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\def\v{\vskip -3pt}
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3.6  = 4.4 Variation of Parameters~~~~~~\hfill
\vskip 5pt
\centerline{Solve $ay''' + by'' + cy' + ey= g(t)$}


{\bf 1)  Find homogeneous solutions:  }%Solve $ay'' + by' + cy = 0$}
\vskip -5pt

Suppose general homogeneous soln to $ay''' + by'' + cy' + ey = 0$ is
\vskip 5pt
\centerline{$y = c_1\phi_1(t) + c_2 \phi_2(t)+ c_3 \phi_3(t)$}


{\bf 2.)  Find a non-homogeneous solution:}
\u
Sect. 3.5 method:  Educated guess
\u
Sect. 3.6 = 4.4:  ~~~{\bf  $y = u_1(t)\phi_1 + u_2(t) \phi_2 + u_3(t) \phi_3$ and}% \hfil \break 

\rightline{ \bf solve for $u_1$, $u_2$, and $u_3$}

3 unknown functions, thus might want 3 equations.  

1 equation will come from plugging into $ay''' + by'' + cy' + ey= g(t)$

We will choose the remaining equations so that we avoid $u_i''$

\def\p{\phi}

$y = u_1 \p_1 + u_2 \p_2+ u_3 \p_3$ implies
\t
\centerline{
$y' = u_1 \p_1' + u_1' \p_1 + u_2 \p_2' + u_2' \p_2+ u_3 \p_3' + u_3' \p_3$}

I.e., $y' = u_1 \p_1' + u_2 \p_2' + u_3 \p_3'+ u_1' \p_1 + u_2' \p_2 + u_3' \p_3$   


 Avoid 2nd derivative $u_i''$: ~~Choose $u_1'\p_1 + u_2' \p_2 + u_3' \p_3 = 0$

Thus $y' = u_1 \p_1' + u_2 \p_2' + u_3 \p_3'$


$y'' = u_1 \p_1'' + u_2 \p_2'' + u_3 \p_3'' + u_1' \p_1' + u_2' \p_2' + u_3' \p_3'$   by product rule.


 Avoid 2nd derivative $u_i''$: ~~Choose $ u_1' \p_1' + u_2' \p_2' + u_3' \p_3' = 0$


Thus $y'' = u_1 \p_1'' + u_2 \p_2'' + u_3 \p_3'' $


$y''' = u_1 \p_1''' + u_2 \p_2''' + u_3 \p_3''' +  u_1' \p_1'' + u_2' \p_2'' + u_3' \p_3'' $  by product rule.

Plug $y = u_1(t)\phi_1 + u_2(t) \phi_2 + u_3(t) \phi_3$ (using above derivatives including simplifications) into $ay''' + by'' + cy' + ey = 0$

After lots of work, this simplifies to $ u_1' \p_1'; + u_2' \p_2'' + u_3' \p_3'' =  {g(t) \over a}$

Thus we have 3 unknowns (the functions, $u_1, u_2, u_3$) and 3 equations (the two equations we choose in order to avoid  $u_i''$) as well as the equation we get by plugging in  $y = u_1(t)\phi_1 + u_2(t) \phi_2 + u_3(t) \phi_3$  into DE and doing lots of simplification.  These 3 equations are

\t
\centerline{ $u_1'\p_1 + u_2' \p_2 + u_3' \p_3 = 0$}
\vskip 5pt
\centerline{ $u_1'\p_1' + u_2' \p_2' + u_3' \p_3' = 0$}
\vskip 5pt
\centerline{\hskip 12pt $u_1'\p_1'' + u_2' \p_2'' + u_3' \p_3'' = {g(t) \over a}$}

\vfill
In matrix form:

\vskip -29pt \hskip 20pt
\centerline{$ \left(\matrix{
\p_1 & \p_2 & \p_3 \cr
\p_1' & \p_2' & \p_3' \cr
\p_1'' & \p_2'' & \p_3''
}\right)
 \left(\matrix{ u_1'\cr u_2' \cr u_3'}\right) = \left(\matrix{0 \cr 0 \cr {g(t) \over a}}\right)$}

Solve using Cramer's rule.

NOTE:  The coefficient matrix is the related to the Wronskian (same matrix, but before you take the determinant).  Thus if $W = $ the coefficient matrix, ${\bf u} = n \times 1$ vector with ith coordinate $u_i$ and ${\bf b}  = n \times 1$ vector where all coordinates are 0 except the last term is ${g(t) \over a}$, then  the matrix formula can be written as
\t

\centerline{$W{\bf u}' = {\bf b}$}

Don't forget to integrate to find $u_i$ after using Cramer's rule to find $u_i'$.


