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

17Calculus Differential Equations - Exam 2

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This page contains a complete differential equations exam with worked out solutions to all problems.
Note - This exam is not from a specific semester. See the list of topics that it covers to determine if it applies to you.

Practice Exam Tips

Each exam page contains a full exam with detailed solutions. Most of these are actual exams from previous semesters used in college courses. You may use these as practice problems or as practice exams. Here are some suggestions on how to use these to help you prepare for your exams.

- Set aside a chunk of full, uninterrupted time, usually an hour to two, to work each exam.
- Go to a quiet place where you will not be interrupted that duplicates your exam situation as closely as possible.
- Use the same materials that you are allowed in your exam (unless the instructions with these exams are more strict).
- Use your calculator as little as possible except for graphing and checking your calculations.
- Work the entire exam before checking any solutions.
- After checking your work, rework any problems you missed and go to the 17calculus page discussing the material to perfect your skills.
- Work as many practice exams as you have time for. This will give you practice in important techniques, experience in different types of exam problems that you may see on your own exam and help you understand the material better by showing you what you need to study.

IMPORTANT -
Exams can cover only so much material. Instructors will sometimes change exams from one semester to the next to adapt an exam to each class depending on how the class performs during the semester while they are learning the material. So just because you do well (or not) on these practice exams, does not necessarily mean you will do the same on your exam. Your instructor may ask completely different questions from these. That is why working lots of practice problems will prepare you better than working just one or two practice exams.
Calculus is not something that can be learned by reading. You have to work problems on your own and struggle through the material to really know calculus, do well on your exam and be able to use it in the future.

Exam Details

Time

1 hour

Questions

11

Total Points

100

Tools

Calculator

no

Formula Sheet(s)

no

Other Tools

none

Instructions:
- This exam is in four main parts, labeled sections 1-4, with different instructions for each section.
- Show all your work.
- For each problem, correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions.
- Correct notation counts (i.e. points will be taken off for incorrect notation).
- Give exact, simplified answers.

Section 1

Find the general solution of each of the following differential equations. Each question in this section is worth 5 points.

\( y'' - 2y' - 3y = 0 \)

Problem Statement

\( y'' - 2y' - 3y = 0 \)

Final Answer

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

Problem Statement

\( y'' - 2y' - 3y = 0 \)

Solution

\( r^2 - 2r - 3 = 0 \)

\( (r-3)(r+1) = 0 \)

\( r=3, ~~~ r = -1 \)

Solutions to the characteristic equation are real and distinct.
\( y(t) = c_1 e^{-t} + c_2 e^{3t} \)

Final Answer

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

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

Problem Statement

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

Final Answer

\( y(t) = c_1 e^t + c_2 t e^t \)

Problem Statement

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

Solution

\( r^2 - 2r + 1 = 0 \)

\( (r-1)^2 = 0 \)

\( r = 1, r = 1 \)

Solutions to the characteristic equation are real and equal.

\( y(t) = c_1 e^t + c_2 t e^t \)

Final Answer

\( y(t) = c_1 e^t + c_2 t e^t \)

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

Problem Statement

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

Final Answer

\( y(t) = e^t [ c_1 \cos(2t) + c_2 \sin(2t) ] \)

Problem Statement

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

Solution

\(\begin{array}{rcl} r^2 - 2r + 5 & = & 0 \\ r^2 - 2r + 1 & = & -5 + 1 \\ (r-1)^2 & = & -4 \\ r-1 & = & \pm 2i \\ r & = & 1 \pm 2i \end{array}\)
Solutions to the characteristic equation are complex.
\( y(t) = e^t [c_1 \cos(2t) + c_2 \sin(2t) ] \)

Final Answer

\( y(t) = e^t [ c_1 \cos(2t) + c_2 \sin(2t) ] \)

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Section 2

Find the general solution to each of the following differential equations using the method of undetermined coefficients. Each question in this section is worth 10 points.

