91Ó°ÊÓ

Given differential equation is y'" - 2y" - y' + 2y = 0.

Let y = e^rt, then y'(t) = re^rt and

$\begin{array}{rcl}y"\left(t\right)& =& r^2{e}^{rt}\\ y\text{'}"& =& r^3{e}^{rt}\end{array}$

Then the auxiliary equation is;

$\begin{array}{rcl}{r}^3{e}^{rt}-2r^2{e}^{rt}-r{e}^{rt}+2e^{rt}& =& 0\\ (r^3-2r^2-r+2)e^{rt}& =& 0\\ r^3-2r^2-r+2& =& 0\end{array}$

The auxiliary equation can be written as;

r³- 2r²- r + 2 = (r-1)(r²- r - 2)

Find the roots of the auxiliary equation.

$r-1=0⇒r=1$

Or

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

Therefore, the general solution is;

y(t) = c₁e^t + c₂e^-t + c₃e^2t

Given boundary conditions are y(0) = 2, y'(0) = 3 and y"(0) = 5

Then,

$\begin{array}{rcl}y\left(0\right)& =& c_1{e}^0+c_2{e}^{-0}+e^{2×0}\\ c_1+c_2+c_3& =& 2...\left(1\right)\end{array}$

And

$\begin{array}{rcl}y\text{'}\left(t\right)& =& c_1{e}^t+c_2{e}^{-t}+2c_3{e}^{2t}\\ y\text{'}\left(0\right)& =& c_1-c_2{e}^0+2c_3{e}^0\\ c_1-c_2+2c_3& =& 3...\left(2\right)\end{array}$

Again using the initial condition;

$\begin{array}{rcl}y"\left(t\right)& =& c_1{e}^t+c_2{e}^{-t}+4c_3{e}^{2t}\\ y"\left(0\right)& =& c_1-c_2{e}^{-0}+4c_3{e}^{2×0}\\ c_1+c_2+4c_3& =& 5...\left(3\right)\end{array}$

Subtract (2) from (1), then we will get

3c₃ = 3

c₃ = 1

Now add (1) and (2), then substitute c₃

$\begin{array}{rcl}2c_1+3c_3& =& 5\\ 2c_1+3\left(1\right)& =& 5\\ 2c_1& =& 2\\ c_1& =& 1\end{array}$

Substitute c₁and c₃ in

1 + c₂ + 1 = 2

c₁ = 0

Therefore, the solution to the given initial problem is y(t) = e^-t+ e^2t.