Date due and handed in Feb. 11,2010
Let \(L\equiv D^{4}+8D^{2}+16\) and let \(L_{A}\equiv D^{2}+1\). Since1 \(L_{A}\left [ -\sin t\right ] =0\), then the differential equation can be written as\begin{align*} L_{A}\left [ L\left [ y\left ( t\right ) \right ] \right ] & =0\\ \left ( D^{2}+1\right ) \left ( D^{4}+8D^{2}+16\right ) & =0\\ \left ( D^{2}+1\right ) \left ( D^{2}+4\right ) \left ( D^{2}+4\right ) & =0 \end{align*}
Hence the characteristic equation is\[ \left ( r^{2}+1\right ) \left ( r^{2}+4\right ) \left ( r^{2}+4\right ) =0 \] And the roots from the particular solution are \(r_{1}=\) \(j\) and \(r_{2}=-j\) and the roots from the homogeneous solution are \(\pm 2j\) and \(\pm 2j\), which we call \(r_{3}=2j\),\(r_{4}=-2j\) and \(r_{5}=2j\) and \(r_{6}=-2j.\) Hence \[ y_{p}\left ( t\right ) =c_{1}e^{-r_{1}t}+c_{2}e^{-r_{2}t}\] and\[ y_{h}\left ( t\right ) =c_{3}e^{-r_{3}t}+c_{4}e^{-r_{4}t}+c_{5}te^{-r_{5}t}+c_{6}te^{-r_{6}t}\] Hence\begin{align*} y_{p}\left ( t\right ) & =c_{1}e^{-jt}+c_{2}e^{jt}\\ & =c_{1}\left ( \cos t-j\sin t\right ) +c_{2}\left ( \cos t+j\sin t\right ) \\ & =\left ( c_{1}+c_{2}\right ) \cos t+\left ( jc_{2}-jc_{1}\right ) \sin t\\ & =C_{1}\cos t+C_{2}\sin t \end{align*}
Where \(C_{1}=\left ( c_{1}+c_{2}\right ) \) and \(C_{2}=\left ( jc_{2}-jc_{1}\right ) \)
and\begin{align*} y_{h}\left ( t\right ) & =c_{3}e^{-2jt}+c_{4}e^{2jt}+c_{5}te^{-2jt}+c_{6}te^{2jt}\\ & =c_{3}\left ( \cos 2t-j\sin 2t\right ) +c_{4}\left ( \cos 2t+j\sin 2t\right ) \\ & +c_{5}t\left ( \cos 2t-j\sin 2t\right ) +c_{6}t\left ( \cos 2t+j\sin 2t\right ) \\ & =\left ( c_{3}+c_{4}\right ) \cos 2t+\left ( -jc_{3}+jc_{4}\right ) \sin 2t+\left ( c_{5}+c_{6}\right ) t\cos 2t+\left ( -jc_{5}+jc_{6}\right ) t\sin 2t\\ & =C_{3}\cos 2t+C_{4}\sin 2t+C_{5}t\cos 2t+C_{6}t\sin 2t \end{align*}
Where \(C_{3}=\left ( c_{3}+c_{4}\right ) ,C_{4}=\left ( -jc_{3}+jc_{4}\right ) ,C_{5}=\left ( c_{5}+c_{6}\right ) ,C_{6}=\left ( -jc_{5}+jc_{6}\right ) \)
Hence we have\begin{equation} y\left ( t\right ) =\overset{y_{p}}{\overbrace{C_{1}\cos t+C_{2}\sin t}}+\overset{y_{h}}{\overbrace{C_{3}\cos 2t+C_{4}\sin 2t+C_{5}t\cos 2t+C_{6}t\sin 2t}}\tag{1} \end{equation} To determine \(C_{1}\) and \(C_{2}\), we insert \(y_{p}\left ( t\right ) \) into the ODE and obtain\begin{align} \left ( D^{4}+8D^{2}+16\right ) y_{p}\left ( t\right ) & =-\sin t\nonumber \\ \left ( D^{4}+8D^{2}+16\right ) \left ( C_{1}\cos t+C_{2}\sin t\right ) & =-\sin t\nonumber \\ C_{1}\left ( D^{4}+8D^{2}+16\right ) \cos t+C_{2}\left ( D^{4}+8D^{2}+16\right ) \sin t & =-\sin t\tag{2} \end{align}
But \(D^{4}\left ( \cos t\right ) =D^{3}\left ( -\sin t\right ) =D^{2}\left ( -\cos t\right ) =D\left ( \sin t\right ) =\cos t\) and \(D^{2}\left ( \cos t\right ) =D\left ( -\sin t\right ) =-\cos t\) and \(D^{4}\left ( \sin t\right ) =D^{3}\left ( \cos t\right ) =D^{2}\left ( -\sin t\right ) =D\left ( -\cos t\right ) =\sin t\) and \(D^{2}\left ( \sin t\right ) =D\left ( \cos t\right ) =-\sin t\), hence (2) becomes\begin{align*} C_{1}\left ( \cos t-8\cos t+16\cos t\right ) +C_{2}\left ( \sin t-8\sin t+16\sin t\right ) & =-\sin t\\ \left ( C_{1}-8C_{1}+16C_{1}\right ) \cos t+\left ( C_{2}-8C_{2}+16C_{2}\right ) \sin t & =-\sin t \end{align*}
