. Use Euler's formula to rewrite $y_b=e^{(\alpha -i\beta )t}$$y_b=e^{(\alpha - i\beta)t}$.

3. Evaluate $y_1=\frac{1}{2}{y}_a+\frac{1}{2}{y}_b$$y_1=\frac{1}{2}y_a+\frac{1}{2}y_b$. Explain why $\frac{1}{2}$$\frac{1}{2}$ is a valid coefficient for $y_a$$y_a$ and $y_b$$y_b$.

4. Evaluate $y_2=\frac{1}{2i}{y}_a-\frac{1}{2i}{y}_b$$y_2=\frac{1}{2i}y_a-\frac{1}{2i}y_b$. Explain why $\frac{1}{2i}$$\frac{1}{2i}$ is a valid coefficient for $y_a$$y_a$ and $y_b$$y_b$.

5. Explain why $y_1$$y_1$ and $y_2$$y_2$ are solutions to $a{y}^{\prime \prime }+b{y}^{\prime }+cy=0$$ay^{\,\prime\prime}+by^{\,\prime}+cy=0$.

6. Explain why $y=c_1{y}_1+c_2{y}_2$$y=c_1y_1+c_2y_2$ is a solution to $a{y}^{\prime \prime }+b{y}^{\prime }+cy=0$$ay^{\,\prime\prime}+by^{\,\prime}+cy=0$.

Investigation 10

Suppose the following are roots to the characteristic equation. Find the homogeneous solution, $y_h(x)$$y_h(x)$.

1. ${\displaystyle r=-2i,2i}$$\displaystyle r=-2i,2i$

2. ${\displaystyle r=0,-3i,3i}$$\displaystyle r=0,-3i,3i$

3. ${\displaystyle r=1,i\sqrt{3},-i\sqrt{3}}$$\displaystyle r=1,i\sqrt{3},-i\sqrt{3}$

4. ${\displaystyle r=-3i,-3i,3i,3i}$$\displaystyle r=-3i,-3i,3i,3i$

5. ${\displaystyle r=1,1,-2i,2i}$$\displaystyle r=1,1,-2i,2i$

6. ${\displaystyle r=1-2i,1+2i}$$\displaystyle r=1-2i,1+2i$

7. ${\displaystyle r=2,2,2-4i,2+4i}$$\displaystyle r=2,2,2-4i,2+4i$

8. ${\displaystyle r=\frac{3-2i}{2},\frac{3+2i}{2}}$$\displaystyle r=\frac{3-2i}{2},\frac{3+2i}{2}$

Investigation 11

Solve the following $n^{\text{th}}$$n^{\text{th}}$-order homogeneous DEs using linear combination of $n$$n$ linearly-independent solutions:

1. ${\displaystyle y^{\prime \prime }-2y^{\prime }-2y=0}$$\displaystyle y^{\,\prime\prime}-2y^{\,\prime}-2y=0$

2. ${\displaystyle y^{\prime \prime }+3y^{\prime }+6y=0}$$\displaystyle y^{\,\prime\prime}+3y^{\,\prime}+6y=0$

3. ${\displaystyle 36y^{\prime \prime }-36y^{\prime }+13y=0}$$\displaystyle 36y^{\,\prime\prime}-36y^{\,\prime}+13y=0$

4. ${\displaystyle y^{\prime \prime }+y^{\prime }+y=0}$$\displaystyle y^{\,\prime\prime}+y^{\,\prime}+y=0$

5. ${\displaystyle y^{\prime \prime \prime }+3y^{\prime }=0}$$\displaystyle y^{\,\prime\prime\prime}+3y^{\,\prime}=0$

6. ${\displaystyle y^{(4)}+y^{\prime \prime }-2y=0}$$\displaystyle y^{\,(4)}+y^{\,\prime\prime}-2y=0$

7. ${\displaystyle y^{(4)}+32y^{\prime \prime }+256y=0}$$\displaystyle y^{\,(4)}+32y^{\,\prime\prime}+256y=0$

8. ${\displaystyle y^{(5)}-5y^{\prime \prime \prime }+4y^{\prime }=0}$$\displaystyle y^{\,(5)}-5y^{\,\prime\prime\prime}+4y^{\,\prime}=0$

9. ${\displaystyle y^{(5)}(t)+y^{(4)}(t)+4y^{\prime \prime \prime }(t)+3y^{\prime \prime }(t)+3y^{\prime }(t)=0}$$\displaystyle y^{\,(5)}(t)+y^{\,(4)}(t)+4y^{\,\prime\prime\prime}(t)+3y^{\,\prime\prime}(t)+3y^{\,\prime}(t)=0$

Investigation 12

Solve the following $n^{\text{th}}$$n^{\text{th}}$-order IVPs using linear combination of $n$$n$ linearly-independent solutions:

1. ${\displaystyle y^{\prime \prime }+4y^{\prime }+5y=0}$$\displaystyle y^{\,\prime\prime}+4y^{\,\prime}+5y=0$; $y(0)=0$$y(0)=0$ and $y^{\prime }(0)=2$$y^{\,\prime}(0) = 2$

2. ${\displaystyle y^{\prime \prime }-2y^{\prime }+2y=0}$$\displaystyle y^{\,\prime\prime}-2y^{\,\prime}+2y=0$; $y(0)=1$$y(0)=1$ and $y^{\prime }(0)=0$$y^{\,\prime}(0) = 0$

3. ${\displaystyle y^{\prime \prime \prime }(t)+3y^{\prime \prime }(t)+6y^{\prime }(t)+8y(t)=0}$$\displaystyle y^{\,\prime\prime\prime}(t)+3y^{\,\prime\prime}(t)+6y^{\,\prime}(t)+8y(t)=0$, $y(0)=1$$y(0)=1$, $y^{\prime }(0)=\sqrt{3}$$y^{\,\prime}(0)=\sqrt{3}$, $y^{\prime \prime }(0)=-2$$y^{\,\prime\prime}(0)=-2$

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Last Modified: Monday, 6 September 2020 13:33 EDT