Note:  If you are likely to forget to divide by $a$, you might want to do so in the beginning (after solving homogeneous).  This is what the book does:


\centerline{ $y''' + {b \over a}y'' + {c\over a}y' + {e\over a}y= {g(t)\over a}$}




\u\u\u\u

\eject

\def\v{\vskip 10pt}
Solve using Cramer's rule

\centerline{$ \left(\matrix{
e^{-t} & e^{2t} & e^{3t} \cr
-e^{-t} & 2e^{2t} & 3e^{3t} \cr
e^{-t} & 4e^{2t} & 9e^{3t}
}\right)
 \left(\matrix{ u_1'\cr u_2' \cr u_3'}\right) = \left(\matrix{0 \cr 0 \cr {8e^{4t} \over 1}}\right)$}

\hr

Per class, $W(e^{-t}, e^{2t}, e^{3t}) =  12 e^{4t}$

\hr

 $u_1' = 
\displaystyle{
\left| 
\matrix{
0 & e^{2t} & e^{3t} \cr
0 & 2e^{2t} & 3e^{3t} \cr
8 e^{4t} & 4e^{2t} & 9e^{3t}
}
 \right|
 \over
12 e^{4t}   }
= 
\displaystyle{8 e^{4t}
\left| 
\matrix{
0 & e^{2t} & e^{3t} \cr
0 & 2e^{2t} & 3e^{3t} \cr
1 & 4e^{2t} & 9e^{3t}
}
 \right|
 \over
12 e^{4t}   }
= 
\displaystyle{8 e^{4t}(0 - 0 + 1
\left| 
\matrix{
e^{2t} & e^{3t} \cr
 2e^{2t} & 3e^{3t} \cr
}
 \right|)
 \over
12 e^{4t}   }
$

$~~~~= 
\displaystyle{8 e^{4t}(3e^{5t} - 2e^{5t})
 \over
12 e^{4t}   }
= 
\displaystyle{2 (e^{5t})
 \over
3  }
= 
\displaystyle{2 e^{5t}
 \over
3  }
$



Thus $u_1 = \int u_1' dt = \int {2 e^{5t}
 \over
3  }dt = {2 e^{5t}
 \over
15 }$
\v

Note:  don't need $+C$ since only need to find one solution for $u_1$.

\hr

Thus $u_2' = 
\displaystyle{
\left| 
\matrix{
e^{-t} &0 & e^{3t} \cr
-e^{-t} &0& 3e^{3t} \cr
e^{-t} &8 e^{4t} & 9e^{3t}
}
 \right|
 \over
12 e^{4t}   }= 
\displaystyle{8 e^{4t} 
\left| 
\matrix{
e^{-t} &0 & e^{3t} \cr
-e^{-t} &0& 3e^{3t} \cr
e^{-t} &1& 9e^{3t}
}
 \right|
 \over
12 e^{4t}   }
= 
\displaystyle{2  (-0 + 0 -1
\left| 
\matrix{
e^{-t}  & e^{3t} \cr
-e^{-t} & 3e^{3t} \cr
}
 \right|)
 \over
3   }
$

$~~~~= 
\displaystyle{-2(3e^{2t} + e^{2t})
 \over
3  }
= 
\displaystyle{-2 (4e^{2t})
 \over
3  }
= -
\displaystyle{8 e^{2t}
 \over
3  }
$



Thus $u_2 = \int u_2' dt = \int (-{8 e^{2t}
 \over
3  })dt = -{4 e^{2t}
 \over
3 }$
\v

\hr

Thus $u_3' = 
\displaystyle{
\left| 
\matrix{
e^{-t} & e^{2t} &0 \cr
-e^{-t} & 2e^{2t} & 0 \cr
e^{-t} & 4e^{2t} & 8 e^{4t}
}
 \right|
 \over
12 e^{4t}   }= 
\displaystyle{ 8 e^{4t}
\left| 
\matrix{
e^{-t} & e^{2t} &0 \cr
-e^{-t} & 2e^{2t} & 0 \cr
e^{-t} & 4e^{2t} & 1
}
 \right|
 \over
12 e^{4t}   }
= 
\displaystyle{ 2(0 - 0 + 1
\left| 
\matrix{
e^{-t} & e^{2t}  \cr
-e^{-t} & 2e^{2t}  \cr
}
 \right|)
 \over
3  }
$

$~~~~= 
\displaystyle{(2e^{t} +e^{t})
 \over
3  }
= 
\displaystyle{2 (3e^{t})
 \over
3  }
= 
\displaystyle{2 e^{t}
  }
$



Thus $u_3 = \int u_3' dt = \int 2 e^{t}dt = {2 e^{t}
}$
\v
\hr

General homogeneous solution:  $y = c_1e^{-t}+ c_2 e^{2t}+ c_3 e^{3t}$

A nonhomogeneous solution:  $y = u_1e^{-t}+ u_2 e^{2t}+ u_3 e^{3t}$

A nonhomogeneous solution:  $y =  ({2 e^{5t}
 \over
15 })e^{-t}+ ( -{4 e^{2t}
 \over
3 } ) e^{2t}+ (2 e^{t}) e^{3t} = {4 \over 5}{e^{4t}}$

Thus the general solution is
$y = c_1e^{-t}+ c_2 e^{2t}+ c_3 e^{3t}+{4 \over 5}{e^{4t}} $



\end