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Differential Equations : Homogeneous Linear Systems

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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,

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

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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 $\begin{bmatrix} -e^t + 2e^{4t}&2e^{t} -2e^{4t} \\ -e^t + e^{4t}&2e^t-e^{4t} \end{bmatrix}\begin{bmatrix} 1\\-1 \end{bmatrix} = \begin{bmatrix} -3e^t + 4e^{4t}\\-3e^t + 2e^{4t} \end{bmatrix}= e^t\begin{bmatrix} -3\\-3 \end{bmatrix} +e^{4t}\begin{bmatrix} 4\\2 \end{bmatrix}$

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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:

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

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Example Question #1 : Homogeneous Linear Systems

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:

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

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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:

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

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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:

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

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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:

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

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Example Question #2 : Homogeneous Linear Systems

Find the general solution to the given system.

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

Possible Answers:

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

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

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

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

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

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,

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

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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.

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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

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

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Differential Equations : Homogeneous Linear Systems

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 #1 : Homogeneous Linear Systems

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 : Homogeneous Linear Systems

Example Question #2 : Homogeneous Linear Systems

Example Question #1 : Homogeneous Linear Systems

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