Toronto Math Forum

I think Cui calculated the homogeneous solution wrong:
$(r-1)(r-1)(r-1) = r^3 -3r^2+3r+1$

Actually:
$r^3 -3r^2+4r-2= (r-1)(r^2-2r-2)$ = 0
$r=1$ or $r=1-i, 1+i$


So, Homogeneous solution is
$y= c_1e^t + c_2e^t\cos(t) + c_3e^t\sin(t)$

Non homogenous part for $10e^t$
We should assume $Y= Ate^t $
$Y’=Ate^t+Ae^t$
$Y’’=Ate^t+2Ae^t$
$Y’’’=Ate^t+3Ae^t$

$Y’’’-3Y’’+4Y’-2Y =(A-3A+4A-2A)te^t+(3A-6A+4A-2A)e^t =-Ae^t =10e^t$
$A =-10, Y=-10te^t$

fixed,thx!

$$Homo: r^3 -2r^2-r+2=0$$
$$(r+1)(r-1)(r-2)=0$$
$$r_1=1, r_2=-1, r_3=2$$
$$\therefore y_c(t)=c_1e^t+c_2e^{-t}+c_3e^{2t}$$
$$W = \left|
\begin {array}{ccc}
  {e^t}&e^{-t}&e^{2t}\\
  e^t& -e^{-t}&2e^{2t}\\
  e^t&e^{-t}& 4e^{2t}
\end {array}
\right| = -6e^{2t}$$
$$W_1 = \left|
\begin {array}{ccc}
  {0}&e^{-t}&e^{2t}\\
  0& -e^{-t}&2e^{2t}\\
  1&e^{-t}& 4e^{2t}
\end {array}
\right| = 3e^{t}$$
$$W_2 = \left|
\begin {array}{ccc}
  {e^t}&0&e^{2t}\\
  e^t&0&2e^{2t}\\
  e^t&1& 4e^{2t}
\end {array}
\right| = e^{3t}$$
$$W_3 = \left|
\begin {array}{ccc}
  {e^t}&e^{-t}&0\\
  e^t& -e^{-t}&0\\
  e^t&e^{-t}& 1
\end {array}
\right| = -2$$
substitute above into the formula:
$$y_p(t)=y_1{\int}{{w_1(s)}{g(s)}\over{W(s)}} ds+y_2{\int}{{w_2(s)}{g(s)}\over{W(s)}}ds+y_3{\int}{{w_2(s)}{g(s)}\over{W(s)}}ds$$
$$=-e^t{\int}{{6e^s}\over{e^s+1}} ds-e^{-t}
{\int}{{2e^{3s}}\over{e^s+1}}ds+e^{2t}{\int}{{4}\over{e^s+1}}ds$$
$$=-6e^tln|e^t+1|-e^{-t}ln|e^t+1|-e^{-t}(e^t-1)^2-e^{2t}4ln|e^{-x}+1| $$
$$y(t)=y_c(t)+y_p(t)$$

