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Differential Equations : Differential Equations

Study concepts, example questions & explanations for Differential Equations

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

Example Questions

Example Question #1 : Initial Value Problems

y''=2

y'(0)=6

y(0)=0

Solve for y

Possible Answers:

y=x^2+6x

None of these answers

y=x^2

y=x^2+6x+6

y=x^2-6x

Correct answer:

y=x^2+6x

Explanation:

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

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

Taking the anti-derivative once, we get:

y'=2x+C Then using the initial condition y'(0)=6

We get that C=6

So y'=2x+6 Then taking the antiderivative one more time, we get:

y=x^2+6x+D and using the initial condition y(0)=0

we get the final answer of:

y=x^2+6x

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Example Question #8 : Initial Value Problems

If is some constant and the initial value of the function, y=ce^x is six, determine the equation.

Possible Answers:

y=6e^x

y=0

y=6e^x+c

y=6ce^x

y=e^x

Correct answer:

y=6e^x

Explanation:

First identify what is known.

The general function is,

y=ce^x

The initial value is six in mathematical terms is,

y(0)=6

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.

y=6e^x

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Example Question #1 : Initial Value Problems

Solve the differential equation for y

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

subject to the initial condition:

y(1)=3

Possible Answers:

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

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

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

y^2=x^3+x^2+6x+1

none of these answers

Correct answer:

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

Explanation:

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

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

2ydy=3x^2+2x+6dx

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

y^2=x^3+x^2+6x+C

Plug in the initial condition y(1)=3

9=1+1+6+C=8+C

Solving for C:

1=C

Which gives us: y^2=x^3+x^2+6x+1

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

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

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Example Question #1 : Initial Value Problems

If is some constant and the initial value of the function, y=ce^x is six, determine the equation.

Possible Answers:

y=e^x

y=0

y=6e^x+c

y=6ce^x

y=6e^x

Correct answer:

y=6e^x

Explanation:

First identify what is known.

The general function is,

y=ce^x

The initial value is six in mathematical terms is,

y(0)=6

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.

y=6e^x

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Example Question #11 : Introduction To Differential Equations

Solve the following initial value problem: y' = 10(y - t) , y(0) = 1 .

Possible Answers:

$y = t + \frac{1}{10} + \frac{9e^{10t}}{10}$

$y = t + 10e^{10t}$

$y = t + \frac{9e^{10t}}{10}$

$y = t + \frac{1}{10} + 10e^{10t}$

$y = t + \frac{1}{10}$

Correct answer:

$y = t + \frac{1}{10} + \frac{9e^{10t}}{10}$

Explanation:

This type of problem will require an integrating factor. First, let's get it into a better form: y' - 10y = -10t . Once we have an equation in the form y' + p(t)y = q(t) , 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 u = -10t 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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Example Question #11 : Introduction To Differential Equations

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 D(t) 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.

Possible Answers:

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

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

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

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

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

Correct answer:

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

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,

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

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Example Question #2 : Mathematical Models

Possible Answers:

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

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

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

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

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

Correct answer:

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

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,

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

Report an Error

Example Question #2 : Mathematical Models

Possible Answers:

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

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

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

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

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

Correct answer:

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

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,

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

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Example Question #1 : Numerical Solutions Of Ordinary Differential Equations

Use Euler's Method to calculate the approximation of y(0.2) where y(x) is the solution of the initial-value problem that is as follows.

y''+xy'+y=0

y(0)=2, y'(0)=3

Possible Answers:

$y(0.2)\approx2.542$

$y(0.2)\approx2.5$

$y(0.2)\approx2.458$

$y(0.2)\approx2.58$

$y(0.2)\approx2.584$

Correct answer:

$y(0.2)\approx2.58$

Explanation:

Using Euler's Method for the function

y''+xy'+y=0

y(0)=2, y'(0)=3

first make the substitution of

$\\y'=u \\u'=-xu-y$

therefore

$\\y_{n+1}=y_n+hu_n \\u_{n+1}=u_n+h[-x_nu_n-y_n]$

where represents the step size.

Let

$\\h=0.1 \\y_0=2 \\u_0=3$

Substitute these values into the previous formulas and continue in this fashion until the approximation for y(0.2) is found.

$\\y_1=y_0+(0.1)u_0=2+(0.1)(3)=2.3 \\u_1=u_0+(0.1)[-x_0u_0-y_0]=3+(0.1)[0(2)-2]=2.8 \\y_2=y_1+(0.1)u_1=2.3+(0.1)(2.8)=2.58 \\u_2=u_1+(0.1)[-x_1u_1-y_1]=2.8+(0.1)[-0.1(2.8)-2.3]=2.542$

Therefore,

$y(0.2)\approx2.58$

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Example Question #1 : Numerical Solutions Of Ordinary Differential Equations

Approximate $y(\pi)$ for $y = \cos(y)(2-y)$ with time steps $h = \frac{\pi}{4}$ and y(0) = 0 .

Possible Answers:

$\frac{5\pi}{8}$

$\frac{\pi}{4}$

$\frac{\pi}{2}$

$\frac{3\pi}{2}$

Correct answer:

$\frac{\pi}{2}$

Explanation:

Approximate $y(\pi)$ for $y = \cos(y)(2-y)$ with time steps $h = \frac{\pi}{4}$ and y(0) = 0 .

The formula for Euler approximations $\mu(t+h) = \mu (t) + h \cdot f(t, \mu(t))$ .

Plugging in, we have

$\mu(\frac{\pi}{4}) = \mu (0) + \frac{\pi}{4}f(0, \mu(0)) = 0 + \frac{\pi}{4}(\cos(0)(2-0)) = \frac{\pi}{2}$

$\mu(\frac{\pi}{2}) = \mu (\frac{\pi}{4}) + \frac{\pi}{4}f(\frac{\pi}{4}, \mu(\frac{\pi}{4})) = \frac{\pi}{2} + \frac{\pi}{4}(\cos(\frac{\pi}{2})(2-\frac{\pi}{2})) = \frac{\pi}{2} + 0 = \frac{\pi}{2}$

Here we can see that we've gotten trapped on a horizontal tangent (a failing of Euler's method when using larger time steps). As the function is not dependent on t, we will continue to move in a horizontal line for the rest of our Euler approximations. Thus $\mu(\pi) = \frac{\pi}{2} \approx y(\pi)$ .

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Differential Equations : Differential Equations

Study concepts, example questions & explanations for Differential Equations

All Differential Equations Resources

Example Questions

Example Question #1 : Initial Value Problems

Example Question #8 : Initial Value Problems

Example Question #1 : Initial Value Problems

Example Question #1 : Initial Value Problems

Example Question #11 : Introduction To Differential Equations

Example Question #11 : Introduction To Differential Equations

Example Question #2 : Mathematical Models

Example Question #2 : Mathematical Models

Example Question #1 : Numerical Solutions Of Ordinary Differential Equations

Example Question #1 : Numerical Solutions Of Ordinary Differential Equations

All Differential Equations Resources

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