Introduction to Differential Equations - Differential Equations

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Question

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

Answer

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.

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Question

Which of the following three equations enjoy both local existence and uniqueness of solutions for any initial conditions?

$2: y' = y^{2/3}$

$3: y' = y^{4/3}$

Answer

By the cauchy-peano theorem, for , as long as is continuous on a closed rectangle around our starting point, we have local existence. All three functions are continuous everywhere, so they enjoy local existence at every starting point.

We can show that the solutions to differential equations are unique by showing that is Lipschitz continuous in y. If $\frac{\partial f}{\partial y}$ is continuous, then this will suffice to show the Lipschitz continuity.

$\frac{\partial \sin(y)}{\partial y} = \cos(y)$

$\frac{\partial y^{2/3}}{\partial y} = \frac{2}{3}y^{-1/3}$

$\frac{\partial y^{4/3}}{\partial y} = \frac{4}{3}y^{1/3}$

Note that the first and third equations are continuous for all y and t, but that the second is not continuous when . More concretely, when , both the equation and the equation $y = \frac{t^3}{27}$ would satisfy the differential equation.

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Question

Which of the following definitions describe an autonomous differential equation.

Answer

By definition, an autonomous differential equation does not depend explicitly on the independent variable. An autonomous differential equation will take the form

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Question

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

Answer

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.

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Question

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

Answer

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.

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Question

Find Order and Linearity of the following differential equation

$\frac{d^3y}{dt^3} + t\frac{d^2y}{dt^2} + (\cos^2{t})y = t^3$

Answer

This equation is third order since that is the highest order derivative present in the equation.

This is equation in linear because and derivatives appear to the first power only. and $\cos^2{t}$ do not affect the linearity.

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Question

If is some constant and the initial value of the function, is six, determine the equation.

Answer

First identify what is known.

The general function is,

The initial value is six in mathematical terms is,

From here, substitute in the initial values into the function and solve for .

$\y=ce^x \6=ce^0 \6=c\cdot 1 \6=c$

Finally, substitute the value found for into the original equation.

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Question

$\frac{dy}{dx}=\frac{4x^2-7x}{3y^2+2}$

Answer

So this is a separable differential equation, but it is also subject to an initial condition. This means that you have enough information so that there should not be a constant in the final answer.

You start off by getting all of the like terms on their respective sides, and then taking the anti-derivative. Your pre anti-derivative equation will look like:

$\int3y^2+2dy=\int4x^2-7dx$

Then taking the anti-derivative, you include a C value:

$y^3+2y=\frac{4}{3}x^3-\frac{7}{2}x^2+C$

Then, using the initial condition given, we can solve for the value of C:

$(1)^3+2(1)=\frac{4}{3}(1)^3-\frac{7}{2}(1)^2+C$

Solving for C, we get

$3=-\frac{13}{6}+C$

$C=\frac{31}{6}$ which gives us a final answer of:

$y^3+2y=\frac{4}{3}x^3-\frac{7}{2}x^2+\frac{31}{6}$

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Question

Solve for y

Answer

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:

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Question

Solve the initial value problem for .

Answer

We have $\frac{y'}{y} = 10$ so that $\int \frac{y'}{y}dt = \int10dt$ and $\ln|y| = 10t + C$ . Solving for y,

$|y| = e^{10t + C} = e^{C}e^{10t}$ and $y = \pm e^{C}e^{10t} = ce^{10t}$

which we can write because $\pm e^{C}$ is just another arbitrary constant.

Plugging in our initial value, we have $5 = ce^{0} = c$ leaving us with a final answer of $y = 5e^{10t}$ .

Note, this type of equation pops up frequently in the course and is potentially good to just memorize. For , we have $y = by^{at}$

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Question

$\frac{dy}{dx}=\frac{e^{2x}}{e^{y}}$

With

Answer

So this is a separable differential equation with a given initial value.

To start off, gather all of the like variables on separate sides.

$e^{y}dy=e^{2x}dx$

Then integrate, and make sure to add a constant at the end

$e^{y}=\frac{1}{2}e^{2x}+C$ To solve for y, take the natural log, ln, of both sides

$y=ln(\frac{1}{}2e^{2x}+C)$ Be careful not to separate this, a log(a+b) can't be separated.

