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Homogeneous Equations with Constant Coefficients (Second Order Linear)

3.1
By definition, y^{\prime\prime}+p(t)y^{\prime}+q(t)y=g(t)
A second order linear equation has constant coefficients if the functions p(t), q(t) and g(t) are constant functions. It is said to be homogeneous if g(t) =0.

In order to solve a second order linear equation, the best way is to translate the given differential equation into a characteristic equation as follows:

y^{\prime\prime}+p(t)y^{\prime}+q(t)y=g(t) \rightarrow r^2+br+c=0 (quadratic equation)

Once we get this quadratic equation, if we can solve it for r, then those roots will make up the solution for the differential equation.
This answer is represented as y=e^{rt}

The General Solution is: y=Ae^{r_1t}+Be^{r_2t}
where r_1 and r_2 are different roots from the characteristic equation.

If the solution of the characteristic equation has a repeated root r, then the general solution is:
y=Ae^{rt}+Bte^{rt}

Example:
Differential equation:
y^{\prime\prime}+5y^{\prime}+14y=0

Characteristic Equation:
r^2+5r-14=0
(r+7)(r-2)=0

Roots:
r_1=-7, r_2=2

General Solution:

y=Ae^{-7t}+Be^{2t}

Example with Repeated Roots:

Differential equation:

y^{\prime\prime}-6y^{\prime}+9y=0

Characteristic Equation:
r^2-6r+9=0
(r-3)(r-3)=0

Roots: r=3

General Solution:
Ae^{3t}+Bte^{3t}

When initial conditions are given, then a Particular Solution can be solved for.

Example:
y^{\prime\prime}-6y^{\prime}+8y=0, y(0)=-7, y^{\prime}(0)=-30
r^2-6r+8=0
(r-4)(r-2)=0
r_1=4, r_2=2

General Solution:

y=Ae^{4t}+Be^{2t}

In order to solve for the given conditions we have to plug in 0 for t, and the equation equal to -7. Keep in mind that the second condition indicates that it has to be applied to y’ for which we have to find the derivative of y.

y=Ae^{4t}+Be^{2t}
y^{\prime}=4Ae^{4t}+2Be^{2t}

Once this is found, we can plug in for the initial conditions:

\textbf{ y(0)=-7}, y=Ae^{4t}+Be^{2t}
\textbf{-7} =Ae^{4\textbf{(0)}}+Be^{2\textbf{(0)}}
\underline{-7=A+B}

y^{\prime}\textbf{ (0)=-30} y^{\prime}=4Ae^{4t}+2Be^{2t}
\textbf{-30} =4Ae^{4\textbf{(0)}}+2Be^{2\textbf{(0)}}
\underline{-30=4A+2B}

With these two equations, we can find the coefficients A and B, by using the elimination method or by solving for one variable and plugging into the other equation:

-7=A+B
-30=4A+2B

A=-7-B

-30=4(\textbf{-7-B})+2B
-30=-28-4B+2B
-30+28=-2B

\textbf{B=1}

then A=-7- \textbf{(1)}

\textbf{A=-8}

Particular Solution:

y=-8e^{4t}+1e^{2t}

Example of Particular Solution with Repeated Roots:

y^{\prime\prime}-18y^{\prime}+81y=0, y(0)=4, y^{\prime}(0)=39
r^2-18r+81=0
(r-9)(r-9)=0

r=9

General Solution:

y=Ae^{9t}+Bte^{9t}

Same steps from the previous example but keep an eye on the derivative of y. The product Rule must be applied.

y=Ae^{9t}+Bte^{9t}
y^{\prime}=9Ae^{9t}+9Bte^{9t}+Be^{9t}

Evaluate with initial conditions:
\textbf{y(0)=4}, y=Ae^{9t}+Bte^{9t}
\textbf{4}=Ae^{9\textbf{(0)}}+9Bte^{9\textbf{(0)}}
\underline{4=A}

y^{\prime} \textbf{(0)=39}, y^{\prime}=9A+9Bte^{9t}+Be^{9t}
\textbf{39}=9A+9B\textbf{(0)}e^{9\textbf{(0)}}+Be^{9\textbf{(0)}}
\underline{39=9A+B}

Find A and B:

\underline{4=A}
\underline{39=9A+B}
39=9\textbf{(4)}+B
39=36+B

\textbf{A=4}
\textbf{B=3}

Particular Solution:
y=4e^{9t}+3te^{9t}

Very useful videos:

I found this video very helpful when trying to solve this type of differential equation. His method of explaining is very clear and recommend you guys to check out more of his videos when solving math problems. Including Calculus I and II.

https://www.youtube.com/watch?v=soU-zRdpsoA&spfreload=10

Khan Academy has always been a great source when providing detailed information. On this video we can find many techniques on how to solve this equation.

https://www.youtube.com/watch?v=UFWAu8Ptth0
https://www.youtube.com/watch?v=SPVqgkOZMAc&spfreload=10

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Posted on April 29, 2015 by in Study Guide

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Second order ordinary homogeneous differential equations with constant coefficients

Posted on March 17, 2015 by | Leave a comment

Here are some video resources showing complete examples (from Khan Academy).  I hope they help!

-Prof. Reitz

Example 1 — in which the characteristic equation has two distinct real roots.

 

Example 2 — in which the characteristic equation has one repeated (real) root.

 

Example 3 — in which the characteristic equation has complex roots.

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Posted in Resources

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