Solve ay′′ + by′ + cy = g(t) , y(0) = y_{0}, y′(0) = y_{1} Step 1: Solve homogeneous eqn ay′′+by′+cy = 0 (**)

Step 1a: Guess solution to (**): Suppose y = e^{rt} y = e^{rt} implies y′ = re^{rt} and y′′ = r^{2}e^{rt} ar^{2}e^{rt} + bre^{rt} + ce^{rt} = 0 implies ar^{2} + br + c = 0, Thus r = \frac{−b±\sqrt{b^{2}−4ac}}{2a}. Let r_{1} = \frac{−b+\sqrt{b^{2}−4ac}}{2a} and r_{2} = \frac{−b−\sqrt{b^{2}−4ac}}{2a}. Thus y = e^{r_{1}t} and y = e^{r_{2}t} are both solutions to (**) We will assume r_{1} ≠ r_{2} so that we have two different solutions. Step 1b: Find general soln to homogeneous eqn (**) Note that (**) is a linear equation. Thus since y = e^{r_{1}t} and y = e^{r_{2}t} are both solutions to (**), y = c_{1}e^{r_{1}t} + c_{2}e^{r_{2}t} is the general solution to (**).

Step 2: Solve ay′′ + by′ + cy = g(t) (*) Step 2a: Guess solution to (*). Step 2b: Use thm below to form general soln to (*). Thm: Suppose c_{1}ϕ_{1}(t) + c_{2}ϕ_{2}(t) is a general solution to

ay′′ + by′ + cy = 0,

If y(t) = ψ(t) is a solution to

ay′′ + by′ + cy = g(t) [*],

Then ψ(t) + c_{1}ϕ_{1}(t) + c_{2}ϕ_{2}(t) is also a solution to [*]. Step 3: If initial value problem: Once general solution is known, can solve initial value problem (i.e., use initial conditions to find c_{1}, c_{2}): General solution: y(t) = ψ(t) + c_{1}ϕ_{1}(t) + c_{2}ϕ_{2}(t) Initial conditions: y(0) = y_{0}, y′(0) = y_{1} Solve the following system of eqns for c_{1} and c_{2}: y_{0} = ψ(0) + c_{1}ϕ_{1}(0) + c_{2}ϕ_{2}(0) y_{1} = ψ′(0) + c_{1}ϕ′_{1}(0) + c_{2}ϕ′_{2}(0)