Toronto Math Forum
MAT2442013F => MAT244 MathTests => Quiz 3 => Topic started by: Victor Ivrii on November 06, 2013, 08:11:53 PM

Find the general solution of the given differential equation.
\begin{equation*}
y'''y''y'+ y = 0.
\end{equation*}

\begin{equation*}
y'''y''y'+ y = 0
\end{equation*}
The responding characteristic equation is $$r^3r^2r+1=0$$
$$(r^3r)(r^21)=0$$
$$r(r^21)(r^21)=0$$
$$(r1)(r^21)=0$$
$$r_1=1, r_2=1, r_3=1$$
So the general solution is $$y=c_1e^t+c_2te^t+c_3e^{t}$$

For the differential equation:
\begin{equation} y'''y''y'+y=0 \end{equation}
We assume that $y = e^{rt}$.
Therefore, we must solve the characteristic equation:
\begin{equation} r^3  r^2  r + 1 = 0 \end{equation}
We find:
$
r^3  r^2  r + 1 = 0 \implies (r^21)(r1) = 0 \implies (r+1)(r1)^2 = 0
$
This means the roots of this equation are:
$
r_1 = 1, r_2=1, r_3=1
$
(We have a repeated root at r = 1)
So the general solution to (1) is:
\begin{equation} y(t) = c_1e^{t} + c_2e^{t}t + c_3e^{t} \end{equation}