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Overview

Second order linear equations become homogeneous when the linear function of y and y’ (which can be written in the form y” + p(t)y’ + q(t)y = g(t)) is equal to zero. In other words when g(t)=0.

This guide will be discussing how to solve homogeneous linear second order differential equation with constant coefficient, which is written in the following form:

y”+by’+cy = 0

The first step is to use the equation above to turn the differential equation into a characteristic    equation. The characteristic equation is written in the following form:

r² +br+c = 0

Second to find the roots, or r₁ and r₂ you can either factor or use the quadratic formula:

 r² = ± -b √b²-4ac

 2a

It is important to remember when to the particular equation above. There will problems where the variable a is not needed in the quadratic formula because there will be no a in the differential equation. In the cases where there is no a variable limit the a variable from the quadratic equation. The quadratic equation will the look like the following:

r² = ± -b √b²-4c

2

Once you get repeated roots , or r₁ and r₂ from the characteristic equation then     y = e^rt is considered a solution of the differential equation.

The next step would be to plug r₁ and r₂ into the general equation:

y = C₁ e^r1t +C₂ e^r2t

Another important thing to realize and remember is that when solving a homogeneous equation for a repeated root the solution will end up cancelling out. In order to avoid this a “t” needs to be place in the general solution. The general solution for the repeated root will then be in the following form:

y = C₁ e^r1t +C₂t e^r2t

Sample Problem

Problem 1:

y”+12y’+36 = 0

Step 1: Turn the differential equation into a characteristic equation

r² + br + c = 0

r² + 12r + 36 = 0

Step 2: Factor the characteristic equation

r² + 12r + 36 = 0

(r + 6) (r + 6) = 0

r₁ = -6 r₂ = -6

Step 3: Use y = e^rt as a solution for r₁ and r₂

y = e^rt

y = e^-6t

Step 4: Plug r₁ and r₂ into the general solution for the repeated roots

y = C₁ e^r1t +C₂t e^r2t

y = e^rt

y = e^-4t

Step 4: Plug r₁ and r₂ into the general solution for the repeated roots

y = C₁ e^r1t +C₂t e^r2t

y = C₁ e^-4t +C₂t e^-4t

y = C₁ e^-6t +C₂t e^-6t

 

Problem 2:

y”+8y’+16 = 0 , y(0) = 2   y’(0)= 6

Step 1: Turn the differential equation into a characteristic equation

r² + br + c = 0

r² + 8r + 16 = 0

Step 2: Factor the characteristic equation

r² + 8r + 16 = 0

(r + 4) (r + 4) = 0

r₁ = -4 r₂ = -4

Step 3: Use y = e^rt as a solution for r₁ and r₂

y = e^rt

y = e^-4t

Step 4: Plug r₁ and r₂ into the general solution for the repeated roots

y = C₁ e^r1t +C₂t e^r2t

y = C₁ e^-4t +C₂t e^-4t

The following three videos are form Khan Academy. I find these videos ever useful . I would recommend you watch them if you are still confused  about Repeated Roots of Second order linear homogeneous equation.