Preliminary version:

Thm 6.2.1: Hint: use integration by parts and let dv = f′(t)

Thm 6.3.2: Hint: Let F (s) = 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

Thm: The Laplace transform is a linear operator. Hint: L(af(t) + bg(t)) = ... OR [ L(af(t)) = ... and L(f(t) + g(t)) = ...]

Thm: L(f) = af′′ + bf′ + cf is linear. Hint: L(k_{1}f(t) + k_{2}g(t)) = ... OR [ L(kf(t)) = ... and L(f(t) + g(t)) = ...]

Thm 3.2.2: If y = ϕ_{1}(t) and y = ϕ_{2}(t) are solutions to y′′ + p(t)y + q(t)y = 0, then y = c_{1}ϕ_{1}(t) + c_{2}ϕ_{2}(t) is also a solution to this homogeneous equation.

Thm 7.4.1: If \overrightarrow{y} = \overrightarrow{ϕ_{1}}(t) and \overrightarrow{y} = \overrightarrow{ϕ_{2}}(t) are solutions to \overrightarrow{y}′ = P (t)\overrightarrow{y}, then \overrightarrow{y} = c_{1}\overrightarrow{ϕ_{1}}(t) + c_{2}\overrightarrow{ϕ_{2}}(t) is also a solution to this homogeneous equation.

Thm 3.5.1: If y = ψ_{1}(t) and y = ψ_{2}(t) are solutions to y′′ + p(t)y + q(t)y = g(t), then y = ψ_{2}(t) − ψ_{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 ψ_{2}(t) − ψ_{1}(t) = c_{1}ϕ_{1} + c_{2}ϕ_{2} for some constants c_{1}, c_{2} where ϕ_{1}, ϕ_{2} are 2 linearly independent solutions to the homogeneous equation.

Hence ψ_{2}(t) = c_{1}ϕ_{1} + c_{2}ϕ_{2} + ψ_{1}(t). I.e., if you can find the general homogeneous solution (c_{1}ϕ_{1}(t) + c_{2}ϕ_{2}(t)) and one nonhomogeneous, 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}ϕ_{1} + c_{2}ϕ_{2} + ψ_{1}(t).