You CAN Ace Differential Equations
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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: second-order linear |
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\(y'' + p(t)y' + q(t)y = g(t)\) |
Homogeneous Equations |
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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 |
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Homogeneous equations with constant coefficients look like \(\displaystyle{ ay'' + by' + cy = 0 }\) where a, b and c are constants. (In the case where we have non-constant coefficients and \(p(t)\) and \(q(t)\) are polynomials is covered a separate page on Cauchy-Euler equations and Chebyshev's equations.)
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.
\(\displaystyle{ ay'' + by' + cy = 0 }\) | |
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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. Solve these differential equations by determining the general solution.
\( y'' - 2y' - 3y = 0 \)
Problem Statement |
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\( y'' - 2y' - 3y = 0 \)
Final Answer |
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\( y(t) = c_1 e^{-t} + c_2 e^{3t} \) |
Problem Statement |
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\( y'' - 2y' - 3y = 0 \)
Solution |
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\( r^2 - 2r - 3 = 0 \) |
\( (r-3)(r+1) = 0 \) |
\( r=3, ~~~ r = -1 \) |
Solutions to the characteristic equation are real and distinct.
\( y(t) = c_1 e^{-t} + c_2 e^{3t} \)
Final Answer |
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\( y(t) = c_1 e^{-t} + c_2 e^{3t} \) |
close solution |
\( y'' - 9y = 0 \)
Problem Statement |
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\( y'' - 9y = 0 \)
Final Answer |
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\( y(t) = c_1e^{3t} + c_2e^{-3t} \) |
Problem Statement |
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\( y'' - 9y = 0 \)
Solution |
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You can ignore the message that comes up near the beginning of the video. It refers to the discussion preceding the video that does not affect the solution.
video by Engineer In Training Exam
Final Answer |
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\( y(t) = c_1e^{3t} + c_2e^{-3t} \) |
close solution |
\( y'' - y' - 6y = 0 \)
Problem Statement |
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\( y'' - y' - 6y = 0 \)
Final Answer |
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\( y(x) = Ae^{-2x} + Be^{3x} \) |
Problem Statement |
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\( y'' - y' - 6y = 0 \)
Solution |
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video by Dr Chris Tisdell
Final Answer |
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\( y(x) = Ae^{-2x} + Be^{3x} \) |
close solution |
\( y'' + y' - 6y = 0 \)
Problem Statement |
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\( y'' + y' - 6y = 0 \)
Final Answer |
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\( y = c_1e^{-3t} + c_2e^{2t} \) |
Problem Statement |
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\( y'' + y' - 6y = 0 \)
Solution |
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video by PatrickJMT
Final Answer |
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\( y = c_1e^{-3t} + c_2e^{2t} \) |
close solution |
\( y'' - 7y' + 10y = 0 \)
Problem Statement |
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\( y'' - 7y' + 10y = 0 \)
Final Answer |
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\( y(x) = c_1e^{5x} + c_2e^{2x} \) |
Problem Statement |
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\( y'' - 7y' + 10y = 0 \)
Solution |
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video by Krista King Math
Final Answer |
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\( y(x) = c_1e^{5x} + c_2e^{2x} \) |
close solution |
\( 4y'' - 4y' - 3y = 0 \)
Problem Statement |
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\( 4y'' - 4y' - 3y = 0 \)
Final Answer |
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\( y(x) = c_1e^{-x/2} + c_2e^{3x/2} \) |
Problem Statement |
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\( 4y'' - 4y' - 3y = 0 \)
Solution |
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video by Krista King Math
Final Answer |
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\( y(x) = c_1e^{-x/2} + c_2e^{3x/2} \) |
close solution |
\( y'' + 3y' - 10y = 0 \)
Problem Statement |
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\( y'' + 3y' - 10y = 0 \)
Solution |
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video by Dr Chris Tisdell
close solution |
Solve these differential equations using the initial conditions to determine the particular solutions.
\( y'' - 4y = 0; y(0)=0, y'(0)=1 \)
Problem Statement |
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\( y'' - 4y = 0; y(0)=0, y'(0)=1 \)
Final Answer |
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\(\displaystyle{ y(t) = \frac{-1}{4} e^{-2t} + \frac{1}{4} e^{2t} }\) |
Problem Statement |
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\( y'' - 4y = 0; y(0)=0, y'(0)=1 \)
Solution |
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\( r^2 - 4 = 0 ~~~ \to ~~~ r^2 = 4 ~~~ \to ~~~ r = \pm 2 \)
\( y = c_1 e^{-2t} + c_2 e^{2t} \)
Use the initial conditions to find the constants \( c_1 \) and \( c_2 \).
