4.3

Example \(\PageIndex{1}\)

Solve \(y'''-4y''+y'+6y=8e^{4t} \)

Solution

The first step is going to be solve this as if it were homogenous. So we have \(r^3-4r^2+r+6=0.\) We are going to guess a solution of \(y=e^{rt}\) so we check if the equation is equal to zero for \(r = \pm 1, \pm 2, \pm 3, \pm 6\). We then see that we can factor \( r^3-4r^2+r+6 = (r+1)(r-2)(r-3)\), so our values of \(r\) are \(-1,2,\) and \(3\). Thus our homogenous generic solution is \(y= c_1e^{-t}+c_2e^{2t}+c_3e^{3t}\).

 

Our next step is to find a non homogenous solution. We are going to guess that the solution \(y = A e^{4 t}.\) Then from a previous section we have \( \psi=u_1\phi_1 +u_2\phi_2 + u_3\phi_3.\) Then we use the Wronskian and find 

 

\[ \notag
w(e^{-t},e^{2t},e^{3t}) =
\left|\begin{array}{ccc}
e^{-t} & e^{2 t} & e^{3 t} \\
-e^{-t} & 2 e^{2 t} & 3 e^{3 t} \\
e^{-t} & 4 e^{2 t} & 9 e^{3t}
\end{array}\right| \xrightarrow[R_2+R_3]{R_3-R_1} \left|\begin{array}{ccc}
e^{-t} & e^{2 t} & e^{3 t} \\
0 & 3 e^{2 t} & 4 e^{3 t} \\
0 & 3 e^{2 t} & 8 e^{3 t}
\end{array}\right|   \xrightarrow[R_3-R_2]{ }
\left|\begin{array}{ccc}
e^{-t} & e^{2 t} & e^{3 t} \\
0 & 3 e^{2 t} & 4 e^{3 t} \\
0 & 0 & 4 e^{3 t}
\end{array}\right| = 12e^{(-1+2+3)t}=12e^{4t}
\]

Note

Alternatively, we could solved the above using Abel's Theorem. We have \(w(e^{-t},e^{2t},e^{3t}) = ce^{-\int p(t)dt} = ce^{-\int(-4)dt} = ce^{\int 4 dt} = ce^{4t}.\) We now evaluate at 0 and have 

\[\notag
w(e^{0},e^{0},e^{0}) =
\left|\begin{array}{ccc}
e^{0} & e^{0} & e^{0} \\
-e^{0} & 2 e^{0} & 3 e^{0} \\
e^{0} & 4 e^{0} & 9 e^{0}
\end{array}\right|=
\left|\begin{array}{ccc}
1 & 1 & 1 \\
-1 & 2  & 3  \\
1 & 4  & 9 
\end{array}\right| \xrightarrow[C_2-C_1]{C_3-C_1}
\left|\begin{array}{ccc}
1 & 0 & 0 \\
-1 & 3  & 4  \\
1 & 3  & 8 
\end{array}\right| = 1(3 \cdot 8 - 3\cdot 4) = 12
\]

Thus we have \(c=12\). The last equality comes from computing the determinant using minors, see the linear algebra review if you want more infromation about this method. 

 

Next we solve for \(w(1),w(2),\) and \(w(3)\). We have 

\[\notag w(1) = \left|\begin{array}{ccc}
0 & e^{2 t} & e^{3 t} \\
0 & 2 e^{2 t} & 3 e^{3 t} \\
1 & 4 e^{2 t} & 9 e^{3 t}
\end{array}\right| = +1(3e^{5t}-2e^{5t})=e^{5t}\]

\[\notag w(2) = \left|\begin{array}{ccc}
e^{-t} & 0 & e^{3 t} \\
-e^{-t} & 0 & 3 e^{3 t} \\
e^{-t} & 1 & 9 e^{3 t}
\end{array}\right| =-1\left|\begin{array}{cc}
e^{-t} & e^{3 t} \\
-e^{-t} & 3 e^{3 t}
\end{array}\right|=-\left(3 e^{2 t}+e^{2 f}\right) = -4e^{2t}\]

\[\notag w(3) = \left|\begin{array}{ccc}
e^{-t} & e^{2 t}& 0 \\
-e^{-t} & 2 e^{2 t} & 0 \\
e^{-t} & 4 e^{2 t} & 1
\end{array}\right| =-1\left|\begin{array}{cc}
e^{-t} & e^{2 t} \\
-e^{-t} & 2 e^{2 t}
\end{array}\right|=-\left(2 e^{t}+e^{t}\right) = 3e^{t}\]

Then we also need to solve for \(u_1, u_2, u_3\), and \( \psi.\)

\[\notag
u_1=\int \frac{g}{a} \frac{w_1 d t}{w}=\int \frac{8 e^{4 t} \cdot e^{5 t}}{12 e^{4 t}} d t = \int\frac{2}{3}e^{5t} dt = \frac{2}{15}e^{5t}
\]

\[ \notag
u_2=\int \frac{2}{a} \frac{w_2}{w} d t=\int \frac{8 e^{4 t} \cdot\left(-4 e^{2 t}\right)}{12 e^{4 t}} d t = -\int \frac{8}{3} e^{2 t} d t=-\frac{4}{3} e^{2 t}
\]

\[ \notag
u_3=\int \frac{g}{a} \frac{\omega_3}{\omega} d t=\int \frac{8 e^{4 t}\left(3 e^t\right)}{12 e^{4 t}} d t =\int 2 e^t d t=2 e^t
\]

\[ \notag
\psi=\left(\frac{2}{15} e^{5 t}\right) e^{-t}+\left(-\frac{4}{5} e^{2 t}\right) e^{2 t}+\left(2 e^t\right) e^{3 t} =\left(\frac{2}{15}-\frac{4}{3}+2\right) e^{4 t}=\frac{2-20+30}{15} e^{4 t} =\frac{12}{15} e^{4 t} =\frac{4}{5} e^{4 t}
\]

Thus we have a general solution of \(y=c_1 e^{-t}+c_2 e^{2 t}+c_3 e^{3 t} +\frac{4}{5} e^{4 t}.\)