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

Problem Statement

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

Final Answer

\( y = c_1 \cos(2x) + c_2 \sin(2x) + (1/8)e^{2x} \)

Problem Statement

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

Solution

First, we need to solve the homogeneous equation \( y'' + 4y = 0 \).
\( r^2+4 = 0 ~~~ \to ~~~ r^2 = -4 ~~~ \to ~~~ r = \pm 2i \)
\( y_h = c_1 \cos(2x) + c_2 \sin(2x) \)
So the form of the non-homogeneous solution is \( y_p(x) = Ae^{2x} \)
\( y'_p(x) = 2Ae^{2x} ~~~~~~ y''_p(x) = 4Ae^{2x} \)
\(\begin{array}{rcl} y''_p + 4y_p & = & e^{2x} \\ 4Ae^{2x} + 4(Ae^{2x}) & = & e^{2x} \\ 8Ae^{2x} & = & e^{2x} \\ 8A & = & 1 \\ A & = & 1/8 \end{array}\)
Combining the two solutions gives us the final answer, i.e. \( y = y_h + y_p \).

Final Answer

\( y = c_1 \cos(2x) + c_2 \sin(2x) + (1/8)e^{2x} \)

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\( y''-4y = e^{2x} \)

Problem Statement

\( y''-4y = e^{2x} \)

Final Answer

\( y = c_1e^{2x} + c_2 e^{-2x} + (x/4)e^{2x} \)

Problem Statement

\( y''-4y = e^{2x} \)

Solution

First, we need to find the solution to the homogeneous equation, \( y'' - 4y = 0 \).

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

characteristic equation

\( r^2 - 4 = 0 \)

\( r^2 = 4 \)

\( r = \pm 2 \)

homogeneous solution

\( y_h(x) = c_1 e^{2x} + c_2 e^{-2x} \)

Now the form of the non-homogeneous (particular) solution is \( y_p(x) = Axe^{2x} \).

\( y_p(x) = Axe^{2x} \)

\( y'_p = Ax(2e^{2x}) + Ae^{2x} = (2Ax+A)e^{2x} \)

\( y''_p = (2Ax+A)(2e^{2x}) + e^{2x}(2A) = (4Ax+2A+2A)e^{2x} = (4Ax+4A)e^{2x} \)

plug these expressions into the original differential equation

\( y''_p - 4y_p = e^{2x} \)

\( (4Ax+4A)e^{2x} - 4(Axe^{2x}) = e^{2x} \)

\( 4Ae^{2x} = e^{2x} \)

\( 4A = 1 \)

\( A = 1/4 \)

particular solution

\( (x/4)e^{2x} \)

Combining the homogeneous and particular solutions gives us the complete solution, \( y = y_h + y_p \)

Final Answer

\( y = c_1e^{2x} + c_2 e^{-2x} + (x/4)e^{2x} \)

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\( y'' - 2y' + y = (2x+1)e^x \)

Problem Statement

\( y'' - 2y' + y = (2x+1)e^x \)

Final Answer

\( \displaystyle{ y = \left[ \frac{x^3}{3} + \frac{x^2}{2} + c_2 x + c_1 \right] e^x }\)

Problem Statement

\( y'' - 2y' + y = (2x+1)e^x \)

Solution

As before, we need to find the solutions to the homogeneous equation.

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

characteristic equation

\( r^2 - 2r + 1 = 0 \)

\( (r-1)^2 = 0 \)

\( r = 1 \) multiplicity 2

homogeneous solution

\( y_h = c_1 e^x + c_2 x e^x \)

So the form of the non-homogeneous (particular) solution is \( y_p = (Ax+B)e^x (x^2) = (Ax^3+Bx^2)e^x \)

\( y_p = (Ax^3+Bx^2)e^x \)

\( y'_p = (Ax^3+Bx^2)e^x + e^x(3Ax^2+2Bx) \)

\( y'_p = (Ax^3 + 3Ax^2 + Bx^2 + 2Bx)e^x \)

\( y''_p = (Ax^3 + 3Ax^2 +Bx^2+2Bx)e^x + e^x(3Ax^2+6Ax+2Bx+2B) \)

\( y''_p = (Ax^3 + 6Ax^2 + Bx^2 +4Bx +2B +6Ax) e^x \)

Now plug these expressions into the original differential equation.

