The discussion of second order linear equations is broken into two main areas based on whether the equation is homogeneous, \(g(t)=0\) or inhomogeneous, \(g(t)\neq0\) (also called nonhomogeneous). We discuss solutions to homogeneous equations on this page. Solutions to inhomogeneous equations are discussed on separate pages for two techniques, undetermined coefficients and variation of parameters.
classification: secondorder linear 
\(y'' + p(t)y' + q(t)y = g(t)\) 

In the homogeneous case (\(g(t)=0\)) where \(p(t)\) and \(q(t)\) are polynomials, check the pages on CauchyEuler equations and Chebyshev's equations.
Homogeneous Equations
Since working with second order equations builds on techniques as we go, we will first consider homogeneous equations. Our second order equations look like \(y'' + p(t)y' + q(t)y = g(t)\) and when they are homogeneous \(g(t)=0\) giving us \(y'' + p(t)y' + q(t)y = 0 \).
Constant Coefficients
Homogeneous equations with constant coefficients look like \(\displaystyle{ ay'' + by' + cy = 0 }\) where a, b and c are constants. We also require that \( a \neq 0 \) since, if \( a = 0 \) we would no longer have a second order differential equation.
When introducing this topic, textbooks will often just pull out of the air that possible solutions are exponential functions. Unfortunately, at this point, you just need to take their word for it. Oh, you can check that what they say is correct. But where they came from will remain something of a mystery for now.
The idea is to find the roots of the polynomial equation \(ar^2+br+c=0\) where a, b and c are the constants from the above differential equation. This equations is called the characteristic equation of the differential equation. If we call the roots to this polynomial \(r_1\) and \(r_2\), then two solutions to the differential equation are
\( y_1 = c_1 e^{r_1t} \) and \( y_2 = c_2 e^{r_2t} \).
Usually we add these two solutions together and write it as one solution and write \( y = c_1 e^{r_1t} + c_2 e^{r_2t} \). This table summarizes these comments.
\(\displaystyle{ ay'' + by' + cy = 0 }\)  

characteristic equation  \(ar^2+br+c=0\) 
roots  \(r_1\) and \(r_2\), real, distinct 
solution  \(\displaystyle{ y = c_1 e^{r_1t} + c_2 e^{r_2t} }\) 
That's the basic idea. Now, if the roots of the polynomial are complex or repeated, there are slight variations of this idea. But in general, that's how this type of differential equation is solved.
Okay, here are some practice problems where the roots are real and distinct. Working these should get you familiar with the technique before moving on to more complex problems.
Practice  Real, Distinct Roots
Solve these differential equations by determining the general solution. If initial conditions are given, use them to determine the particular solution.
\( y''  9y = 0 \)
Problem Statement 

\( y''  9y = 0 \)
Final Answer 

\( y(t) = c_1e^{3t} + c_2e^{3t} \)
Problem Statement 

\( y''  9y = 0 \)
Solution 

You can ignore the message that comes up near the beginning of the video. It refers to the discussion preceding the video clip that does not affect the solution.
video by Engineer In Training Exam 

Final Answer 

\( y(t) = c_1e^{3t} + c_2e^{3t} \) 
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\( y''  y'  6y = 0 \)
Problem Statement 

\( y''  y'  6y = 0 \)
Final Answer 

\( y(x) = Ae^{2x} + Be^{3x} \)
Problem Statement 

\( y''  y'  6y = 0 \)
Solution 

video by Dr Chris Tisdell 

Final Answer 

\( y(x) = Ae^{2x} + Be^{3x} \) 
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\( y'' + y'  6y = 0 \)
Problem Statement 

\( y'' + y'  6y = 0 \)
Final Answer 

\( y = c_1e^{3t} + c_2e^{2t} \)
Problem Statement 

\( y'' + y'  6y = 0 \)
Solution 

video by PatrickJMT 

Final Answer 

\( y = c_1e^{3t} + c_2e^{2t} \) 
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\( y''  7y' + 10y = 0 \)
Problem Statement 

\( y''  7y' + 10y = 0 \)
Final Answer 

\( y(x) = c_1e^{5x} + c_2e^{2x} \)
Problem Statement 

\( y''  7y' + 10y = 0 \)
Solution 

video by Krista King Math 

Final Answer 

\( y(x) = c_1e^{5x} + c_2e^{2x} \) 
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\( 4y''  4y'  3y = 0 \)
Problem Statement 