\( y(0) = c_1 + c_2 = 0 ~~~ \to ~~~ c_1 = -c_2 \)
\(\begin{array}{rcl}
y'(t) & = & -2c_1 e^{-2t} + 2 c_2 e^{2t} \\
y'(0) & = & -2c_1 + 2c_2 = 1 \\
& & -2(-c_2) + 2c_2 = 1 \\
& & 4c_2 = 1 \\
& & c_2 = 1/4 \\
& & c_1 = -c_2 = -1/4
\end{array}\)
Final Answer |
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\(\displaystyle{ y(t) = \frac{-1}{4} e^{-2t} + \frac{1}{4} e^{2t} }\) |
close solution |
\( y'' - 4y' - 5y = 0 \); \( y(0)=1, y'(0)=0 \)
Problem Statement |
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\( y'' - 4y' - 5y = 0 \); \( y(0)=1, y'(0)=0 \)
Final Answer |
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\( y(x) = (1/6)e^{5x} + (5/6)e^{-x} \) |
Problem Statement |
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\( y'' - 4y' - 5y = 0 \); \( y(0)=1, y'(0)=0 \)
Solution |
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video by Dr Chris Tisdell
Final Answer |
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\( y(x) = (1/6)e^{5x} + (5/6)e^{-x} \) |
close solution |
\( y'' - 18y' + 77y = 0 \); \( y(0)=4, y'(0)=8 \)
Problem Statement |
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\( y'' - 18y' + 77y = 0 \); \( y(0)=4, y'(0)=8 \)
Final Answer |
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\( y(x) = 9e^{7x} - 5e^{11x} \) |
Problem Statement |
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\( y'' - 18y' + 77y = 0 \); \( y(0)=4, y'(0)=8 \)
Solution |
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video by Krista King Math
Final Answer |
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\( y(x) = 9e^{7x} - 5e^{11x} \) |
close solution |
\( 2y'' - 11y' + 12y = 0 \); \( y(0)=5, y'(0)=15 \)
Problem Statement |
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\( 2y'' - 11y' + 12y = 0 \); \( y(0)=5, y'(0)=15 \)
Final Answer |
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\( y(x) = 2e^{3x/2} + 3e^{4x} \) |
Problem Statement |
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\( 2y'' - 11y' + 12y = 0 \); \( y(0)=5, y'(0)=15 \)
Solution |
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video by Krista King Math
Final Answer |
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\( y(x) = 2e^{3x/2} + 3e^{4x} \) |
close solution |
Real Repeated Roots |
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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 in-depth discussion with great examples using a spring-mass-damped 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. Solve these differential equations by determining the general solution. If initial conditions are given, use them to determine the particular solution.
\( y'' - 2y' + y = 0 \)
Problem Statement |
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\( y'' - 2y' + y = 0 \)
Final Answer |
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\( y(t) = c_1 e^t + c_2 t e^t \) |
Problem Statement |
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\( y'' - 2y' + y = 0 \)
Solution |
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\( r^2 - 2r + 1 = 0 \) |
\( (r-1)^2 = 0 \) |
\( r = 1, r = 1 \) |
Solutions to the characteristic equation are real and equal.
\( y(t) = c_1 e^t + c_2 t e^t \)
Final Answer |
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\( y(t) = c_1 e^t + c_2 t e^t \) |
close solution |
\( y'' - 4y' + 4y = 0 \)
Problem Statement |
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\( y'' - 4y' + 4y = 0 \)
Final Answer |
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\( y(x) = e^{2x}(A+Bx) \) |
Problem Statement |
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\( y'' - 4y' + 4y = 0 \)
Solution |
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video by Dr Chris Tisdell
Final Answer |
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\( y(x) = e^{2x}(A+Bx) \) |
close solution |
\( y'' + 22y' + 121y = 0 \); \( y(0)=2, y'(0)=-25 \)
Problem Statement |
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\( y'' + 22y' + 121y = 0 \); \( y(0)=2, y'(0)=-25 \)
Final Answer |
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\( y(x) = e^{-11x}(2-3x) \) |
Problem Statement |
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\( y'' + 22y' + 121y = 0 \); \( y(0)=2, y'(0)=-25 \)
Solution |
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video by Krista King Math
Final Answer |
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\( y(x) = e^{-11x}(2-3x) \) |
close solution |
\( y'' + 6y' + 9y = 0 \)
Problem Statement |
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\( y'' + 6y' + 9y = 0 \)
Final Answer |
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\( y(x) = e^{-3x}(A+Bx) \) |
Problem Statement |
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\( y'' + 6y' + 9y = 0 \)
Solution |
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video by Dr Chris Tisdell
Final Answer |
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\( y(x) = e^{-3x}(A+Bx) \) |
close solution |
\( y'' - 8y' + 16y = 0 \)
Problem Statement |
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\( y'' - 8y' + 16y = 0 \)
Final Answer |
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\( y = e^{4t}(c_1+c_2x) \) |
Problem Statement |
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\( y'' - 8y' + 16y = 0 \)
Solution |
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Here are two videos solving this problem worked by two different instructors.