\( y''_p - 2y'_p + y_p \)

\( (Ax^3 + 6Ax^2 + Bx^2 +4Bx +2B +6Ax) e^x \)

\( -2(Ax^3 + 3Ax^2 +Bx^2+2Bx)e^x + (Ax^3 + Bx^2)e^x \)

\( (6Ax + 2B)e^x = (2x+1)e^x \)

Equating coefficients in the last equation yields

\( 2B = 1 ~~~ \to ~~~ B = 1/2 \)

\( 6A = 2 ~~~ \to ~~~ A = 1/3 \)

particular solution

\(\displaystyle{ y_p = \frac{x^3}{3} + \frac{x^2}{2} }\)

The final answer combines the results from the homogeneous and non-homgeneous solutions, \( y = y_h + y_p \).

Final Answer

\( \displaystyle{ y = \left[ \frac{x^3}{3} + \frac{x^2}{2} + c_2 x + c_1 \right] e^x }\)

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Section 3

Solve the following problems.

Use variation of parameters to find the general solution of \( y'' - 4y' + 4y = \sec^2(x)e^{2x} \)

Problem Statement

Use variation of parameters to find the general solution of \( y'' - 4y' + 4y = \sec^2(x)e^{2x} \)

Final Answer

\( y = [ -\ln|\cos(x)| + c_2 x + c_1 ] e^{2x} \)

Problem Statement

Use variation of parameters to find the general solution of \( y'' - 4y' + 4y = \sec^2(x)e^{2x} \)

Solution

Homogeneous solution first. \( r^2 - 4r + 4 = (r-2)^2 = 0 ~~~ \to ~~~ r = +2 \) multiplicity \(2\)

\( y_h = c_1 e^{2x} + c_2 x e^{2x} \)

Now, find the Wronskian.
\(\displaystyle{W = \begin{vmatrix} y_{h1} & y_{h2} \\ y'_{h1} & y'_{h2} \end{vmatrix} = }\) \(\displaystyle{\begin{vmatrix} e^{2x} & xe^{2x} \\ 2e^{2x} & (2x+1)e^{2x} \end{vmatrix} = }\) \(\displaystyle{ (2x+1)e^{4x} - 2xe^{4x} = e^{4x} }\)

Using variation of parameters, our non-homogeneous solution looks like \( y=u_1(x)e^{2x} + u_2(x)xe^{2x} \).
Setting up the equations for \( u'_1(x) \) and \( u'_2(x) \), we have

\(\displaystyle{u'_1 = \frac{\begin{vmatrix} 0 & xe^{2x} \\ \sec^2 xe^{2x} & (2x+1)e^{2x} \end{vmatrix} }{W} = \frac{-x\sec^2 x e^{4x}}{e^{4x}} = -x\sec^2 x }\)

\(\displaystyle{u'_2 = \frac{\begin{vmatrix} e^{2x} & 0 \\ 2e^{2x} & \sec^2xe^{2x} \end{vmatrix} }{W} = \frac{\sec^2xe^{4x}}{e^{4x}} = \sec^2 x }\)

So we need to solve the two equations
\(\displaystyle{ u'_1 = -x \sec^2 x ~~~ \to ~~~ u_1 = -\int{x\sec^2 x ~ dx} }\)

\(\displaystyle{ u'_2 = \sec^2 x ~~~ \to ~~~ u_2 = \int{\sec^2 x~dx} }\)

The first integral is solved in the practice problems on the integration by parts page. So we just give the result here.
\( \displaystyle{ u_1 = -\int{x\sec^2 x ~ dx} = -x\tan(x)-\ln|\cos(x)|+k_1 }\)

The second one is easy, since \( (\tan(x))' = \sec^2(x) \)

\(\displaystyle{ u_2 = \int{\sec^2 x~dx} = \tan(x)+k_2 }\)

To get our final answer, we substitute these results into the equation \( y=u_1e^{2x} + u_2xe^{2x} \) and simplify.
We also let \( c_1 = A+k_1\) and \( c_2 = B+k_2 \) to combine the constants.