\( 4y''  4y'  3y = 0 \)
Final Answer 

\( y(x) = c_1e^{x/2} + c_2e^{3x/2} \)
Problem Statement 

\( 4y''  4y'  3y = 0 \)
Solution 

video by Krista King Math 

Final Answer 

\( y(x) = c_1e^{x/2} + c_2e^{3x/2} \) 
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\( y'' + 3y'  10y = 0 \)
Problem Statement 

\( y'' + 3y'  10y = 0 \)
Solution 

video by Dr Chris Tisdell 

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\( y''  4y'  5y = 0 \); \( y(0)=1, y'(0)=0 \)
Problem Statement 

\( y''  4y'  5y = 0 \); \( y(0)=1, y'(0)=0 \)
Final Answer 

\( y(x) = (1/6)e^{5x} + (5/6)e^{x} \)
Problem Statement 

\( y''  4y'  5y = 0 \); \( y(0)=1, y'(0)=0 \)
Solution 

video by Dr Chris Tisdell 

Final Answer 

\( y(x) = (1/6)e^{5x} + (5/6)e^{x} \) 
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\( y''  18y' + 77y = 0 \); \( y(0)=4, y'(0)=8 \)
Problem Statement 

\( y''  18y' + 77y = 0 \); \( y(0)=4, y'(0)=8 \)
Final Answer 

\( y(x) = 9e^{7x}  5e^{11x} \)
Problem Statement 

\( y''  18y' + 77y = 0 \); \( y(0)=4, y'(0)=8 \)
Solution 

video by Krista King Math 

Final Answer 

\( y(x) = 9e^{7x}  5e^{11x} \) 
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\( 2y''  11y' + 12y = 0 \); \( y(0)=5, y'(0)=15 \)
Problem Statement 

\( 2y''  11y' + 12y = 0 \); \( y(0)=5, y'(0)=15 \)
Final Answer 

\( y(x) = 2e^{3x/2} + 3e^{4x} \)
Problem Statement 

\( 2y''  11y' + 12y = 0 \); \( y(0)=5, y'(0)=15 \)
Solution 

video by Krista King Math 

Final Answer 

\( y(x) = 2e^{3x/2} + 3e^{4x} \) 
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Real Repeated Roots
So, if the roots are real and repeated, we call this root \(r_1 = r_2 = r\). From the above discussion, we have one solution \( y_1 = c_1 e^{rt} \). The second solution is obtained by multiplying the first solution by t to get
\( y_2 = c_2 te^{rt} \). (The reduction of order page contains an explanation of where this comes from.) So the combined solution is \( y = c_1 e^{rt} + c_2 te^{rt} \).
Okay, before we go on to discuss complex roots, let's watch this video discussing these two cases so far. This is a very indepth discussion with great examples using a springmassdamped system. It will be well worth your time to watch it carefully.
video by MIT OCW 

Now let's work some practice problems with real, repeated roots.
Practice  Real, Repeated Roots
Solve these differential equations by determining the general solution. If initial conditions are given, use them to determine the particular solution.
\( y''  4y' + 4y = 0 \)
Problem Statement 

\( y''  4y' + 4y = 0 \)
Final Answer 

\( y(x) = e^{2x}(A+Bx) \)
Problem Statement 

\( y''  4y' + 4y = 0 \)
Solution 

video by Dr Chris Tisdell 

Final Answer 

\( y(x) = e^{2x}(A+Bx) \) 
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\( y'' + 22y' + 121y = 0 \); \( y(0)=2, y'(0)=25 \)
Problem Statement 

\( y'' + 22y' + 121y = 0 \); \( y(0)=2, y'(0)=25 \)
Final Answer 

\( y(x) = e^{11x}(23x) \)
Problem Statement 

\( y'' + 22y' + 121y = 0 \); \( y(0)=2, y'(0)=25 \)
Solution 

video by Krista King Math 

Final Answer 

\( y(x) = e^{11x}(23x) \) 
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\( y'' + 6y' + 9y = 0 \)
Problem Statement 

\( y'' + 6y' + 9y = 0 \)
Final Answer 

\( y(x) = e^{3x}(A+Bx) \)
Problem Statement 

\( y'' + 6y' + 9y = 0 \)
Solution 

video by Dr Chris Tisdell 

Final Answer 

\( y(x) = e^{3x}(A+Bx) \) 
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\( y''  8y' + 16y = 0 \)
Problem Statement 

