We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Differential Equations : System of Linear First-Order Differential Equations

Study concepts, example questions & explanations for Differential Equations

Create An Account

All Differential Equations Resources

1 Diagnostic Test 29 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : System Of Linear First Order Differential Equations

Find the general solution to the given system.

$\\\frac{dx}{dt}=x+2y \\\\\frac{dy}{dt}=2x+y$

Possible Answers:

$X=c_1\binom{1}{-1}e^{3t}+c_2\binom{-1}{3}e^{-t}$

$X=c_1\binom{1}{3}e^{3t}+c_2\binom{1}{1}e^{-t}$

$X=c_1\binom{1}{1}e^{3t}+c_2\binom{-1}{3}e^{-t}$

$X=c_1\binom{1}{1}e^{3t}+c_2\binom{1}{3}e^{-t}$

$X=c_1\binom{-1}{1}e^{3t}+c_2\binom{1}{3}e^{-t}$

Correct answer:

$X=c_1\binom{1}{1}e^{3t}+c_2\binom{-1}{3}e^{-t}$

Explanation:

To find the general solution to the given system

$\\\frac{dx}{dt}=x+2y \\\\\frac{dy}{dt}=2x+y$

first find the eigenvalues and eigenvectors.

$det(A-\lambda I )=\begin{vmatrix} 1-\lambda&2 \\ 2&1-\lambda \end{vmatrix}$

$\\=(1-\lambda)(1-\lambda)-(2)(2) \\=1-2\lambda+\lambda^2-4 \\=\lambda^2-2\lambda-3 \\=(\lambda-3)(\lambda+1)$

Therefore the eigenvalues are

$\lambda_1=3, \lambda_2=-1$

Now calculate the eigenvectors

$(A-\lamda I)K=0$

For

$\lambda_1=3$

$\\(1-3)k_1+2k_2 \\2k_1+(1-3)k_2$

Thus,

$k_1=k_2 \rightarrow K_1\binom{1}{1}$

For

$\lambda_1=-1$

$\\(2--1)k_1+1k_2 \\1k_1+(2--1)k_2$

Thus

$k_1=-\frac{1}{3}k_2\rightarrow K_2\binom{-1}{3}$

Therefore,

$X_1=\binom{1}{1}e^{3t};\ X_2\binom{-1}{3}e^{-t}$

Now the general solution is,

$\\X=c_1X_1+c_2X_2 \\\\X=c_1\binom{1}{1}e^{3t}+c_2\binom{-1}{3}e^{-t}$

Report an Error

Example Question #1 : Homogeneous Linear Systems

Solve the initial value problem $\mathbf{x'} = \begin{bmatrix} 3&-2 \\ -1&2 \end{bmatrix}\mathbf{x}$ . Where $\mathbf{x}_0 = \begin{bmatrix} 1\\-1 \end{bmatrix}$

Possible Answers:

$\mathbf{x} = e^t\begin{bmatrix} 3\\1 \end{bmatrix} +e^{4t}\begin{bmatrix} 5\\6 \end{bmatrix}$

$\mathbf{x} = e^{-t}\begin{bmatrix} 5\\2 \end{bmatrix} +e^{3t}\begin{bmatrix} -1\\3 \end{bmatrix}$

$\mathbf{x} = e^{-t}\begin{bmatrix} -2\\1 \end{bmatrix} +e^{3t}\begin{bmatrix} 2\\2 \end{bmatrix}$

$\mathbf{x} = e^t\begin{bmatrix} -3\\-3 \end{bmatrix} +e^{4t}\begin{bmatrix} 4\\2 \end{bmatrix}$

$\mathbf{x} = e^t\begin{bmatrix} 1\\4 \end{bmatrix} +e^{4t}\begin{bmatrix} 5\\2 \end{bmatrix}$

Correct answer:

$\mathbf{x} = e^t\begin{bmatrix} -3\\-3 \end{bmatrix} +e^{4t}\begin{bmatrix} 4\\2 \end{bmatrix}$

Explanation:

To solve the homogeneous system, we will need a fundamental matrix. Specifically, it will help to get the matrix exponential. To do this, we will diagonalize the matrix. First, we will find the eigenvalues which we can do by calculating the determinant of $\lambda I - A$ .