video by Dr Chris Tisdell
video by PatrickJMT
Final Answer |
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\( y = e^{4t}(c_1+c_2x) \) |
close solution |
\( y'' + 20y' + 100y = 0 \)
Problem Statement |
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\( y'' + 20y' + 100y = 0 \)
Final Answer |
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\( y(x) = e^{-10x}(c_1+c_2x) \) |
Problem Statement |
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\( y'' + 20y' + 100y = 0 \)
Solution |
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video by Krista King Math
Final Answer |
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\( y(x) = e^{-10x}(c_1+c_2x) \) |
close solution |
Complex Roots |
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\( e^{\pm i\mu} = \cos(\mu) \pm i \sin(\mu) \) |
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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
\(\displaystyle{ e^{it} = \sum_{n=0}^{\infty}{\frac{(it)^n}{n!}} = \sum_{n=0}^{\infty}{\frac{(-1)^nt^{2n}}{(2n)!}} +
i\sum_{n=1}^{\infty}{\frac{(-1)^{n-1}t^{2n-1}}{(2n-1)!}} }\)
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 spring-mass-damped system video.
video by MIT OCW
Okay, let's work some practice problems with complex roots. Solve these differential equations by determining the general solution. If initial conditions are given, use them to determine the particular solution.
\( y''-2y'+5y=0 \)
Problem Statement |
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\( y''-2y'+5y=0 \)
Final Answer |
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\( y(t) = e^t [ c_1 \cos(2t) + c_2 \sin(2t) ] \) |
Problem Statement |
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\( y''-2y'+5y=0 \)
Solution |
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\(\begin{array}{rcl}
r^2 - 2r + 5 & = & 0 \\
r^2 - 2r + 1 & = & -5 + 1 \\
(r-1)^2 & = & -4 \\
r-1 & = & \pm 2i \\
r & = & 1 \pm 2i
\end{array}\)
Solutions to the characteristic equation are complex.
\( y(t) = e^t [c_1 \cos(2t) + c_2 \sin(2t) ] \)
Final Answer |
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\( y(t) = e^t [ c_1 \cos(2t) + c_2 \sin(2t) ] \) |
close solution |
\( y'' - 6y' + 10y = 0 \)
Problem Statement |
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\( y'' - 6y' + 10y = 0 \)
Final Answer |
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\( y(t) = e^{3t}(c_1\cos t + c_2\sin t) \) |
Problem Statement |
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\( y'' - 6y' + 10y = 0 \)
Solution |
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video by Michel vanBiezen
Final Answer |
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\( y(t) = e^{3t}(c_1\cos t + c_2\sin t) \) |
close solution |
\( y'' + 4y' + 13y = 0 \)
Problem Statement |
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\( y'' + 4y' + 13y = 0 \)
Solution |
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video by Michel vanBiezen
close solution |
\( y'' - 2y' + 5y = 0 \)
Problem Statement |
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\( y'' - 2y' + 5y = 0 \)
Final Answer |
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\( y(x) = e^{x} \left[ A\cos(2x) + B\sin(2x) \right] \) |
Problem Statement |
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\( y'' - 2y' + 5y = 0 \)
Solution |
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video by Dr Chris Tisdell
Final Answer |
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\( y(x) = e^{x} \left[ A\cos(2x) + B\sin(2x) \right] \) |
close solution |
\( y'' - 6y' + 13y = 0 \)
Problem Statement |
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\( y'' - 6y' + 13y = 0 \)
Final Answer |
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\( y = e^{3x}(c_1\cos(2x) + c_2\sin(2x)) \) |
Problem Statement |
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\( y'' - 6y' + 13y = 0 \)
Solution |
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video by PatrickJMT
Final Answer |
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\( y = e^{3x}(c_1\cos(2x) + c_2\sin(2x)) \) |
close solution |
\( y'' + 4y' + 20y = 0 \); \( y(0)=9, y'(0)=10 \)
Problem Statement |
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\( y'' + 4y' + 20y = 0 \); \( y(0)=9, y'(0)=10 \)
Final Answer |
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\( y(x) = e^{-2x} \left[ 9\cos(4x) + 7\sin(4x) \right] \) |
Problem Statement |
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\( y'' + 4y' + 20y = 0 \); \( y(0)=9, y'(0)=10 \)
Solution |
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video by Krista King Math
Final Answer |
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\( y(x) = e^{-2x} \left[ 9\cos(4x) + 7\sin(4x) \right] \) |
close solution |
\( y'' + 2y' + 17y = 0 \)
Problem Statement |
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\( y'' + 2y' + 17y = 0 \)
Solution |
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video by Dr Chris Tisdell
close solution |
Three Case Summary |
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The following table summaries the three cases based on the types of roots.