Hence by comparing coefficients, we see that\begin{align*} C_{2}-8C_{2}+16C_{2} & =-1\\ C_{1}-8C_{1}+16C_{1} & =0 \end{align*}
Or\begin{align*} 9C_{2} & =-1\\ 9C_{1} & =0 \end{align*}
Hence \(C_{2}=\frac{-1}{9}\) and \(C_{1}=0\), therefore the particular solution is\begin{align*} y_{p}\left ( t\right ) & =C_{1}\cos t+C_{2}\sin t\\ & =\frac{-1}{9}\sin t \end{align*}
Substitute the above into (1), we obtain\[ y\left ( t\right ) =\frac{-\sin t}{9}+C_{3}\cos 2t+C_{4}\sin 2t+C_{5}t\cos 2t+C_{6}t\sin 2t \] Which is what we are required to show. Book uses different names for the constants I used. This can be easily changed: Let \(C_{3}=C_{1}\), Let \(C_{4}=C_{2}\), Let \(C_{5}=C_{3}\) and let \(C_{6}=C_{4}\), the above can be written as\[ y\left ( t\right ) =C_{1}\cos 2t+C_{2}\sin 2t+C_{3}t\cos 2t+C_{4}t\sin 2t-\frac{\sin t}{9}\]
We need to solve \(\left ( D^{3}-2D^{2}+D-2\right ) y\left ( t\right ) =0\) subject to the initial conditions \(y\left ( 0\right ) =y^{\prime }\left ( 0\right ) =y^{\prime \prime }\left ( 0\right ) =1\). The characteristic equation is\[ r^{3}-2r^{2}+r-2=0 \] By trial and error, we see that \begin{align*} \left ( r-2\right ) \left ( r-j\right ) \left ( r+j\right ) & =\left ( r-2\right ) \left ( r^{2}+1\right ) \\ & =r^{3}-2r^{2}+r-2 \end{align*}
Therefore, the roots are \(r_{1}=2,r_{2}=j,r_{3}=-j\), hence the solution can be written as\begin{align*} y\left ( t\right ) & =c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}+c_{3}e^{r_{3}t}\\ & =c_{1}e^{2t}+c_{2}e^{jt}+c_{3}e^{-jt}\\ & =c_{1}e^{2t}+c_{2}\left ( \cos t+j\sin t\right ) +c_{3}\left ( \cos t-j\sin t\right ) \\ & =c_{1}e^{2t}+\left ( c_{2}+c_{3}\right ) \cos t+\left ( jc_{2}-jc_{3}\right ) \sin t \end{align*}
Let \(c_{2}+c_{3}=C_{2}\) and let \(jc_{2}-jc_{3}=C_{3}\), the above can be written as\begin{equation} y\left ( t\right ) =C_{1}e^{2}t+C_{2}\cos t+C_{3}\sin t \tag{1} \end{equation} Now to find the constants \(C_{i}\) we apply the boundary conditions. The first boundary condition \(y\left ( 0\right ) =1\) yields\begin{equation} y\left ( 0\right ) =1=C_{1}+C_{2} \tag{2} \end{equation} Now\[ y^{\prime }\left ( t\right ) =2C_{1}e^{2t}-C_{2}\sin t+C_{3}\cos t \] And the second boundary condition \(y^{\prime }\left ( 0\right ) =1\) yields\begin{equation} y^{\prime }\left ( 0\right ) =1=2C_{1}+C_{3} \tag{3} \end{equation} and\[ y^{\prime \prime }\left ( t\right ) =4C_{1}e^{2t}-C_{2}\cos -C_{3}\sin t \] and the third boundary condition \(y^{\prime \prime }\left ( 0\right ) =1\) yields\begin{equation} y^{\prime \prime }\left ( 0\right ) =1=4C_{1}-C_{2} \tag{4} \end{equation} So we have 3 equations to solve for \(C_{1},C_{2},C_{3}\). Add (2) and (4), we obtain \(2=5C_{1}\), hence\[ C_{1}=\frac{2}{5}\] Hence from (2) we obtain \(C_{2}=1-\frac{2}{5}\)\[ C_{2}=\frac{3}{5}\] and from (3) we obtain
\(C_{3}=1-2C_{1}=1-\frac{4}{5}\), hence\[ C_{3}=\frac{1}{5}\] Hence the solution is from (1) is found to be\begin{align*} y\left ( t\right ) & =C_{1}e^{2t}+C_{2}\cos t+C_{3}\sin t\\ & =\frac{2}{5}e^{2t}+\frac{3}{5}\cos t+\frac{1}{5}\sin t\\ & =\frac{1}{5}\left ( 2e^{2t}+3\cos t+\sin t\right ) \end{align*}
Which is the answer we are asked to show.