$|\lambda I - A| = \begin{vmatrix} \lambda - 3&2 \\ 1&\lambda - 2 \end{vmatrix} = (\lambda -3)(\lambda -2) - 2 = \lambda^2 - 5\lambda +4 = (\lambda -4)(\lambda -1)$

Finding the eigenspaces, for lambda = 1, we have

$\begin{bmatrix} -2&2 &\vline &0 \\ 1&-1 &\vline & 0 \end{bmatrix}$

Adding -1/2 Row 1 to Row 2 and dividing by -1/2, we have $\begin{bmatrix} 1&-1 &\vline &0 \\ 0&0 &\vline & 0 \end{bmatrix}$ which means x_1 + x_2 = 0

Thus, we have an eigenvector of $\begin{bmatrix} 1\\1 \end{bmatrix}$ .

For lambda = 4

$\begin{bmatrix} 1&-2 &\vline &0 \\ -1&2 &\vline & 0 \end{bmatrix}$

Adding Row 1 to Row 2, we have

$\begin{bmatrix} 1&-2 &\vline &0 \\ 0&0 &\vline & 0 \end{bmatrix}$

So x_1 - 2x_2 = 0 with an eigenvector $\begin{bmatrix} 2\\1 \end{bmatrix}$ .

Thus, we have $P = \begin{bmatrix} 1&2 \\ 1&1 \end{bmatrix}$ and $D = \begin{bmatrix} 1&0 \\ 0&4 \end{bmatrix}$ . Using the inverse formula for 2x2 matrices, we have that $P^{-1} = \begin{bmatrix} -1&2 \\ 1&-1 \end{bmatrix}$ . As we know that $e^{tA} = Pe^{tD}P^{-1}$ , we have $e^{tA} = \begin{bmatrix} 1&2 \\ 1&1 \end{bmatrix}\begin{bmatrix} e^t&0 \\ 0&e^{4t} \end{bmatrix}\begin{bmatrix} -1&2 \\ 1&-1 \end{bmatrix}= \begin{bmatrix} -e^t + 2e^{4t}&2e^{t} -2e^{4t} \\ -e^t + e^{4t}&2e^t-e^{4t} \end{bmatrix}$

The solution to a homogenous system of linear equations is simply to multiply the matrix exponential by the intial condition. For other fundamental matrices, the matrix inverse is needed as well.

Thus, our final answer is $\begin{bmatrix} -e^t + 2e^{4t}&2e^{t} -2e^{4t} \\ -e^t + e^{4t}&2e^t-e^{4t} \end{bmatrix}\begin{bmatrix} 1\\-1 \end{bmatrix} = \begin{bmatrix} -3e^t + 4e^{4t}\\-3e^t + 2e^{4t} \end{bmatrix}= e^t\begin{bmatrix} -3\\-3 \end{bmatrix} +e^{4t}\begin{bmatrix} 4\\2 \end{bmatrix}$

Report an Error

Example Question #1 : Homogeneous Linear Systems

Solve the homogenous equation:

y''-6y'+8y=0

With the initial conditions:

y(0)=2

y'(0)=5

Possible Answers:

$y(t)=C_{1}e^{-4t}+C_{2}e^{-2t}$

none of these answers

$y(t)=\frac{e^{-4t}}{2}+\frac{3e^{-2t}}{2}$

$y(t)=\frac{e^{4t}}{2}+\frac{3e^{2t}}{2}$

$y(t)=C_{1}e^{4t}+C_{2}e^{2t}$

Correct answer:

$y(t)=\frac{e^{4t}}{2}+\frac{3e^{2t}}{2}$

Explanation:

So this is a homogenous, second order differential equation. In order to solve this we need to solve for the roots of the equation. This equation can be written as:

r^2-6r+8=0 Which, using the quadratic formula or factoring gives us roots of $r_{1}=4$ and $r_{2}=2$

The solution of homogenous equations is written in the form:

$y(t)=C_{1}e^{r_{1}t}+C_{2}e^{r_{2}t}$ so we don't know the constants, but can substitute the values we solved for the roots:

$y(t)=C_{1}e^{4t}+C_{2}e^{2t}$

We have two initial values, one for y(t) and one for y'(t), both with t=0\

So:

$y(0)=2=C_{1}+C_{2}$

$y'(t)=4C_{1}e^{4t}+2C_{2}e^{2t}$ so: $y'(0)=5=4C_{1}+2C_{2}$

We can solve for $C_{2}$ : $2-C_{1}=C_{2}$ Then plug into the other equation to solve for $C_{1}$

$5=4C_{1}+2(2-C_{1})=4C_{1}+4-2C_{1}=2C_{1}+4$

So, solving, we get: $C_{1}=\frac{1}{2}$ Then $C_2=\frac{3}{2}$

This gives a final answer of:

$y(t)=\frac{e^{4t}}{2}+\frac{3e^{2t}}{2}$

Report an Error

Example Question #2 : System Of Linear First Order Differential Equations

Solve the second order differential equation:

3y''-3y'-6y=0

Subject to the initial values:

y(0)=4

y'(0)=2

Possible Answers:

none of these answers

$y(t)=2e^{2t}+2e^{-t}$

$y(t)=3e^{2t}+e^{-t}$

$y(t)=2e^{2t}-2e^{-t}$

$y(t)=3e^{2t}-e^{-t}$

Correct answer:

$y(t)=2e^{2t}+2e^{-t}$

Explanation:

So this is a homogenous, second order differential equation. In order to solve this we need to solve for the roots of the equation. This equation can be written as:

3r^2-3r-6=0 Which, using the quadratic formula or factoring gives us roots of $r_{1}=2$ and $r_{2}=-1$

The solution of homogenous equations is written in the form:

$y(t)=C_{1}e^{r_{1}t}+C_{2}e^{r_{2}t}$ so we don't know the constants, but can substitute the values we solved for the roots:

$y(t)=C_{1}e^{2t}+C_{2}e^{-t}$

We have two initial values, one for y(t) and one for y'(t), both with t=0

So:

$y(0)=4=C_{1}+C_{2}$

$y'(t)=2C_{1}e^{2t}-C_{2}e^{-t}$ so: $y'(0)=2=2C_{1}-C_{2}$

We can solve $4-C_{1}=C_{2}$ Then plug into the other equation to solve for $C_{1}$

$2=2C_{1}-(4-C_{1})=2C_{1}+C_{1}-4=3C_{1}-4$

So, solving, we get: $C_{1}=2$ Then C_2=2

This gives a final answer of:

$y(t)=2e^{2t}+2e^{-t}$

Report an Error

Example Question #1 : Homogeneous Linear Systems

Solve the differential equation for y:

y'-6y=0

Subject to the initial condition:

y(0)=10

Possible Answers:

$y(t)=-10e^{6t}$

$y(t)=-6e^{10t}$

$y(t)=6e^{10t}$

$y(t)=10e^{6t}$

$y(t)=10e^{-6t}$

Correct answer:

$y(t)=10e^{6t}$

Explanation:

So this is a homogenous, first order differential equation. In order to solve this we need to solve for the roots of the equation. This equation can be written as:

r-6=0 gives us a root of $r_{1}=6$

The solution of homogenous equations is written in the form:

$y(t)=C_{1}e^{r_{1}t}$ so we don't know the constant, but can substitute the values we solved for the root:

$y(t)=C_{1}e^{6t}$

We have one initial values, for y(t) with t=0

So:

$y(0)=10=C_{1}$

This gives a final answer of:

$y(t)=10e^{6t}$

Report an Error

Example Question #1 : System Of Linear First Order Differential Equations

Solve the third order differential equation:

y'''+y''-12y'=0

y(0)=1

y'(0)=2

y''(0)=3

Possible Answers:

$y(t)=\frac{7e^t}{12}+\frac{11e^{3t}}{21}+\frac{3e^{-4t}}{28}$

$y(t)=\frac{7}{12}+\frac{11e^{3t}}{21}-\frac{3e^{-4t}}{28}$

$y(t)=\frac{7e^t}{12}+\frac{11e^{3t}}{21}-\frac{3e^{-4t}}{28}$

$y(t)=7e^t+11e^{3t}-3e^{-4t}}$

none of these answers

Correct answer:

$y(t)=\frac{7}{12}+\frac{11e^{3t}}{21}-\frac{3e^{-4t}}{28}$

Explanation:

So this is a homogenous, third order differential equation. In order to solve this we need to solve for the roots of the equation. This equation can be written as:

r^3+r^2-12r=0 Which, using the cubic formula or factoring gives us roots of $r_{1}=0$ , $r_{2}=3$ and $r_{3}=-4$

The solution of homogenous equations is written in the form:

$y(t)=C_{1}e^{r_{1}t}+C_{2}e^{r_{2}t}+C_3e^{r_3t}$ so we don't know the constants, but can substitute the values we solved for the roots:

$y(t)=C_{1}+C_{2}e^{3t}+C_3e^{-4t}$

We have three initial values, one for y(t), one for y'(t), and for y''(t) all with t=0

So:

$y(0)=1=C_{1}+C_{2}+C_3$

$y'(t)=3C_{2}e^{3t}-4C_{3}e^{-4t}$ so: $y'(0)=2=3C_{2}-4C_{3}$

y''(0)=3=9C_2+16C_3

So this can be solved either by substitution or by setting up a 3X3 matrix and reducing. Once you do either of these methods, the values for the constants will be: $C_{1}=\frac{7}{12}$ Then $C_2=\frac{11}{21}$ and $C_3=\frac{-3}{28}$

This gives a final answer of:

$y(t)=\frac{7}{12}+\frac{11e^{3t}}{21}-\frac{3e^{-4t}}{28}$

Report an Error

Example Question #1 : System Of Linear First Order Differential Equations

Solve the differential equation:

y'''-4y''-4y'+16y=0

Subject to the initial conditions:

y(0)=2

y'(0)=2

y''(0)=3

Possible Answers:

$y(t)=\frac{17e^{2t}}{8}+\frac{7e^{-2t}}{24}+\frac{5e^{4t}}{12}$

$y(t)=\frac{17e^{2t}}{8}-\frac{7e^{-2t}}{24}+\frac{5e^{4t}}{12}$

$y(t)=-\frac{17e^{2t}}{8}+\frac{7e^{-2t}}{24}-\frac{5e^{4t}}{12}$

$y(t)=\frac{17e^{2t}}{8}+\frac{7e^{-2t}}{24}-\frac{5e^{4t}}{12}$

$y(t)=-\frac{17e^{2t}}{8}-\frac{7e^{-2t}}{24}-\frac{5e^{4t}}{12}$

Correct answer:

$y(t)=\frac{17e^{2t}}{8}+\frac{7e^{-2t}}{24}-\frac{5e^{4t}}{12}$

Explanation:

So this is a homogenous, third order differential equation. In order to solve this we need to solve for the roots of the equation. This equation can be written as:

r^3-4r^2-4r+16=0 Which, using the cubic formula or factoring gives us roots of $r_{1}=2$ , $r_{2}=-2$ and $r_{3}=4$

The solution of homogenous equations is written in the form:

$y(t)=C_{1}e^{r_{1}t}+C_{2}e^{r_{2}t}+C_3e^{r_3t}$ so we don't know the constants, but can substitute the values we solved for the roots:

$y(t)=C_{1}e^{2t}+C_{2}e^{-2t}+C_3e^{4t}$

We have three initial values, one for y(t), one for y'(t), and for y''(t) all with t=0

So:

$y(0)=2=C_{1}+C_{2}+C_3$

$y'(t)=2C_{1}e^{2t}-2C_{2}e^{-2t}+4C_3e^{4t}$ $y'(0)=2=2C_{1}-2C_{2}+4C_3$

y''(0)=3=4C_1+4C_2+16C_3

So this can be solved either by substitution or by setting up a 3X3 matrix and reducing. Once you do either of these methods, the values for the constants will be: $C_{1}=\frac{17}{8}$ Then $C_2=\frac{7}{24}$ and $C_3=\frac{-5}{12}$

This gives a final answer of:

$y(t)=\frac{17e^{2t}}{8}+\frac{7e^{-2t}}{24}-\frac{5e^{4t}}{12}$

Report an Error

Example Question #2 : System Of Linear First Order Differential Equations

Find the general solution to the given system.

