Q36E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Although the real general solution form (9) is convenient, it is also possible to use the form (21) ${\mathit{d}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{(}\mathbf{\alpha }\mathbf{+}\mathbf{i\beta }\mathbf{)}\mathbf{t}}\mathbf{+}{\mathit{d}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{(}\mathbf{\alpha }\mathbf{-}\mathbf{i\beta }\mathbf{)}\mathbf{t}}$ to solve initial value problems, as illustrated in Example 1. The coefficients ${\mathit{d}}_{\mathbf{1}} \mathbf{\text{and}} {\mathit{d}}_{\mathbf{2}}$ are complex constants.

1. Use the form (21) to solve Problem 21. Verify that your form is equivalent to the one derived using (9).
2. Show that, in general, ${\mathit{d}}_{\mathbf{1}} \mathbf{\text{and}} {\mathit{d}}_{\mathbf{2}}$ in (21) must be complex conjugates so that the solution is real.

Step-by-Step Solution

Answer

1. The general solution is $y\left(t\right)=\left[1-\frac{3i}{2}\right]e^{(-1+i)t}+\left[1+\frac{3i}{2}\right]e^{(-1-i)t}$.
2. Thus $d_1 \text{and} d_2$  are complex conjugates.

Step 1: Given data

The equation (9) is $y\left(t\right)=c_1{e}^{\alpha t}\cos \beta t{c}_2{e}^{\alpha t}\sin \beta t$.

Step 2: Solve the problem (21) d 1 e ( α + i β ) t + d 2 e ( α - i β ) t .

The differential equation is $y\text{'}\text{'}+2y\text{'}+2y=0$.

The auxiliary equation is:

 $$\begin{gathered} r^2+2r+2=0 \\ r=-1\pm i \end{gathered}$$

The general equation is $y\left(t\right)=c_1{e}^{(-1+i)t}+c_2{e}^{(-1-i)t}$.

Apply initial conditions $y\left(0\right)=0,y\text{'}\left(0\right)=2$.

$$\begin{gathered} y\left(0\right)=c_1=\left[1-\frac{3i}{2}\right] \\ y\text{'}\left(0\right)=c_2=\left[1+\frac{3i}{2}\right] \end{gathered}$$

 

The general solution is $y\left(t\right)=\left[1-\frac{3i}{2}\right]e^{(-1+i)t}+\left[1+\frac{3i}{2}\right]e^{(-1-i)t}$.

Step 3: Check the solution.

Let $z=\alpha +i\beta$ then the conjugate of z is $z^{*}=\alpha -i\beta$ 

 

Thus, $d_1 \text{and} d_2$ are complex conjugates.