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\underbar{Preliminary version:}
\vskip -5pt
{\it Thm 6.2.1:  }
Hint:  use integration by parts and let $dv = f'(t)$


%%{\it Thm 6.3.1}

%%Hint:  $ \int_0^\infty  h(t)dt =  \int_0^c h(t)dt +  \int_c^\infty h(t)dt$ and use %%$u$-substitution 
%%(let $u = t-c$).

{\it Thm 6.3.2:  }
 Hint: Let $F(s) = {\cal L}(f(t)) = ......$.  To calculate $F(s-c)$ 
evaluate $F(s)$ at $s-c$  (i.e. replace $s$ with $s-c$).
 Use definition of Laplace transform 
to evaluate 


{\it Thm:}  The Laplace transform is a linear operator.
\hfil \break
Hint:  ${\cal   L}(af(t) + bg(t)) = ...$  OR 
[ ${\cal   L}(af(t)) = ...$  and
 ${\cal   L}(f(t) + g(t)) = ...$]


{\it Thm:}  $L(f) = af'' + bf' + cf$ is linear.
\hfil \break
Hint:  ${  L}(k_1f(t) + k_2g(t)) = ...$  OR 
[ ${ L}(kf(t)) = ...$  and
 ${L}(f(t) + g(t)) = ...$]

{\it Thm 3.2.2:}  If $y = \phi_1(t)$ and $y = \phi_2(t)$ are solutions to $y'' + p(t)y + q(t)y = 0$, then 
$y = c_1\phi_1(t) +  c_2\phi_2(t)$ is also a solution to this homogeneous equation.

{\it Thm 7.4.1:}  If $\overrightarrow{y} = \overrightarrow{\phi_1}(t)$ and $\overrightarrow{y} = 
\overrightarrow{\phi_2}(t)$ are solutions to $\overrightarrow{y}' = P(t)\overrightarrow{y}$, then 
$\overrightarrow{y} = c_1\overrightarrow{\phi_1}(t) +  c_2\overrightarrow{\phi_2}(t)$ is also a solution 
to this homogeneous equation.

{\it Thm 3.5.1:}  If $y = \psi_1(t)$ and $y = \psi_2(t)$ are solutions to $y'' + p(t)y + q(t)y = g(t)$, 
then $y = \psi_2(t) - \psi_1(t)$ is  a solution to the homogeneous equation
$y'' + p(t)y + q(t)y = 0$.

Note 3.5.1 plus previous thms imply $\psi_2(t) - \psi_1(t) = c_1\phi_1 + c_2\phi_2$ for some constants 
$c_1$, $c_2$ where $\phi_1, \phi_2$ are 2 linearly independent solutions to the homogeneous equation.

Hence $\psi_2(t)  = c_1\phi_1 + c_2\phi_2 + \psi_1(t)$.  I.e., if you can find the general homogeneous 
solution ($c_1\phi_1(t) +  c_2\phi_2(t)$) and one non-homogeneous, then you can find all non-homogeneous 
solutions to $y'' + p(t)y + q(t)y = g(t)$.  I.e., this non-homogeneous equation has general solution: $y = 
c_1\phi_1 + c_2\phi_2 + \psi_1(t)$.





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