Final Answer

\( y = [ -\ln|\cos(x)| + c_2 x + c_1 ] e^{2x} \)

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Determine the second fundamental solution of \( x^2y'' - 6y = 0 \), given \( y_1 = x^3 \)

Problem Statement

Determine the second fundamental solution of \( x^2y'' - 6y = 0 \), given \( y_1 = x^3 \)

Final Answer

\( y_2 = x^{-2} \) is the second fundamental solution.

Problem Statement

Determine the second fundamental solution of \( x^2y'' - 6y = 0 \), given \( y_1 = x^3 \)

Solution

We will use the technique of Reduction of Order, where \( y_2 = vy_1 \) and \(v\) is a function of \(x\).

\( y_2 = vy_1 \)

\( y'_2 = v(3x^2) + x^3 v' \)

\( y''_2 = v(6x) + 3x^2 v + x^3 v'' + 3x^2 v' \)

\( y''_2 = x^3 v'' + 6x^2 v' + 6xv \)

\( x^2 y''_2 - 6y_2 = x^2[ x^3 v'' + 6x^2 v' + 6xv] - 6vx^3 \)

\( x^5 v'' + 6x^4 v' = 0 \)

Let \( u = v' \).
\(\begin{array}{rcl} x^5 u' + 6x^4u & = & 0 \\ x^5 u' & = & -6x^4 u \\ \displaystyle{\frac{u'}{u}} & = & \displaystyle{\frac{-6}{x}} \\ \displaystyle{\frac{du}{u}} & = & \displaystyle{\frac{-6}{x}dx} \\ \ln|u| & = & -6\ln|x| + c_1 \\ u & = & c_1 x^{-6} \\ v' & = & c_1 x^{-6} \\ dv & = & c_1 x^{-6}dx \\ \displaystyle{v = \frac{c_1 x^{-5}}{-5} + k} \end{array}\)

\(\displaystyle{ y_2 = v x^3 = \left[ \frac{c_1 x^{-5}}{-5} + k \right] x^3 = \frac{c_1x^{-2}}{-5} + kx^3 }\)

let \( c = c_1 / (-5) \)

\( y_2 = c/x^2 + kx^3 \)

Since \( kx^3 \) is a multiple of \( y_1 \) it is already included in \( y_1 \), so we just drop that term.

Final Answer

\( y_2 = x^{-2} \) is the second fundamental solution.

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Section 4

Solve the following initial value problems.

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

Problem Statement

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

Final Answer

\(\displaystyle{ y(t) = \frac{-1}{4} e^{-2t} + \frac{1}{4} e^{2t} }\)

Problem Statement

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

Solution

\( r^2 - 4 = 0 ~~~ \to ~~~ r^2 = 4 ~~~ \to ~~~ r = \pm 2 \)
\( y = c_1 e^{-2t} + c_2 e^{2t} \)
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

\(\displaystyle{ y(t) = \frac{-1}{4} e^{-2t} + \frac{1}{4} e^{2t} }\)

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\( y'' + 4y = e^{2x}; y(0)=1, y'(0)=3 \)

Problem Statement

\( y'' + 4y = e^{2x}; y(0)=1, y'(0)=3 \)

Final Answer

\(\displaystyle{ y(x) = \frac{7}{8}\cos(2x) + \frac{11}{8}\sin(2x) + \frac{1}{8}e^{2x} }\)

Problem Statement

\( y'' + 4y = e^{2x}; y(0)=1, y'(0)=3 \)

Solution

From the previous problem, we have the general solution \(\displaystyle{ y = c_1 \cos(2x) + c_2 \sin(2x) + \frac{1}{8} e^{2x}}\)
Using the initial conditions to find the constants \(c_1\) and \(c_2\), we have
\( y(0) = c_1 + 1/8 = 1 ~~~ \to ~~~ c_1 = 1-1/8 = 7/8 \)

\(\displaystyle{ y'(x) = -2c_1 \sin(2x) + 2c_2 \cos(2x) + \frac{1}{8}(2)e^{2x} }\)

