Solve the initial value problem y''' - y' = 0 , with y(0)=3…

Solve the initial value problem $y''' - y' = 0$, with $y(0)=3,\qquad y'(0)=-1,\qquad y''(0)=1$.

Step 1: Characteristic equation

The equation is linear, homogeneous, with constant coefficients. Substituting $y=e^{rx}$ gives: \[r^3 - r = 0 \;\Longrightarrow\; r(r-1)(r+1)=0 \;\Longrightarrow\; r = 0,\ 1,\ -1.\]

Three distinct real roots, so the general solution is: \[y = C_1 + C_2e^{x} + C_3e^{-x}. \tag{1}\]

Step 2: Derivatives

\[ y' = C_2e^{x} - C_3e^{-x}, \qquad y'' = C_2e^{x} + C_3e^{-x}. \]

Step 3: Apply the initial conditions

At $x=0$, (1) and its derivatives give the linear system: \[C_1+C_2+C_3=3, \qquad C_2-C_3=-1, \qquad C_2+C_3=1.\]

Adding the last two equations: $2C_2=0\Rightarrow C_2=0$. Subtracting them: $2C_3=2\Rightarrow C_3=1$. Then from the first equation, $C_1 = 3-C_2-C_3 = 3-0-1=2$.

Step 4: Particular solution

The solution is recorded in the result below.

Step 5: Verification

Differentiate three times: \[y' = -e^{-x}, \qquad y'' = e^{-x}, \qquad y''' = -e^{-x}.\]

Then $y'''-y' = -e^{-x}-(-e^{-x}) = 0\ \checkmark$. At $x=0$: $y(0)=2+1=3\ \checkmark$, $y'(0)=-1\ \checkmark$, $y''(0)=1\ \checkmark$. All three conditions and the equation itself are satisfied.