$\\\frac{dx}{dt}=x+2y \\\\\frac{dy}{dt}=2x+y$

Possible Answers:

$X=c_1\binom{-1}{1}e^{3t}+c_2\binom{1}{3}e^{-t}$

$X=c_1\binom{1}{1}e^{3t}+c_2\binom{-1}{3}e^{-t}$

$X=c_1\binom{1}{1}e^{3t}+c_2\binom{1}{3}e^{-t}$

$X=c_1\binom{1}{-1}e^{3t}+c_2\binom{-1}{3}e^{-t}$

$X=c_1\binom{1}{3}e^{3t}+c_2\binom{1}{1}e^{-t}$

Correct answer:

$X=c_1\binom{1}{1}e^{3t}+c_2\binom{-1}{3}e^{-t}$

Explanation:

To find the general solution to the given system

$\\\frac{dx}{dt}=x+2y \\\\\frac{dy}{dt}=2x+y$

first find the eigenvalues and eigenvectors.

$det(A-\lambda I )=\begin{vmatrix} 1-\lambda&2 \\ 2&1-\lambda \end{vmatrix}$

$\\=(1-\lambda)(1-\lambda)-(2)(2) \\=1-2\lambda+\lambda^2-4 \\=\lambda^2-2\lambda-3 \\=(\lambda-3)(\lambda+1)$

Therefore the eigenvalues are

$\lambda_1=3, \lambda_2=-1$

Now calculate the eigenvectors

$(A-\lamda I)K=0$

For

$\lambda_1=3$

$\\(1-3)k_1+2k_2 \\2k_1+(1-3)k_2$

Thus,

$k_1=k_2 \rightarrow K_1\binom{1}{1}$

For

$\lambda_1=-1$

$\\(2--1)k_1+1k_2 \\1k_1+(2--1)k_2$

Thus

$k_1=-\frac{1}{3}k_2\rightarrow K_2\binom{-1}{3}$

Therefore,

$X_1=\binom{1}{1}e^{3t};\ X_2\binom{-1}{3}e^{-t}$

Now the general solution is,

$\\X=c_1X_1+c_2X_2 \\\\X=c_1\binom{1}{1}e^{3t}+c_2\binom{-1}{3}e^{-t}$

Report an Error

Example Question #2 : Homogeneous Linear Systems

When substituted into the homogeneous linear system $\mathbf{x'} = A\mathbf{x}$ for , which of the following matrices will have a saddle point equilibrium in its phase plane?

Possible Answers:

$\begin{bmatrix} 3 & 5\\ 0& 1 \end{bmatrix}$

$\begin{bmatrix} -1 & -1\\ 3& -3 \end{bmatrix}$

$\begin{bmatrix} 3 & 0\\ 9& 1 \end{bmatrix}$

$\begin{bmatrix} -4 & 2\\ -7& 5 \end{bmatrix}$

$\begin{bmatrix} -2 & 0\\ 7& -1 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -4 & 2\\ -7& 5 \end{bmatrix}$

Explanation:

A saddle point phase plane results from two real eigenvalues of different signs. Three of these matrices are triangular, which means their eigenvalues are on the diagonal. For these three, the eigenvalues are real, but both the same sign, meaning they don't have saddles. For the remaining two, we'll need to find the eigenvalues using the characteristic equations.

For $\begin{bmatrix} -1 & -1\\ 3& -3 \end{bmatrix}$ , we have

$\begin{vmatrix} \lambda + 1 & 1 \\ - 3& \lambda + 3 \end{vmatrix} = (\lambda + 1)(\lambda + 3) + 3 = \lambda^2 + 4\lambda + 6=0$

The discriminant to this is $4^2 - 4 \cdot 1 \cdot 6 = -8$ , so the solutions are non-real. Thus, this matrix doesn't yield a saddle point.

For $\begin{bmatrix} -4 & 2\\ -7& 5 \end{bmatrix}$ we have,

$\begin{vmatrix} \lambda + 4 & -2 \\ 7& \lambda - 5 \end{vmatrix} = (\lambda +4)(\lambda - 5) + 14 = \lambda^2 -\lambda - 6=(\lambda - 3)(\lambda + 2) = 0$

We see that this matrix yields two real eigenvalues with different signs. Thus, it is the correct choice.