Plug in the initial condition to get:

$4=ln(\frac{1}{2}e^{4}+C)$ So raising e to the power of both sides:

$e^{4}=\frac{1}{2}e^{4}+C$ Solving for C:

$\frac{1}{2}e^{4}=C$ giving us a final answer of:

$y=ln(\frac{e^{2x}}{2}+\frac{e^{4}}{2})$

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Question

Solve the separable differential equation

$\frac{dy}{dx}=\frac{sec(y)}{csc(x)}$

with the initial condition $y(\frac{\pi }{4})=\frac{\pi }{4}$

Answer

So this is a separable differential equation with a given initial value.

To start off, gather all of the like variables on separate sides.

Notice that when you divide sec(y) to the other side, it will just be cos(y),

and the csc(x) on the bottom is equal to sin(x) on the top.

Integrating, we get:

so we can plug pi/4 into both x and y:

$\frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{2}+C$ this gives us a C value of $\sqrt{2}$

$sin(y)=-cos(x)+\sqrt{2}$

In order to solve for y, we just need to take the arcsin of both sides:

$y=arcsin(-cos(x)+\sqrt{2})$

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Question

Solve the differential equation

$-2y\frac{dy}{dx}+x^2+2x^4=0$

Subject to:

Answer

So if you rearrange this equation, you will arrive at a separable differential equation by adding the $-2y\frac{dy}{dx}$ to the other side:

$x^2+2x^4=2y\frac{dy}{dx}$ Now, to solve this, multiply the dx to the other side and take the anti-derivative:

$\int x^2+2x^4dx=\int 2ydy$

Then, after the anti-derivative, make sure to add the constant C:

$\frac{x^3}{3}+\frac{2x^5}{5}+C=y^2$ Now, plug in the initial condition that y(0)=0, which will give you a C=0 as well. Then just take the square root, and you arrive at:

$y=\sqrt{\frac{x^3}{3}+\frac{2x^5}{5}}$ Then, to get the correct answer, simplify by factoring out an and pulling it outside of the square root to get:

$y=x\sqrt{\frac{x}{3}+\frac{2x^{3}}{5}}$

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Question

Solve the differential equation for y

$\frac{dy}{dx}=\frac{3x^2+2x+6}{2y}$

subject to the initial condition:

Answer

So this is a separable differential equation with a given initial value.

To start off, gather all of the like variables on separate sides.

Then integrate, and make sure to add a constant at the end

Plug in the initial condition

Solving for C:

Which gives us:

Then taking the square root to solve for y, we get:

$y=\sqrt{x^3+x^2+6x+1}$

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Question

If is some constant and the initial value of the function, is six, determine the equation.

Answer

First identify what is known.

The general function is,

The initial value is six in mathematical terms is,

From here, substitute in the initial values into the function and solve for .

$\y=ce^x \6=ce^0 \6=c\cdot 1 \6=c$

Finally, substitute the value found for into the original equation.

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Question

If is some constant and the initial value of the function, is six, determine the equation.

Answer

First identify what is known.

The general function is,

The initial value is six in mathematical terms is,

From here, substitute in the initial values into the function and solve for .

$\y=ce^x \6=ce^0 \6=c\cdot 1 \6=c$

Finally, substitute the value found for into the original equation.

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Question

Solve the following initial value problem: , .

Answer

This type of problem will require an integrating factor. First, let's get it into a better form: . Once we have an equation in the form , we find $\mu = e^{\int p dt} = e^{\int -10 dt} = e^{-10t}$ (where the constant of integration is omitted because we only need one, arbitrary integrating factor).

Once we do this, we can see that $(\mu y)' = \mu y' + \mu'y = e^{-10t}y' - 10e^{-10t}y = e^{-10t}(y' - 10t) = e^{-10t}\cdot-10t$

This is simply due to product rule, and then at the end, substitution of the original equation. Thus, as we know that $(\mu y)' = \mu q$ , we can just integrate both sides to find y.

$e^{-10t}y = \int-10te^{-10t}dt$ . A quick application of integration by parts with and $dv = e^{-10t}$ tells us that the right hand side is $te^{-10t} + \frac{e^{-10t}}{10}+C$ . Dividing both sides by mu, we are left with $y = t + \frac{1}{10} + \frac{C}{e^{-10t}}$ . Plugging in the initial condition, we find $1 = \frac{1}{10} + C$ giving us $C = \frac{9}{10}$ . Thus, we have a final answer of $y = t + \frac{1}{10} + \frac{9e^{10t}}{10}$ .

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Question

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.

Answer

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,

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

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Question

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.

Answer

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,

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

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Question

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.

Answer

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,

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

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