Solve the initial value problem:

y"' - y' = 0; y(0) = 2,

y'(0) = 3, y"(0) = -1

Given differential equation is y"'-y' = 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;

role="math" localid="1655813037917" $\begin{array}{rcl}{r}^3{e}^{rt}-r{e}^{rt}& =& 0\\ (r^3-r)e^{rt}& =& 0\\ r^3-r& =& 0\\ r(r^2-1)& =& 0\\ r(r-1)(r+1)& =& 0\end{array}$

Therefore, the solutions are;

r = 0,-1,1

Therefore, the general solution is:

y(t) = c₁ + c₂e^-t + c₃e^t

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

Using the given initial conditions;

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

And

y'(t) = -c₂e^-t+ c₃e^t

y'(0) = -c₂e⁰ + c₃e⁰

-c₂ + c₃ = 3 ...(2)

Again using the given condition;

y"(t) = -c₂e^-t+ c₃e^t

y"(0) = -c₂e⁰ + c₃e⁰

c₂ + c₃ = -1 ...(3)

Add (1) and (2), then one will get

$\begin{array}{rcl}2c_3& =& 2\\ c_3& =& 1\end{array}$

Substitute c₃ in (3)

role="math" localid="1655814462532" $\begin{array}{rcl}{c}_2+1& =& -1\\ c_2& =& -2\end{array}$

Substitute c₂ and c₃ in (1)

$\begin{array}{rcl}{c}_1-2+1& =& 2\\ c_1& =& 3\end{array}$

Therefore, the solution to the given initial problem is y(t) = 3 - 2e^-t+ e^t.