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.
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.
Recommended Books on Amazon (affiliate links)  

Join Amazon Student  FREE TwoDay Shipping for College Students 
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 14, 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 

\( r^2  2r  3 = 0 \) 
\( (r3)(r+1) = 0 \) 
\( r=3, ~~~ r = 1 \) 
Solutions to the characteristic equation are real and distinct.
\( y(t) = c_1 e^{t} + c_2 e^{3t} \)
Final Answer 

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

Log in to rate this practice problem and to see it's current rating. 

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

\( r^2  2r + 1 = 0 \) 
\( (r1)^2 = 0 \) 
\( r = 1, r = 1 \) 
Solutions to the characteristic equation are real and equal.
\( y(t) = c_1 e^t + c_2 t e^t \)
Final Answer 

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

Log in to rate this practice problem and to see it's current rating. 

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

\(\begin{array}{rcl} r^2  2r + 5 & = & 0 \\ r^2  2r + 1 & = & 5 + 1 \\ (r1)^2 & = & 4 \\ r1 & = & \pm 2i \\ r & = & 1 \pm 2i \end{array}\)
Solutions to the characteristic equation are complex.
\( y(t) = e^t [c_1 \cos(2t) + c_2 \sin(2t) ] \)
Final Answer 

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

Log in to rate this practice problem and to see it's current rating. 

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 nonhomogeneous 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 

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

Log in to rate this practice problem and to see it's current rating. 

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

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

Log in to rate this practice problem and to see it's current rating. 

\( y''  2y' + y = (2x+1)e^x \)
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 
\( r^2  2r + 1 = 0 \) 
\( (r1)^2 = 0 \) 
\( r = 1 \) multiplicity 2 
homogeneous solution 
\( y_h = c_1 e^x + c_2 x e^x \) 
So the form of the nonhomogeneous (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 nonhomgeneous solutions, \( y = y_h + y_p \).
Final Answer 

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

Log in to rate this practice problem and to see it's current rating. 

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. \( r^2  4r + 4 = (r2)^2 = 0 ~~~ \to ~~~ r = +2 \) 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 nonhomogeneous 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 

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

Log in to rate this practice problem and to see it's current rating. 

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 \( u = v' \).
\(\begin{array}{rcl} x^5 u' + 6x^4u & = & 0 \\ x^5 u' & = & 6x^4 u \\ \displaystyle{\frac{u'}{u}} & = & \displaystyle{\frac{6}{x}} \\ \displaystyle{\frac{du}{u}} & = & \displaystyle{\frac{6}{x}dx} \\ \lnu & = & 6\lnx + c_1 \\ u & = & c_1 x^{6} \\ v' & = & c_1 x^{6} \\ dv & = & c_1 x^{6}dx \\ \displaystyle{v = \frac{c_1 x^{5}}{5} + k} \end{array}\)
\(\displaystyle{ y_2 = v x^3 = \left[ \frac{c_1 x^{5}}{5} + k \right] x^3 = \frac{c_1x^{2}}{5} + kx^3 }\)
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.
close solution

Log in to rate this practice problem and to see it's current rating. 

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 

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

Log in to rate this practice problem and to see it's current rating. 

\( y'' + 4y = e^{2x}; y(0)=1, y'(0)=3 \)
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
\( y(0) = c_1 + 1/8 = 1 ~~~ \to ~~~ c_1 = 11/8 = 7/8 \)
\(\displaystyle{ y'(x) = 2c_1 \sin(2x) + 2c_2 \cos(2x) + \frac{1}{8}(2)e^{2x} }\)
\( y'(0) = 2c_2 + 1/4 = 3 ~~~ \to ~~~ 2c_2 = 31/4 = 11/4 ~~~ \to ~~~ c_2 = 11/8 \)
Final Answer 

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

Log in to rate this practice problem and to see it's current rating. 

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

\(\begin{array}{rcl} r^2+r+1.25 & = & 0 \\ r^2+r+1/4 & = & 5/4+1/4 \\ (r+1/2)^2 & = & 1 \\ r+1/2 & = & \pm i \\ r & = & 1/2 \pm i \end{array}\)
\(y_h=e^{x/2}[c_1\cos(x)+c_2\sin(x)]\)
\(\begin{array}{rcl} y_p & = & A\cos(x)+B\sin(x) \\ y'_p & = & A\sin(x)+B\cos(x) \\ y''_p & = & A\cos(x)B\sin(x) \end{array}\)
\( \displaystyle{y''_p+y'_p+\frac{5}{4}y} \) 
\( \displaystyle{[A\cos(x)B\sin(x)]+[A\sin(x)+B\cos(x)]+\frac{5}{4}[A\cos(x)+B\sin(x)]} \) 
\( [A+B+5A/4]\cos(x)+[BA+5B/4]\sin(x)=5\sin(x) \) 
Equating coefficients, we have two equations and two unknowns.
cosine term  sine term  

\(B+A/4=0\)  \(A+B/4=5\)  
\(4B+A=0\)  \(4A+B=20\)  
\(B=20+4A\)  
\(4(20+4A)+A=0\)  
\(80+16A+A=0\)  
\(17A=80\)  
\(A=80/17 \)  
\(B=20+4(80/17)\)  
\(B=(340320)/17\)  
\(B=20/17\) 
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 \(\displaystyle{y_{ss}(x)=\frac{80}{17}\cos(x)+\frac{20}{17}\sin(x)}\)
close solution

Log in to rate this practice problem and to see it's current rating. 

You CAN Ace Differential Equations
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  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle 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
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
To bookmark this page and practice problems, log in to your account or set up a free account.
Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
 
The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free. 