The ODE is\[ \left ( D^{4}-D\right ) y\left ( t\right ) =t^{2}\] Hence \(L\equiv D^{4}-D\) and \(L_{A}=D^{3}\) since \(D^{3}\left ( t^{2}\right ) =D^{2}\left ( 2t\right ) =D\left ( 2\right ) =0\), then the above ODE can be written as\[ D^{3}\left ( D^{4}-D\right ) y\left ( t\right ) =0 \] And the characteristic equation is\begin{align*} r^{3}\left ( r^{4}-r\right ) & =0\\ r^{3}r\left ( r^{3}-1\right ) & =0 \end{align*}
Hence, for the roots that are related to the particular solution are \(r_{1}=r_{2}=r_{3}=0.\,\)
And the roots that are related to the homogenous solution are \(r_{4}=0\) (notice now that this root is repeated 4 times now), and the roots of \(\left ( r^{3}-1\right ) =0\) which are the cubic roots of unity and can be found as follows\begin{align*} r^{3} & =1\\ r^{3} & =e^{2\pi j}\\ r & =e^{\frac{2\pi }{3}j}\ \ \ \ \ \ \end{align*}
Hence the 3 roots of unity are \(1,e^{\frac{2\pi }{3}j},e^{\frac{4\pi }{3}j}\), therefore the first root of unity \(1\), and the second root of unity is \(e^{\frac{2\pi }{3}j}=\cos \left ( \frac{2}{3}\pi \right ) +j\sin \left ( \frac{2}{3}\pi \right ) =\) \(-\frac{1}{2}+j\frac{1}{2}\sqrt{3}\)and the third root of unity is \(e^{\frac{4\pi }{3}j}=\cos \left ( \frac{4}{3}\pi \right ) +j\sin \left ( \frac{4}{3}\pi \right ) =\) \(-\frac{1}{2}-j\frac{1}{2}\sqrt{3}\)
Hence \(r_{5}=1\), \(r_{6}=-\frac{1}{2}+j\frac{\sqrt{3}}{2},r_{7}=-\frac{1}{2}-j\frac{\sqrt{3}}{2}\), in otherwords, the solution is\[ y\left ( t\right ) =\overset{y_{p}\left ( t\right ) }{\overbrace{c_{1}e^{r_{1}t}+c_{2}te^{r_{2}t}+c_{3}t^{2}e^{r_{3}t}}}+\overset{y_{h}\left ( t\right ) }{\overbrace{c_{4}t^{3}e^{r_{4}t}+c_{5}e^{r_{5}t}+c_{6}e^{r_{6}t}+c_{7}e^{r_{7}t}}}\] We now substitute the values of \(r_{i}\) we found and obtain\begin{align*} y\left ( t\right ) & =\overset{y_{p}\left ( t\right ) }{\overbrace{c_{1}+c_{2}t+c_{3}t^{2}}}+\overset{y_{h}\left ( t\right ) }{\overbrace{c_{4}t^{3}+c_{5}e^{t}+c_{6}e^{\left ( -\frac{1}{2}+j\frac{1}{2}\sqrt{3}\right ) t}+c_{7}e^{\left ( -\frac{1}{2}-j\frac{1}{2}\sqrt{3}\right ) t}}}\\ & =c_{1}+c_{2}t+c_{3}t^{2}+c_{4}t^{3}+c_{5}e^{t}+c_{6}e^{-\frac{1}{2}t}e^{j\frac{\sqrt{3}}{2}t}+c_{7}e^{-\frac{1}{2}t}e^{-j\frac{\sqrt{3}}{2}t}\\ & =c_{1}+c_{2}t+c_{3}t^{2}+c_{4}t^{3}+c_{5}e^{t}+e^{-\frac{1}{2}t}\left ( c_{6}e^{j\frac{\sqrt{3}}{2}t}+c_{7}e^{-j\frac{\sqrt{3}}{2}t}\right ) \\ & =c_{1}+c_{2}t+c_{3}t^{2}+c_{4}t^{3}+c_{5}e^{t}+e^{-\frac{1}{2}t}\left ( c_{6}\left [ \cos \frac{\sqrt{3}}{2}t+j\sin \frac{\sqrt{3}}{2}t\right ] +c_{7}\left [ \cos \frac{\sqrt{3}}{2}t-j\sin \frac{\sqrt{3}}{2}t\right ] \right ) \end{align*}