differential equation |
\(\displaystyle{ ay'' + by' + cy = 0 }\) | |
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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 | \(r_1 = r_2 = r\) |
\( y = c_1 e^{rt} + c_2 te^{rt} \) |
complex roots | \(r = \alpha \pm i\beta \) |
\( y = e^{\alpha t}[ A\cos(\beta t) +\) \( B\sin(\beta t) ] \) |
Higher-Order Linear Equations |
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Higher-order linear equations work exactly like second-order, just with additional roots. Here are some practice problems to demonstrate this. There is nothing new here, just more terms in the equations.
Solve these higher-order differential equations. First, calculate the general solution and, if initial conditions are given, calculate the particular solution as well.
\( y^{(3)}-4y' = 0 \)
Problem Statement |
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Find the general solution to \( y^{(3)}-4y' = 0 \)
Final Answer |
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\( y = A + Be^{-2t} + Ce^{2t} \) |
Problem Statement |
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Find the general solution to \( y^{(3)}-4y' = 0 \)
Solution |
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\( r^3 - 4r = r(r^2-4) = 0 \) |
\( r=0; ~~~ r^2-4=0 \) |
\( r^2 = 4; ~~~ r = \pm 2 \) |
\( y = A + Be^{-2t} + Ce^{2t} \) |
Final Answer |
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\( y = A + Be^{-2t} + Ce^{2t} \) |
close solution |
\( y^{(4)} + 2y'' + y = 0 \)
Problem Statement |
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Find the general solution to \( y^{(4)} + 2y'' + y = 0 \)
Final Answer |
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\( y = (A+Bt)\cos(t) + (C+Dt)\sin(t) \) |
Problem Statement |
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Find the general solution to \( y^{(4)} + 2y'' + y = 0 \)
Solution |
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\(\begin{array}{rcl}
r^4 + 2r^2 + 1 & = & 0 \\
(r^2+1)^2 & = & 0 \\
r^2 + 1 & = & 0 ~~~ \text{multiplicity 2} \\
r^2 & = & -1 \\
r = \pm i
\end{array}\)
\( y = (A+Bt)\cos(t) + (C+Dt)\sin(t) \)
Final Answer |
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\( y = (A+Bt)\cos(t) + (C+Dt)\sin(t) \) |
close solution |
\( y^{(3)} - 3y''+3y' - y = 0 \)
Problem Statement |
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Find the general solution to \( y^{(3)} - 3y''+3y' - y = 0 \)
Final Answer |
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\( y = (A+Bt+Ct^2)e^t \) |
Problem Statement |
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Find the general solution to \( y^{(3)} - 3y''+3y' - y = 0 \)
Solution |
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\(\begin{array}{rcl}
r^3 - 3r^2 + 3r - 1 & = & 0 \\
(r-1)(r^2-2r+1) & = & 0 \\
(r-1)(r-1)^2 & = & 0 \\
(r-1)^3 & = & 0 \\
r-1 & = & 0 ~~~ \text{X}3^* \\
r & = & 1
\end{array}\)
^{*}Multiplicity 3
\( y = (A+Bt+Ct^2)e^t \)
Final Answer |
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\( y = (A+Bt+Ct^2)e^t \) |
close solution |
Solve the initial value problem \( y^{(4)} + 13y'' + 36y = 0; ~~~ y(0) = 5, y'(0) = y''(0) = y^{(3)} = 0 \)
Problem Statement |
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Solve the initial value problem \( y^{(4)} + 13y'' + 36y = 0; ~~~ y(0) = 5, y'(0) = y''(0) = y^{(3)} = 0 \)
Final Answer |
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\( y = 9\cos(2t) - 4\cos(3t) \) |
Problem Statement |
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Solve the initial value problem \( y^{(4)} + 13y'' + 36y = 0; ~~~ y(0) = 5, y'(0) = y''(0) = y^{(3)} = 0 \)
Solution |
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For this solution, we can apply the techniques of second order linear equations with constant coefficients. So our characteristic polynomial is \( r^4+13r^2+36=0 \to (r^2+4)(r^2+9)=0 \to r=\pm2i,\pm3i \).
This gives us the general form of the solution as \( y = A\cos(2t) + B\sin(2t) + C\cos(3t) + D\sin(3t) \).
Taking derivatives and solving for the coefficients, gives us \( A=9, B=0, C=-4, D=0 \).
Final Answer |
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\( y = 9\cos(2t) - 4\cos(3t) \) |
close solution |
After working the practice problems on this page, you are ready to start on the inhomogeneous case on the undetermined coefficients page.