Report an Error

Example Question #1 : Homogeneous Linear Systems

Find the general solution to the system of ordinary differential equations

$\widehat{x}^{'}(t) = A \widehat{x}(t)$

where

$A = \begin{bmatrix} 1 & 3\\ 12 & 1 \end{bmatrix}$

Possible Answers:

$\widehat{x}(t) = C_1e^{-5t}\begin{bmatrix} 1\\ 2 \end{bmatrix} + C_2e^{7t}\begin{bmatrix} 1\\ -2 \end{bmatrix}$

$\widehat{x}(t) = C_1e^{-5t}\begin{bmatrix} 1\\ 2 \end{bmatrix} + C_2e^{7t}\begin{bmatrix} 1\\ 2 \end{bmatrix}$

$\widehat{x}(t) = C_1e^{5t}\begin{bmatrix} 1\\ 2 \end{bmatrix} + C_2e^{-7t}\begin{bmatrix} 1\\ -2 \end{bmatrix}$

None of the other answers.

$\widehat{x}(t) = C_1e^{-5t}\begin{bmatrix} 1\\ -2 \end{bmatrix} + C_2e^{7t}\begin{bmatrix} 1\\ 2 \end{bmatrix}$

Correct answer:

$\widehat{x}(t) = C_1e^{-5t}\begin{bmatrix} 1\\ -2 \end{bmatrix} + C_2e^{7t}\begin{bmatrix} 1\\ 2 \end{bmatrix}$

Explanation:

Finding the eigenvalues and eigenvectors of with the characteristic equation of the matrix

$(1-\lambda)^2 - 36 = 0 \implies \lambda = -5, 7$

The corresponding eigenvalues are, respectively

$\begin{bmatrix} 1\\ -2 \end{bmatrix}$ and $\begin{bmatrix} 1\\ 2 \end{bmatrix}$

This gives us that the general solution is

$\widehat{x}(t) = C_1e^{-5t}\begin{bmatrix} 1\\ -2 \end{bmatrix} + C_2e^{7t}\begin{bmatrix} 1\\ 2 \end{bmatrix}$

Report an Error

View Differential Equations Tutors

Muhammad
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Mechanical Engineering.

View Differential Equations Tutors

Omar
Certified Tutor

Arizona State University, Bachelors, Civil Engineering.

View Differential Equations Tutors

Yoonsik
Certified Tutor

Seoul National University, Bachelor of Science, Physics. University of Pennsylvania, Doctor of Philosophy, Atomic and Molecul...

All Differential Equations Resources

1 Diagnostic Test 29 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

GRE Tutors in Denver, French Tutors in Dallas Fort Worth, Physics Tutors in Phoenix, Computer Science Tutors in Denver, Math Tutors in Miami, Biology Tutors in Phoenix, GMAT Tutors in San Diego, MCAT Tutors in Atlanta, Reading Tutors in Chicago, SAT Tutors in Denver

Popular Courses & Classes

GMAT Courses & Classes in New York City, MCAT Courses & Classes in Atlanta, ISEE Courses & Classes in Los Angeles, MCAT Courses & Classes in Los Angeles, MCAT Courses & Classes in Dallas Fort Worth, Spanish Courses & Classes in Miami, Spanish Courses & Classes in Denver, SAT Courses & Classes in Denver, SAT Courses & Classes in Boston, GRE Courses & Classes in Dallas Fort Worth

Popular Test Prep

MCAT Test Prep in Washington DC, LSAT Test Prep in Washington DC, ISEE Test Prep in Boston, SAT Test Prep in Seattle, SAT Test Prep in New York City, GMAT Test Prep in Atlanta, SSAT Test Prep in San Francisco-Bay Area, LSAT Test Prep in Philadelphia, MCAT Test Prep in Atlanta, MCAT Test Prep in New York City

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Differential Equations : System of Linear First-Order Differential Equations

Study concepts, example questions & explanations for Differential Equations

All Differential Equations Resources

Example Questions

Example Question #1 : System Of Linear First Order Differential Equations

Example Question #1 : Homogeneous Linear Systems

Example Question #1 : Homogeneous Linear Systems

Example Question #2 : System Of Linear First Order Differential Equations

Example Question #1 : Homogeneous Linear Systems

Example Question #1 : System Of Linear First Order Differential Equations

Example Question #1 : System Of Linear First Order Differential Equations

Example Question #2 : System Of Linear First Order Differential Equations

Example Question #2 : Homogeneous Linear Systems

Example Question #1 : Homogeneous Linear Systems

All Differential Equations Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: