\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus - Second(and Higher)-Order, Linear 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

\(y'' + p(t)y' + q(t)y = g(t)\)

In the homogeneous case (\(g(t)=0\)) where \(p(t)\) and \(q(t)\) are polynomials, check the pages on Cauchy-Euler 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.

Solve \( y'' - 9y = 0 \)

Problem Statement

Solve \( y'' - 9y = 0 \)

Final Answer

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

Problem Statement

Solve \( y'' - 9y = 0 \)

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 to this problem.

2459 video

Final Answer

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

close solution

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Solve \( y'' - y' - 6y = 0 \)

Problem Statement

Solve \( y'' - y' - 6y = 0 \)

Final Answer

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

Problem Statement

Solve \( y'' - y' - 6y = 0 \)

Solution

629 video

video by Dr Chris Tisdell

Final Answer

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

close solution

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Solve \( y'' + y' - 6y = 0 \)

Problem Statement

Solve \( y'' + y' - 6y = 0 \)

Final Answer

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

Problem Statement

Solve \( y'' + y' - 6y = 0 \)

Solution

633 video

video by PatrickJMT

Final Answer

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

close solution

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Solve \( y'' - 7y' + 10y = 0 \)

Problem Statement

Solve \( y'' - 7y' + 10y = 0 \)

Final Answer

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

Problem Statement

Solve \( y'' - 7y' + 10y = 0 \)

Solution

636 video

video by Krista King Math

Final Answer

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

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Solve \( 4y'' - 4y' - 3y = 0 \)

Problem Statement

Solve \( 4y'' - 4y' - 3y = 0 \)

Final Answer

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

Problem Statement

Solve \( 4y'' - 4y' - 3y = 0 \)

Solution

637 video

video by Krista King Math

Final Answer

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

close solution

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Solve \( y'' + 3y' - 10y = 0 \)

Problem Statement

Solve \( y'' + 3y' - 10y = 0 \)

Solution

1849 video

video by Dr Chris Tisdell

close solution

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

Problem Statement

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

Final Answer

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

Problem Statement

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

Solution

630 video

video by Dr Chris Tisdell

Final Answer

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

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

Problem Statement

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

Final Answer

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

Problem Statement

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

Solution

639 video

video by Krista King Math

Final Answer

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

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

Problem Statement

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

Final Answer

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

Problem Statement

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

Solution

642 video

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 in-depth discussion with great examples using a spring-mass-damped system. It will be well worth your time to watch it carefully.

MIT OCW - Solving Second-order Linear ODE's with Constant Coefficients: The Three Cases [50mins]

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.

Solve \( y'' - 4y' + 4y = 0 \)

Problem Statement

Solve \( y'' - 4y' + 4y = 0 \)

Final Answer

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

Problem Statement

Solve \( y'' - 4y' + 4y = 0 \)

Solution

628 video

video by Dr Chris Tisdell

Final Answer

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

close solution

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

Problem Statement

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

Final Answer

\( y(x) = e^{-11x}(2-3x) \)

Problem Statement

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

Solution

638 video

video by Krista King Math

Final Answer

\( y(x) = e^{-11x}(2-3x) \)

close solution

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

Problem Statement

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

Final Answer

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

Problem Statement

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

Solution

631 video

video by Dr Chris Tisdell

Final Answer

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

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Solve \( y'' - 8y' + 16y = 0 \)

Problem Statement

Solve \( y'' - 8y' + 16y = 0 \)

Final Answer

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

Problem Statement

Solve \( y'' - 8y' + 16y = 0 \)

Solution

Here are two videos solving this problem worked by two different instructors.

634 video

video by Dr Chris Tisdell

634 video

video by PatrickJMT

Final Answer

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

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

Problem Statement

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

Final Answer

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

Problem Statement

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

Solution

641 video

video by Krista King Math

Final Answer

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

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Complex Roots

Euler's Formula

\( e^{\pm i\mu} = \cos(\mu) \pm i \sin(\mu) \)

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
\(\displaystyle{ e^{it} = \sum_{n=0}^{\infty}{\frac{(it)^n}{n!}} = \sum_{n=0}^{\infty}{\frac{(-1)^nt^{2n}}{(2n)!}} + i\sum_{n=1}^{\infty}{\frac{(-1)^{n-1}t^{2n-1}}{(2n-1)!}} }\)
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.

MIT OCW - Complex Numbers and Complex Exponentials [45mins-28secs]

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.

MIT OCW - Continuation: Complex Characteristic Roots; Undamped and Damped Oscillations [46mins-23secs]

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.

Solve \( y'' - 6y' + 10y = 0 \)

Problem Statement

Solve \( y'' - 6y' + 10y = 0 \)

Final Answer

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

Problem Statement

Solve \( y'' - 6y' + 10y = 0 \)

Solution

2148 video

video by Michel vanBiezen

Final Answer

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

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

Problem Statement

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

Solution

2149 video

video by Michel vanBiezen

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Solve \( y'' - 2y' + 5y = 0 \)

Problem Statement

Solve \( y'' - 2y' + 5y = 0 \)

Final Answer

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

Problem Statement

Solve \( y'' - 2y' + 5y = 0 \)

Solution

632 video

video by Dr Chris Tisdell

Final Answer

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

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Solve \( y'' - 6y' + 13y = 0 \)

Problem Statement

Solve \( y'' - 6y' + 13y = 0 \)

Final Answer

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

Problem Statement

Solve \( y'' - 6y' + 13y = 0 \)

Solution

635 video

video by PatrickJMT

Final Answer

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

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

Problem Statement

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

Final Answer

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

Problem Statement

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

Solution

640 video

video by Krista King Math

Final Answer

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

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

Problem Statement

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

Solution

1850 video

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

\(r = \alpha \pm i\beta \)

\( y = e^{\alpha t}[ A\cos(\beta t) +\) \( B\sin(\beta t) ] \)

Higher-Order Linear Equations

Higher-order linear equations work exactly like first and 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. 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 - Higher-Order

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

3541 video

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.

3542 video

video by blackpenredpen

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

Problem Statement

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

Solution

3543 video

video by blackpenredpen

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

Problem Statement

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

Solution

3544 video

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

Problem Statement

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

Solution

3545 video

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

Problem Statement

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

Solution

3546 video

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

Problem Statement

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

Solution

3547 video

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You CAN Ace Differential Equations

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Precalculus

Engineering

Circuits

Semiconductors

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Homogeneous Equations

Constant Coefficients - Practice

Real Repeated Roots - Practice

Complex Roots - Practice

Three Case Summary

Higher-Order Linear Equations - Practice

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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.

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