\( y''  8y' + 16y = 0 \)
Final Answer 

\( y = e^{4t}(c_1+c_2x) \)
Problem Statement 

\( y''  8y' + 16y = 0 \)
Solution 

Here are two videos solving this problem worked by two different instructors.
video by Dr Chris Tisdell 

video by PatrickJMT 

Final Answer 

\( y = e^{4t}(c_1+c_2x) \) 
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\( y'' + 20y' + 100y = 0 \)
Problem Statement 

\( y'' + 20y' + 100y = 0 \)
Final Answer 

\( y(x) = e^{10x}(c_1+c_2x) \)
Problem Statement 

\( y'' + 20y' + 100y = 0 \)
Solution 

video by Krista King Math 

Final Answer 

\( y(x) = e^{10x}(c_1+c_2x) \) 
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Complex Roots
\( e^{\pm i\mu} = \cos(\mu) \pm i \sin(\mu) \) 

When the roots in the above discussion are complex, they are in the form
\(r_1 = \alpha + i\beta \) and \(r_2 = \alpha  i\beta \) and our solution looks like
\(\displaystyle{ y = c_1 e^{(\alpha + i\beta) t} + c_2 e^{(\alpha  i\beta)t} }\). This form is fine but it will look better if we use Eulers Formula to put it into another form.
From infinite series we know that
\(\displaystyle{ e^x = \sum_{n=0}^{\infty}{ \left[ \frac{x^n}{n!} \right] }, ~~~ \infty < x < \infty }\).
This is the Taylor series for \(e^x\) about \(x=0\).
If we let \( x = it \), then this sum becomes
\(\displaystyle{ e^{it} = \sum_{n=0}^{\infty}{\frac{(it)^n}{n!}} = \sum_{n=0}^{\infty}{\frac{(1)^nt^{2n}}{(2n)!}} +
i\sum_{n=1}^{\infty}{\frac{(1)^{n1}t^{2n1}}{(2n1)!}} }\)
Notice that we have separated the real and imaginary parts of the series (see note below). The first series is the Taylor series for \(\cos(t)\) about \(t=0\) and the second series is the Taylor series for \(\sin(t)\) about \(t=0\). So this gives us Euler's Formula
\( e^{it} = \cos(t) + i \sin(t).\)
Now that we have Euler's Formula, we can solve homogeneous equations with constant coefficients when the characteristic equation has complex roots, just as we did when the roots were real and not equal.
Note  When substituting \( x=it \) we have moved from the real domain to the complex plane. For the sake of argument we will assume this jump is valid without proof since this discussion is only meant to give you a feel for Euler's Formula.
Here is a video explaining this in more detail.
video by MIT OCW 

After some manipulation, we can write this solution as \( y = e^{\alpha t}( A\cos(\beta t) + B\sin(\beta t) ) \). This video shows the details. It is a continuation of the springmassdamped system video.
video by MIT OCW 

Okay, let's work some practice problems with complex roots.
Practice  Complex Roots
Solve these differential equations by determining the general solution. If initial conditions are given, use them to determine the particular solution.
\( y''  6y' + 10y = 0 \)
Problem Statement 

\( y''  6y' + 10y = 0 \)
Final Answer 

\( y(t) = e^{3t}(c_1\cos t + c_2\sin t) \)
Problem Statement 

\( y''  6y' + 10y = 0 \)
Solution 

video by Michel vanBiezen 

Final Answer 

\( y(t) = e^{3t}(c_1\cos t + c_2\sin t) \) 
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\( y'' + 4y' + 13y = 0 \)
Problem Statement 

\( y'' + 4y' + 13y = 0 \)
Solution 

video by Michel vanBiezen 

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\( y''  2y' + 5y = 0 \)
Problem Statement 

\( y''  2y' + 5y = 0 \)
Final Answer 

\( y(x) = e^{x} \left[ A\cos(2x) + B\sin(2x) \right] \)
Problem Statement 

\( y''  2y' + 5y = 0 \)
Solution 

video by Dr Chris Tisdell 

Final Answer 

\( y(x) = e^{x} \left[ A\cos(2x) + B\sin(2x) \right] \) 
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\( y''  6y' + 13y = 0 \)
Problem Statement 

