Q17E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

$\mathit{y}\mathbf{"}\mathbf{-}\mathbf{6}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{:}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{25}}{\mathbf{3}}$

Step-by-Step Solution

Answer

The solution is $y\left(t\right)=2e^{3t}+\frac{7}{3}t{e}^{3t}$.

Step 1: Find the solution of the differential equation.

The given differential equation is y" -6y' + 9y = 0.

The auxiliary equation is;

$\begin{array}{rcl}{r}^2-6r^1+9r^0& =& 0\\ r^2-6r+9& =& 0\\ r^2-3r-3r+9& =& 0\\ r(r-3)-3(r-3)& =& 0\\ {(r-3)}^2& =& 0r=3,3\end{array}$

Therefore, the solution is y(t) = c₁e^3t + c₂te^3t.

Step 2: Apply initial conditions .

The initial conditions are $y\left(0\right)=2,y\text{'}\left(0\right)=\frac{25}{3}$.

Therefore,

$\begin{array}{rcl}y\left(0\right)& =& c_1{e}^0+c_2\left(0\right)e^0\\ c_1& =& 2\end{array}$

And

$$\begin{gathered} y\text{'}\left(t\right)=3c_1{e}^{3t}+c_2(3t{e}^{3t}+e^{3t}) \\ y\text{'}\left(0\right)=3c_1{e}^0+c_2(0+e^0) \\ 3c_1+c_2=\frac{25}{3} \end{gathered}$$

Solving for c₁,c₂ then;

$$\begin{gathered} c_1=2 \\ {c}_2=\frac{7}{3} \end{gathered}$$

Therefore, the solution is $y\left(t\right)=2e^{3t}+\frac{7}{3}t{e}^{3t}$.