Hence\[ y\left ( t\right ) =c_{1}+c_{2}t+c_{3}t^{2}+c_{4}t^{3}+c_{5}e^{t}+e^{-\frac{1}{2}t}\left ( \left [ c_{6}+c_{7}\right ] \cos \frac{\sqrt{3}}{2}t+\left [ jc_{6}-jc_{7}\right ] \sin \frac{\sqrt{3}}{2}t\right ) \] Let \(\left [ c_{6}+c_{7}\right ] =C_{6}\) and let \(jc_{6}-jc_{7}=C_{7}\) the above becomes\begin{equation} y\left ( t\right ) =\overset{y_{p}\left ( t\right ) }{\overbrace{c_{1}+c_{2}t+c_{3}t^{2}}}+\overset{y_{h}\left ( t\right ) }{\overbrace{c_{4}t^{3}+c_{5}e^{t}+e^{-\frac{t}{2}}\left ( C_{6}\cos \frac{\sqrt{3}}{2}t+C_{7}\sin \frac{\sqrt{3}}{2}t\right ) }}\tag{1} \end{equation} Now plug \(y_{p}\left ( t\right ) \) back in the original ODE we obtain\begin{align*} \left ( D^{4}-D\right ) y_{p}\left ( t\right ) & =t^{2}\\ \left ( D^{4}-D\right ) \left ( c_{1}+c_{2}t+c_{3}t^{2}\right ) & =t^{2}\\ D^{4}\left ( c_{1}+c_{2}t+c_{3}t^{2}\right ) -D\left ( c_{1}+c_{2}t+c_{3}t^{2}\right ) & =t^{2}\\ D^{3}\left ( c_{2}+2c_{3}t\right ) -\left ( c_{2}+2c_{3}t\right ) & =t^{2}\\ D^{2}\left ( 2c_{3}\right ) -\left ( c_{2}+2c_{3}t\right ) & =t^{2}\\ -\left ( c_{2}+2c_{3}t\right ) & =t^{2} \end{align*}
Hence we see that \(c_{2}=0\) and \(c_{3}=0\), then (1) simplifies to \begin{equation} y\left ( t\right ) =c_{1}+c_{4}t^{3}+c_{5}e^{t}+e^{-\frac{t}{2}}\left ( C_{6}\cos \frac{\sqrt{3}}{2}t+C_{7}\sin \frac{\sqrt{3}}{2}t\right ) \tag{2} \end{equation} To find \(c_{4}\,\), we substitute \(y\left ( t\right ) \) found above, into the ode, hence\begin{align*} \left ( D^{4}-D\right ) y\left ( t\right ) & =t^{2}\\ \left ( D^{4}-D\right ) \left [ c_{1}+c_{4}t^{3}+c_{5}e^{t}+e^{-\frac{t}{2}}\left ( C_{6}\cos \frac{\sqrt{3}}{2}t+C_{7}\sin \frac{\sqrt{3}}{2}t\right ) \right ] & =0 \end{align*}
Now, since we only care about finding \(c_{4}\), we can just apply \(D\) on that, hence\begin{align*} D^{4}\left [ \cdots +c_{4}t^{3}+\cdots \right ] -D\left [ \cdots +c_{4}t^{3}+\cdots \right ] & =t^{2}\\ D^{3}\left [ \cdots +3c_{4}t^{2}+\cdots \right ] -\left [ \cdots +3c_{4}t^{2}+\cdots \right ] & =t^{2}\\ D^{2}\left [ \cdots +6c_{4}t+\cdots \right ] -\left [ \cdots +3c_{4}t^{2}+\cdots \right ] & =t^{2}\\ D\left [ \cdots +6c_{4}+\cdots \right ] -\left [ \cdots +3c_{4}t^{2}+\cdots \right ] & =t^{2}\\ -\left [ \cdots +3c_{4}t^{2}+\cdots \right ] & =t^{2} \end{align*}
By comparing coefficients, we see that \(c_{4}=-\frac{1}{3}\) then (1) becomes\[ y\left ( t\right ) =c_{1}+c_{5}e^{t}+e^{-\frac{t}{2}}\left ( C_{6}\cos \frac{\sqrt{3}}{2}t+C_{7}\sin \frac{\sqrt{3}}{2}t\right ) -\frac{1}{3}t^{3}\] Which is what we are asked to show.
