Q16E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

y"-4y'-5y = 0 : y(-1) = 3,y'(-1) = 9

Step-by-Step Solution

Answer

The solution is $y\left(t\right)=-\frac{2}{3}{e}^{t-1}+\frac{5}{3}{e}^{5+5t}$.

Step 1: Find the solution of the differential equation.

The given differential equation is y" - 4y' + 3y = 0.

The auxiliary equation is;

$\begin{array}{rcl}{r}^2-4r^1-53t^0& =& 0\\ r^2-4r-5& =& 0\\ r^2+r-5r-5& =& 0\\ r(r+1)-5(r+1)& =& 0\\ (r+1)(r-5)& =& 0\\ r& =& -1,5\end{array}$

Therefore, the solution is y(t) = c₁e^-t + c₂e^5t.

Step 2: Apply initial conditions .

The initial conditions are y(-1) = 3,y'(-1) = 9.

Therefore,

$\begin{array}{rcl}y(-1)& =& c_1{e}^1+c_2{e}^{-5}\\ c_1{e}^1+c_2{e}^{-5}& =& 3\end{array}$

And

$\begin{array}{rcl}y\text{'}\left(t\right)& =& -c_1{e}^{-t}+5c_2{e}^{5t}\\ y\text{'}(-1)& =& -c_1{e}^1+5c_2{e}^{-5}\\ -c_1{e}^1+5c_2{e}^{-5}& =& 9\end{array}$

Solving for c₁,c₂ then,

$$\begin{gathered} c_1=-\frac{2}{3}{e}^{-1} \\ {c}_2=\frac{5}{3}{e}^5 \end{gathered}$$

Therefore, the solution is $y\left(t\right)=-\frac{2}{3}{e}^{t-1}+\frac{5}{3}{e}^{5+5t}$.