Tag Archives: Homogeneous Equations with Constant Coefficients

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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Complex Roots(of the Characteristic Equation)

Posted on April 26, 2015 by | Leave a comment

Complex Roots relate to the topic of Second order Linear Homogeneous equations with constant coefficients. The Second Order linear refers to the equation having the setup formula of y”+p(t)y’ + q(t)y = g(t). Constant coefficients  are the values in front of the derivatives of y and y itself. Homogeneous means the equation is equal to zero.So a homogeneous equation would look like

y”+by’ + cy = 0 or y”+p(t)y’ + q(t)y = 0.

When solving the general solution of the homogeneous equation, we have to look for the characteristic equation which is r^2 + br + c = 0. But how would you get there? To start we are given the differential equation of

y = e^rt

which can be used to look for the characteristic equation. To start we look for the derivative of the differential equation then replace y and its derivatives with what we found which the homogeneous equation would look like

(r^2)e^rt + bre^rt + ce^rt = 0

You can see there is a common value which is e^rt you can factor out. Once you factor it out, you can divide e^rt and this is the characteristic equation,

r^2 + br + c = 0

The general solution of the homogeneous equation will have 3 different results depending on the roots which are found from the characteristic equation. This is where Complex Roots come into play. Complex Roots is when the roots have an imaginary number or i which looks like r =  a ± i*b. The general solution will be

 y = C1e^(a + ib)t + C2e^(a – ib)t

But there is a problem when you want to solve for a particular solution how would you get rid of the i? So the general solution is still not done. With the general solution, y = C1e^(a + ib)t + C2e^(a – ib)t you can right away separate the roots which will become y = C1e^at * e^(ibt) + C2e^at * e^( –ibt). Next you factor out e^a which makes the equation y = e^at(C1e^(ibt) + C2e^(-ibt). This is where the Euler’s Formula is introduced which says that

e^it = cos(t) + isin(t)

So the equation is written as such

y = e^at(C1(cos(bt) + isin(bt)) + C2(cos(-bt) + isin(-bt)))

This is where your knowledge of trig identities comes to play. Two trig identities which can be used are

sin(-x) = -sin(x), cos(-x) = -cos(x)

This would replace the negative values in the sin and cos but we are left with the complex part i. To get rid of the complex part, the equation is rearranged so that the constants can be created into a new constant that will work with real and imaginary numbers.

y = e^at((C1 + C2)cos(bt) + (C1i – C2i)isin(bt)))

y = e^at(Acos(bt) + Bsin(bt)), general solution

Sample Problem:                                                                                                                                y” + 4y’ + 5y = 0, y(0) = 1, y'(0) = 0                                                                                      y=e^(rt), y’ = re^(rt), y” = (r^2)e^(rt)                                                                                     e^(rt)*(r^2 + 4r +5) = 0

Since the roots are the easy to separate, we use the quadratic formula to look for the roots which you will end up with                                                                                            r = -2±i

The General Solution will equal to

y = e^(-2t)*(Acos(t) + Bsin(t))

For the particular solution, we solve for A and B using y(0) = 1 and y'(0) = 0.   y(0) = 1                                                                                                                                                      1 = e^(-2(0))*(Acos(0) + Bsin(0))                                                                                                1 = A

y'(0) = 0                                                                                                                                                  y’ = -2Ae^(-2t)cos(t) – Ae^(-2t)sin(t) – 2Be^(-2t)sin(t) + Be^(-2t)cos(t)                    0 = -2(1)e^(-2(0))cos(0) – (1)e^(-2(0))sin(0) – 2Be^(-2(0))sin(0) + Be^(-2(0))cos(0)                                                                                                                                 0 = -2 + B                                                                                                                                                B = 2

The Particular Solution:

y = e^(-2t)*(cos(t) + 2sin(t))

Videos to help with Studies:

Complex Roots Part 1

Complex Roots Part 2

Complex Roots Part 3

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