Scalar Form of Gen Solution for Second-Order Lin Hom Equations

Essentially, finding the Scalar Form of the General Solution for Second-Order, Linear, Homogeneous Equations is similar to guessing the right solution. It is more formulaic and requires less work than does finding the Vector Form of the General Solution. 

In case you do not understand what we mean when we distinguish the "Scalar Form" from the "Vector Form," we will provide an explanation.
If you viewed our other page, General Solution for Second-Order, Linear, Homogeneous Equations you would see that the "Vector Form" of the solution requires that you include the eigenvectors as part of the solution. However, for the "Scalar Form" you do not have to include eigenvectors. 

Take for example the following solution:
     `bb(Y)(t) = c_(1)e^(-5t) ((1),(-5)) + c_(2)e^(-2t) ((1),(-2))`

This is considered the "Vector Form" of the General Solution, and if you are tested on Differential Equations it is more likely to be the form of a solution that an examiner is looking for.
     `bb(Y)(t)=c_(1)e^(-5t) + c_(2)e^(-2t)` 

Above is the "Scalar Form" of the General Solution and if you are looking for a formulaic way of finding solutions to Differential Equations, it is most likely what you are looking for. 

In the equation above, we are taking the Second-Order Linear, Homogeneous Equation that we are given and we are formulating the characteristic polynomial, the roots of the characteristic polynomial (eigenvalues), and the general solution without converting the Second-Order Equation into a system of First-Order Equations as we did on the other page. 

Just as we saw when classifying harmonic oscillators with damping, we know the following given a harmonic oscillator equation of the form:
     `m frac(d^(2) y)(d t^(2))+ b frac(dy)(dt) + ky =0`

• The characteristic polynomial is `ms^2 +bs +k= 0.`
• The roots of the characteristic polynomial are `frac( -b pm sqrt(b^(2)- 4mk))(2m).`
  • If `b=0` then the harmonic oscillator is said to be undamped, and the eigenvalues are `pm i sqrt(k/m)` (`sqrt(k/m)` is often referred to as `omega`).
  • If `b^(2)-4km < 0` there are complex eigenvalues and the harmonic oscillator is said to be underdamped.
  • If `b^(2)-4km > 0` there are real eigenvalues and the harmonic oscillator is said to be overdamped.
  • If `b^(2)-4km = 0` there are repeated eigenvalues and the harmonic oscillator is said to be critically damped.

For an undamped harmonic oscillator, the scalar form of the general solution is
     `bb(Y)(t) = c_(1)cos( omega t) + c_(2) sin( omega t)`
where `omega = sqrt(k/m).`

For an underdamped harmonic oscillator the form of the general solution is
     `bb(Y)(t) = c_(1)e^((-b/2) t) cos( sqrt(abs(b^2-4mk))/2 t) + c_(2)e^((-b/2) t) sin( sqrt(abs(b^2-4mk))/2  t)`
where there is a complex eigenvalue, and the form of the solution is based on Euler's formula.

For an overdamped harmonic oscillator the form of the general solution is
     `bb(Y)(t) = c_(1)e^(lambda_(1)) + c_(2)e^(lambda_(2))`
where there are two real eigenvalues. 

For a critically damped harmonic oscillator the form of the general solution is
     `bb(Y)(t)= c_(1)e^(lambda) + c_(2) t e^(lambda)`
where there is one repeated eigenvalue.
 

Sources

Blanchard, Paul, Robert L. Devaney, and Glen R. Hall. Differential Equations. 3rd ed. Belmont, CA: Thomson Brooks/Cole, 2006. Print.