Q19E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

y" + 2y' + y = 0 : y(0) = 1, y'(0) = -3

Step-by-Step Solution

Answer

The solution is y(t) = e^(-t) -2e^(-t).

Step 1: Find the solution of the differential equation.

The given differential equation is y" + 2y' + y = 0.

The auxiliary equation is r² + 2r + 1 = 0

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

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

Step 2: Apply initial conditions .

The initial conditions are y(0) = 1, y'(0) =-3.

Therefore,

$\begin{array}{rcl}y\left(0\right)& =& c_{}{e}^0\_c_2\left(0\right)e^0\\ c_1& =& 1\end{array}$

And

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

Solving for c₁,c₂ then;

$$\begin{gathered} c_1=1 \\ {c}_2=-2 \end{gathered}$$

Therefore, the solution is y(t) = e^(-t) - 2e^(-1).