$$det(A-\lambda I) = \left|
\begin {array}{ccc}
  {3- \lambda}&2&4\\
  2& {- \lambda}&2\\
  4&2&{3- \lambda}
\end {array}
\right| = 0$$
$${-\lambda}^3 + 6 {\lambda}^2+  15{\lambda}+8 = 0 $$
$$By\ long\ devision\ method, we\ get\ -({\lambda+1})^{2}(\lambda-8) = 0$$
$$\quad\therefore \lambda = -1,-1,8$$
$when\ \lambda = 8:$
$$(A-\lambda I)x = 0$$
$$\left[
\begin {array}{ccc}
  -5&2&4\\
  2&-8&2\\
  4&2&-5
\end {array}
\right]x = 0
$$
$$By\ row\ operation, we\ get: \left[
\begin {array}{ccc}
  1&0&-1\\
  0&2&-1\\
  0&0&0
\end {array}
\right]\left[
\begin {array}{c}
  x_1\\
  x_2\\
  x_3
\end {array}
\right]= 0$$
$let x_3 = t:$
$$x_1= t$$
$$2x_2= t$$
$we\ have$:
$$\left[
\begin {array}{c}
  2\\
  1\\
  2
\end {array}
\right]$$
$When \lambda = -1:$
$$\left[
\begin {array}{ccc}
  4&2&4\\
  2&1&2\\
  4&2&4
\end {array}
\right] x=0$$
$$By\ row\ operation, we\ get: \left[
\begin {array}{ccc}
  2&1&2\\
  0&0&0\\
  0&0&0
\end {array}
\right]\left[
\begin {array}{c}
  x_1\\
  x_2\\
  x_3
\end {array}
\right]=0$$
$let\ x_3=t,x_2=s:$
$$2x_1=-s-2t$$
$$x_1={{-1}\over {2}}s-t$$
$$x_2=s$$
$$x_3=t$$
$we\ have: $
$$\left[
\begin {array}{c}
  {{-1}\over {2}}s-t\\
  s\\
  t
\end {array}
\right] = t\left[
\begin {array}{c}
  -1\\
  0\\
  1
\end {array}
\right] +s \left[
\begin {array}{c}
  {{-1}\over {2}}\\
  1\\
  0
\end {array}
\right]  $$
$\quad\therefore we\ have\ eigenvector: \left[
\begin {array}{c}
  {{-1}\over {2}}\\
  1\\
  0
\end {array}
\right] , \left[
\begin {array}{c}
  -1\\
  0\\
  1
\end {array}
\right] $
$$\quad\therefore x(t) = c_1e^{8t}\left(
\begin {array}{c}
  2\\
  1\\
  2
\end {array}
\right)+c_2e^{-t} \left(
\begin {array}{c}
  {{-1}\over2}\\
  1\\
  0
\end {array}
\right)+c_3e^{-t}\left(
\begin {array}{c}
  -1\\
  0\\
  1
\end {array}
\right)$$

$Since\ f(z)\ has\ a\ zero\ of\ order\ m\ at \ z_{0}$,
$$\quad\therefore f(z) = (z-z_{0})^mg(z),where g'(z_{0})\ne0$$
$$\quad\therefore f'(z) = (z-z_{0})^mg'(z)+m(z-z_{0})^{m-1}g(z)$$
$$ f'(z) = (z-z_{0})^m(g'(z)+m(z-z_{0})^{-1}g(z)) $$
$$\quad\therefore {{f'(z)}\over {f(z)}} ={{(z-z_{0})^m(g'(z)+m(z-z_{0})^{-1}g(z))}\over {(z-z_{0})^mg(z)}}  $$
$$= {{g'(z)}\over {g(z)}}+m(z-z_{0}^{-1}) $$
$$\quad\therefore Res({{f'}\over f}, z_{0})=m$$

$$let\ h(z) = 0, g(z) = 1- cos(z)$$
$$As \ z _0 = 0, h(0) = 1\ne0; order = 0$$
$$g(z) = 1- cos(z)$$
$$g'(z) = sin(z), g'(0) = 0$$
$$g''(z) = cos(z), g''(0) = 1\ne0, \quad\therefore order = 2 $$
$$\quad\therefore 2-0= 2   \quad\therefore \ order\ of \ pole = 2$$
$$\quad\therefore{{1}\over1-cos(z)} = a_{-2}z^{-2} + a_{-1}z^{-1} + a_{0} + a_{1}z^{1} + a_{2}z^{2} +... $$
$${{1}\over 1-(1-{{z^{2}\over {2 !}}+{z^{4}\over 4!}+...})} = a_{-2}z^{-2} + a_{-1}z^{-1} + a_{0} + a_{1}z^{1} + a_{2}z^{2} +...  $$
$$1= ({z^{2}\over 2!} - {z^{4}\over 4!} + {z^{6}\over {6!}}... )(a_{-2}z^{-2} + a_{-1}z^{-1} + a_{0} + a_{1}z^{1} + a_{2}z^{2} +...  ) $$
$Corresponding\ coefficients \ equal\ with\ each\ other:$
$${{a_{-2}}\over {2!}} = 1  \quad\therefore a_{-2} = 2 $$
$${{a_{-1}}\over {2!}} = 0  \quad\therefore a_{-1} = 0 $$
$${{a_{0}}\over {2!}} - {{1}\over4!}a_{-2} = 0  \quad\therefore a_{0} = {{1}\over6} $$
$${{a_{0}}\over {2!}} - {{a_{-1}}\over {4!}}  = 0 \quad\therefore a_{1} = 0 $$
$$similar\ like \ the\ above\ procedure\ we\ can\ get \ a_{2} = {{1}\over120} $$
$$\quad\therefore {{2}\over {z^2}} + {{1}\over6} + {{1}\over120} {z^2} +...,\   Res(f;0) = 0  $$