Tag Archives: Second Order Linear

Second order linear homogeneous equations with constant coefficients

Posted on April 30, 2015 by | 1 comment

You have probably seen first order linear equations. Now we”ll move to second order linear equations. What makes it linear? A differential equation is linear if it is a linear function of the variables y, y’, y” and so on.

The standard form of the second order linear equation is

y” + p(t)y’ + q(t)y = g(t)

where p(t), q(t), and g(t) are constant coefficients.
There can also be a constant coefficient in front of the y”.
When g(t) = 0, we can call the differential equation homogeneous, otherwise it is a non-homogeneous.
We can write the second order linear homogeneous equation can be written as:

ay”+by’+cy=0

From that, we can write that characteristic equation:

ar^2+br+c=0

From the characteristic equation, we can find the roots. The roots can come out to be real roots, repeated roots, or complex roots. But we will only focus on real roots.
How do we know if the roots are real roots? We can find out by find out the discriminant. To find out the discriminant, we use b^2-4ac. For the roots to be real, it must follow this condition:

b^2-4ac > 0

After finding out the roots, we can plug it into the general solution which is:

y(t)=C1e^(r1*t)+C2e^(r2*t)

Now we can figure out the derivative of the general solution which we’ll use later.

y'(t)=(r1)C1e^(r1*t)+(r2)C2e^(r2*t))

With problems like these, we will have initial condition, y(0) and y'(0), which will be used to find C1 and C2 to determine the particular solution.
You plug in the initial to determine C1 and C2.
Once you determined C1 and C2, plug them back into the general solution to get the particular solution satisfying the initial condition.
The initial conditions gives us a unique solution.
Example 1:      y” + 5y’ + 6y = 0         y(0) = 2     y'(0) = 3
We start out with the characteristic equation which is:

r^2 + 5r + 5 = 0

Now find the roots anyway you feel comfortable with.
One way is by using the quadratic formula.
Now we got

r1 = (-2)
r2 = (-3)

We can plug it into the general solution

y(t)=C1e^(-2t)+C2e^(-3t)

Now that we have to figure out the derivative of general solution which will be:

y'(t)=(-2)C1e^(-2t)-3C2e^(-3t)

Now we can plug in the initial values which are y(0) = 2     y'(0) = 3
For the initial condition y(0) = 2:

2=C1e^(-2*0)+C2e^(-3*0)
2=C1+C2

For the initial condition y'(0) = 3:

3=(-2)C1e^(-2*0)-3C2e^(-3*0)
3=-2C1-3C2

Now that we have 2 equations, we can use them to solve for C1 and C2:

2=C1+C2
3=-2C1-3C2

Multiply the first equation by 2 so the C1 can cancel out.

4=2C1+2C2
3=-2C1-3C2
7=-1C2
C2=(-7)

Now that we know C2, we can plug it back into any of the two equations to find C1.

2=(-7)+C1
C1=9

Lastly we plug the C1 and C2 into the general solution to determine the particular solution.

y(t)=9e^(-2t)-7e^(-3t)

Videos to help with Second order linear homogeneous equations with constant coefficients:
This video is a transition from first order linear differential equations to second order linear differential equations.


This video goes into more detail when the roots are repeated or complex. It only goes up the general solution


In this video, he does a whole example using initial conditions.

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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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