\( y'(0) = 2c_2 + 1/4 = 3 ~~~ \to ~~~ 2c_2 = 3-1/4 = 11/4 ~~~ \to ~~~ c_2 = 11/8 \)

Final Answer

\(\displaystyle{ y(x) = \frac{7}{8}\cos(2x) + \frac{11}{8}\sin(2x) + \frac{1}{8}e^{2x} }\)

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\( y'' + y' + 1.25y = 5\sin(x); y(0) = y'(0) = 0 \)

Problem Statement

\( y'' + y' + 1.25y = 5\sin(x); y(0) = y'(0) = 0 \)

Final Answer

\(\displaystyle{y(x)=e^{-x/2}\left[\frac{80}{17}\cos(x)-\right.}\) \(\displaystyle{\left.\frac{20}{17}\sin(x)\right]-}\) \(\displaystyle{\frac{80}{17}\cos(x)+}\) \(\displaystyle{\frac{20}{17}\sin(x)}\)
steady state occurs as \(x\to\infty\); so \(e^{-x/2}\to 0\) and \(\displaystyle{y_{ss}(x)=\frac{-80}{17}\cos(x)+\frac{20}{17}\sin(x)}\)

Problem Statement

\( y'' + y' + 1.25y = 5\sin(x); y(0) = y'(0) = 0 \)

Solution

\(\begin{array}{rcl} r^2+r+1.25 & = & 0 \\ r^2+r+1/4 & = & -5/4+1/4 \\ (r+1/2)^2 & = & -1 \\ r+1/2 & = & \pm i \\ r & = & -1/2 \pm i \end{array}\)
\(y_h=e^{-x/2}[c_1\cos(x)+c_2\sin(x)]\)
\(\begin{array}{rcl} y_p & = & A\cos(x)+B\sin(x) \\ y'_p & = & -A\sin(x)+B\cos(x) \\ y''_p & = & -A\cos(x)-B\sin(x) \end{array}\)

\( \displaystyle{y''_p+y'_p+\frac{5}{4}y} \)

\( \displaystyle{[-A\cos(x)-B\sin(x)]+[-A\sin(x)+B\cos(x)]+\frac{5}{4}[A\cos(x)+B\sin(x)]} \)

\( [-A+B+5A/4]\cos(x)+[-B-A+5B/4]\sin(x)=5\sin(x) \)

Equating coefficients, we have two equations and two unknowns.

cosine term

   

sine term

\(B+A/4=0\)\(-A+B/4=5\)
\(4B+A=0\)\(-4A+B=20\)
\(B=20+4A\)
\(4(20+4A)+A=0\)
\(80+16A+A=0\)
\(17A=-80\)
\(A=-80/17 \)
\(B=20+4(-80/17)\)
\(B=(340-320)/17\)
\(B=20/17\)

So far, we have \(\displaystyle{y=e^{-x/2}[c_1\cos(x)+c_2\sin(x)]-\frac{80}{17}\cos(x)+\frac{20}{17}\sin(x)}\)
Now we use the initial conditions to determine the constants \(c_1\) and \(c_2\).

\(y(0)=0\)

\(y(0)=1(c_1+0)-(80/17)(1)+0=0 ~~~ \to ~~~ c_1=80/17\)

\(y'(0)=0\)

\(y'=e^{-x/2}[-c_1\sin(x)+c_2\cos(x)]+(80/17)\sin(x)+(20/17)\cos(x)\)
\(y'(0)=1(0+c_2)+0+20/17=0 ~~~ \to ~~~ c_2=-20/17\)

Final Answer

\(\displaystyle{y(x)=e^{-x/2}\left[\frac{80}{17}\cos(x)-\right.}\) \(\displaystyle{\left.\frac{20}{17}\sin(x)\right]-}\) \(\displaystyle{\frac{80}{17}\cos(x)+}\) \(\displaystyle{\frac{20}{17}\sin(x)}\)
steady state occurs as \(x\to\infty\); so \(e^{-x/2}\to 0\) and \(\displaystyle{y_{ss}(x)=\frac{-80}{17}\cos(x)+\frac{20}{17}\sin(x)}\)

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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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