Q18E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 13–20, solve the given initial value problem.

z" - 2z' -2z = 0 : z(0) = 0, z'(0) = 3

Step-by-Step Solution

Answer

The solution is $z\left(t\right)=\frac{\sqrt{3}}{2}{e}^{\left(1+\sqrt{3}\right)t}-\frac{\sqrt{3}}{2}{e}^{\left(1+\sqrt{3}\right)t}$.

Step 1: Find the solution of the differential equation.

The given differential equation is z" - 2z' - 2z= 0.

The auxiliary equation is r² - 2r -2 = 0

Find the roots of the auxiliary equation.

$\begin{array}{rcl}{r}^2-2r-2& =& 0\\ r& =& \frac{2\pm \sqrt{4^2-4\left(1\right)(-2)}}{2\left(1\right)}\\ r& =& \frac{2\pm \sqrt{16+8}}{2}\\ r& =& 1\pm \sqrt{3}\end{array}$

Therefore, the solution is $z\left(t\right)=c_1{e}^{\left(1+\sqrt{3}\right)t}+c_2{e}^{\left(1+\sqrt{3}\right)t}$.

Step 2: Apply initial conditions .

The initial conditions are z(0) = 0, z'(0) = 3.

Therefore,

$\begin{array}{rcl}z\left(0\right)& =& c_1{e}^0+c_2{e}^0\\ c_1+c_2& =& 0\end{array}$

And 

$\begin{array}{rcl}z\text{'}\left(t\right)& =& \left(1+\sqrt{3}\right)c_1{e}^{(1+\sqrt{3)t}}+\left(1-\sqrt{3}\right)c_2{e}^{\left(1-\sqrt{3}\right)t}\\ z\text{'}\left(0\right)& =& 1+\sqrt{3}{c}_1{e}^0+\left(1-\sqrt{3}\right)c_2{e}^0\\ 1+\sqrt{3}{c}_1+\left(1-\sqrt{3}\right)c_2& =& 3\end{array}$

Solving for c₁,c₂ then,

$$\begin{gathered} c_1=\frac{\sqrt{3}}{2} \\ {c}_2=-\frac{\sqrt{3}}{2} \end{gathered}$$

Thus, the solution is $z\left(t\right)=\frac{\sqrt{3}}{2}{e}^{\left(1+\sqrt{3}\right)t}-\frac{\sqrt{3}}{2}{e}^{\left(1-\sqrt{3}\right)t}$.