D Module - Second Order ODEs

Question 1

Theorem: Solutions to a second order homogenous ODE

Answer 1

If y1(t) and y2(t) are solutions to this ODE and aren’t constant multiples of each other, all other solutions can be written as:

y(t) = c1y1(t) + c2y2(t)

for some coefficients c1 and c2. For every choice of 2 coefficients, we get a solution.

Question 2

Theorem: Solutions to a second order homogenous ODE: Characteristic Equation

Answer 2

Consider a second order homogenous equation with condtant coefficients: ay’‘+by’+cy = 0. The characteristic equation is ar^2 + br + c = 0.

1. If b^2 - 4ac = 0, the characteristic equation has only one root and the general solution is of the form:
y(t) = c1e^rt + c2te^rt
where y1(t) = c1e^rt and y2(t) = c2te^rt

2. If b^2 - 4ac > 0, the characteristic equation has 2 distinct real roots and the general solution is of the form:
y(t) = c1e^r1t + c2e^r2t
where y1(t) = c1e^r2t and y2(t) = c2e^r2t

3. If b^2 - 4ac < 0, the characteristic equation has no distinct real roots and must take one of the r values and use Euler’s Formula: e^(ix) = cos(x) + isin(x). The general solution will be of the form:

y(t) = e^xt(c1cos(i) + c2sin(i))
where x = the real part of r and i = the imaginary part of r

Question 3

Theorem: Solutions to a second order non-homogenous ODE

Answer 3

Consider ay’’ + by’ + c = f(t)

The complimentary solution yc(t) is the solution to the homogenous version of the ODE. It is the solution to:
ay’’ + by’ + c = 0

The particular solution yp(t) is the solution to the non-homogenous version of the ODE. It is the solution to:
ay’’ + by’ + c = 0

Then the general solution y(t) = yc(t) + yp(t)

Question 4

Strategies: Solving yp(t)

Answer 4

Consider ay’’ + by’ + c = f(t)

To solve yp(t), we must “guess” a solution for f(t), similarly to how we guess y = e^rt when solving homogenous equations.

1. Try to guess something that is as similar to it’s own derivatives as possible. But DO NOT only consider this
2. REMEMBER to consider the derivatives of the guess. The guess must cancel out it’s own derivative.
3. The guess CANNOT be apart of the complimentary solution. If it is, multiply the guess by x(or t).