\( y''  6y' + 13y = 0 \)
Final Answer 

\( y = e^{3x}(c_1\cos(2x) + c_2\sin(2x)) \)
Problem Statement 

\( y''  6y' + 13y = 0 \)
Solution 

video by PatrickJMT 

Final Answer 

\( y = e^{3x}(c_1\cos(2x) + c_2\sin(2x)) \) 
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\( y'' + 4y' + 20y = 0 \); \( y(0)=9, y'(0)=10 \)
Problem Statement 

\( y'' + 4y' + 20y = 0 \); \( y(0)=9, y'(0)=10 \)
Final Answer 

\( y(x) = e^{2x} \left[ 9\cos(4x) + 7\sin(4x) \right] \)
Problem Statement 

\( y'' + 4y' + 20y = 0 \); \( y(0)=9, y'(0)=10 \)
Solution 

video by Krista King Math 

Final Answer 

\( y(x) = e^{2x} \left[ 9\cos(4x) + 7\sin(4x) \right] \) 
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\( y'' + 2y' + 17y = 0 \)
Problem Statement 

\( y'' + 2y' + 17y = 0 \)
Solution 

video by Dr Chris Tisdell 

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Three Case Summary
The following table summaries the three cases based on the types of roots. In this table, \(y\) is a function of \(t\) and \(a\), \(b\) and \(c\) are constants. We use the variable \(r\) in the characteristic (sometimes called auxiliary) equation but you may see other instructors use different variables. Check with your instructor to see what they expect.
Also, some instructors use hyperbolic functions in their answers. From our experience, this is pretty rare but there are several practice problems with video solutions that use this notation.
differential equation 
\(\displaystyle{ ay'' + by' + cy = 0 }\)  

characteristic equation 
\(ar^2+br+c=0\)  
real distinct roots  \(r_1, r_2\) 
\(\displaystyle{ y = c_1 e^{r_1t} + c_2 e^{r_2t} }\) 
real repeated roots  \(r = r_1 = r_2\) 
\( y = c_1 e^{rt} + c_2 te^{rt} \) 
complex roots  \(r = \alpha \pm i\beta \) 
\( y = e^{\alpha t}[ A\cos(\beta t) +\) \( B\sin(\beta t) ] \) 
HigherOrder Linear Equations
Higherorder linear equations work exactly like first and secondorder, just with additional roots. Here are some practice problems to demonstrate this. There is nothing new here, just more terms in the equations. You need to factor into linear and/or quadratic terms and apply the techniques described above.
After working the practice problems on this page, you will be ready to start on the inhomogeneous case on the undetermined coefficients page.
Practice  HigherOrder
Solve these differential equations by determining the general solution. If initial conditions are given, use them to determine the particular solution.
\( y''' + y''  6y' + 4 = 0 \)
Problem Statement 

Solve \( y''' + y''  6y' + 4 = 0 \)
Solution 

video by blackpenredpen 

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\( y^{(4)} + 2y'' + y = 0 \)
Problem Statement 

Solve \( y^{(4)} + 2y'' + y = 0 \)
Solution 

In this video, he writes the fourth derivative, \( y^{(4)} \), as \( y^{iv} \) using roman numerals. We have never seen it written this way before. So, although it is clear what he is doing, we recommend that you do not adopt this notation. However, as usual, check with your instructor to see what they expect.
video by blackpenredpen 

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\( y^{(6)} + y = 0 \)
Problem Statement 

Solve \( y^{(6)} + y = 0 \)
Solution 

video by blackpenredpen 

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\( 2y'''  6y''  5y' + 15y = 0 \)
Problem Statement 

Solve \( 2y'''  6y''  5y' + 15y = 0 \)
Solution 

video by Houston Math Prep 

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\( y''' + 8y'' + 12y' = 0 \)
Problem Statement 

Solve \( y''' + 8y'' + 12y' = 0 \)
Solution 

video by Houston Math Prep 

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\( y^{(4)} + y'''  7y''  y' +6y = 0 \)
Problem Statement 

Solve \( y^{(4)} + y'''  7y''  y' +6y = 0 \)
Solution 

video by Houston Math Prep 

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\( y^{(4)}  3y''  28y = 0 \)
Problem Statement 

Solve \( y^{(4)}  3y''  28y = 0 \)
Solution 

video by Houston Math Prep 

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Practice Instructions
Solve these differential equations by determining the general solution. If initial conditions are given, use them to determine the particular solution.