Study Guide: Second Order Linear Homogeneous Equations with Constant Coefficients

Posted on April 29, 2015 by | Leave a comment

What is Second Order Linear Homogeneous Equations with Constants Coefficients?

A second order homogeneous equation with constant coefficients is written as:

Ay''+By'+Cy=0 where A, B, C are constant and A\neq0

In order to solve these equation, we would convert the differential equation into a characteristic equation which is something like this:

Ar^2+Br+C=0

Then we use the quadratic equation to find the roots for the equation:

r_{1,2}= \frac{-B\pm\sqrt{B^2-4AC}}{2A}

There are three cases will be distinguish by using the discriminant (D=B^2-4AC), after getting the discriminant then we apply the roots in to their own general solution, such that:

  1. D>0 (there will be two distinct real roots, r_1 and r_2)
    y(t)= K_1e^{r_1t} + K_2e^{r_2t}
  2. D=0 (there will be one repeated root, r_1=r_2)
    y(t)= K_1e^{r_1t} + K_2te^{r_2t}
  3. D<0 (there will be two complex roots, r_{1,2}=\alpha \pm i\beta)
    y(t)= K_1e^{\alpha t}\cos(\beta t) + K_2e^{\alpha t}\sin(\beta t)

Sample Problem:
Find the solution for the following equation:

y''-6y'+9y=0, y(o)=-5, y'(0)=-4

Step 1: Converting the differential equation into characteristic equation and distinguish A, B and C.

r^2-6r+9y=0
where A=1,B=-6,C=9

Step 2: Distinguish which case will the equation be with the discriminant.

D=(-6)^2-4(1)(9)
D=36-36
D=0

In this case, we will be looking into an equation with a repeated root. So our general solution will look something like this:

y(t)= K_1e^{r_1t} + K_2te^{r_2t}

Step 3: Using the quadratic equation to solve for r_{1,2} or you can solve it by factoring if it’s possible(in this case factoring is possible).
By factoring:

r^2-6r+9y=0
(r-3)(r-3)=0
r_1=r_2=3

By quadratic equation:

r_{1,2}= \frac{-(-6)\pm\sqrt{(-6)^2-4(1)(9)}}{2(1)}

r_{1,2}= \frac{6\pm\sqrt{36-36}}{2}

r_{1,2}= \frac{6}{2}

r_{1,2}= 3

Step 4:After solving for the roots, now we apply the roots into the general solution.

y(t) = K_1e^{3t} + K_2te^{3t}

Step 5:Then we solve for the constant(K_1, K_2)using the initial values that are given to us.

y(o)=-5, y'(0)=-4

y(t) = K_1e^{3t} + K_2te^{3t}

Also we need to find the first derivatives of the equation.

y'(t) = 3K_1e^{3t} + K_2e^{3t} + 3K_2te^{3t}

By using the initial values we will then able to solve for K_1,K_2.

y(0) = K_1e^{3(0)} + K_2(0)e^{3(0)}=-5
K_1=-5

y'(0) = 3(-5)e^{3(0)} + K_2e^{3(0)} + 3K_2(0)e^{3(0)}=-4
K_2=11

Lastly we apply the constant back into the equation and we will have the final solution for this problem.

y(t) = -5e^{3t} + 11te^{3t}

Videos for reference:
Here are some videos that I think will be useful to look at.
Introduction of Second Order Linear Homogeneous Equations with Constant Coefficients:

Second Order Linear Homogeneous Equations with Constant Coefficients with two distinct real roots:

Second Order Linear Homogeneous Equations with Constant Coefficients with repeated roots:

Second Order Linear Homogeneous Equations with Constant Coefficients with two complex roots:

*I don’t own these videos, all credits goes to the you-tuber Houston Math Prep.*

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