Differential Equations

Solve the second order differential equation:

Subject to the initial values:

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

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

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

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

none of these answers

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:

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

This gives a final answer of:

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

Suppose a dog is carrying a virus returns to a isolated doggy day care of 40 dogs. Determine the differential equation for the number of dogs who have contracted the virus if the rate at which it spreads is proportional to the number of interactions between the dogs with the virus and the dogs that have not yet come in contact with the virus.

$\frac{dD}{dt}=kD(40-D)$

$\frac{dD}{dt}=kD$

$\frac{dD}{dt}=k(40-D)$

$\frac{dD}{dt}=k40-D^2$

$\frac{dD}{dt}=kP$

Explanation

This question is asking a population dynamic type of scenario.

The total population in terms of time and where is the constant rate of proportionality, is described by the following differential equation.

$\frac{dP}{dt}=kP$

For this particular function it's known that the population is in the form to represent the dogs. Since the virus spreads based on the interactions between the dogs who have the virus and those who have not yet contracted it, the population will be the product of the number of dogs with the virus and those without the virus.

Therefore the differential equation becomes,

$\frac{dD}{dt}=kD(40-D)$

Solve for y

None of these answers

Explanation

So this is a separable differential equation. We can think of as $\frac{d^2y}{dx^2}$

and as $\frac{dy}{dx}$ .

Taking the anti-derivative once, we get:

Then using the initial condition

We get that

So Then taking the antiderivative one more time, we get:

and using the initial condition

we get the final answer of:

State the order of the given differential equation and determine if it is linear or nonlinear.

Third ordered, linear

Third ordered, nonlinear

Second ordered, linear

Second ordered, nonlinear

Fourth ordered, linear

Explanation

This problem contains two questions that need to be solved for: order of the differential equation and whether it is linear or nonlinear.

To determine the order of the differential equation, look for the highest derivative in the equation.

For this particular function recall that,

$\y'=\frac{dy}{dx} \\y''=\frac{d^2y}{dx^2} \\y'''=\frac{d^3y}{dx^3}$

therefore the highest derivative is three which makes the equation a third ordered differential equation.

The second part of this problem is to determine if the equation is linear or nonlinear. For a differential equation to be linear two characteristics must hold true:

1. The dependent variable and all its derivatives have a power involving one.

2. The coefficients depend on the independent variable .

Looking at the given function,

it is seen that all the variable and all its derivatives have a power involving one and all the coefficients depend on therefore, this differential equation is linear.

To answer this problem completely, the differential equation is a linear, third ordered equation.

Solve the differential equation:

Subject to the initial conditions:

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

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

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

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

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

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:

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:

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

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

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:

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

Solve the differential equation for y:

Subject to the initial condition:

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

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

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

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

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

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:

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:

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

$\frac{dD}{dt}=kD(40-D)$

$\frac{dD}{dt}=kD$

$\frac{dD}{dt}=k(40-D)$

$\frac{dD}{dt}=k40-D^2$

$\frac{dD}{dt}=kP$

Explanation

This question is asking a population dynamic type of scenario.

The total population in terms of time and where is the constant rate of proportionality, is described by the following differential equation.

$\frac{dP}{dt}=kP$

Therefore the differential equation becomes,

$\frac{dD}{dt}=kD(40-D)$

Solve the homogenous equation:

With the initial conditions:

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

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

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

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

none of these answers

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:

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:

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

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

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

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

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

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

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

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

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

Solve the third order differential equation:

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

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

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

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

none of these answers

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:

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: $y'(0)=2=3C_{2}-4C_{3}$

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 $C_3=\frac{-3}{28}$