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Differential Equations : Introduction to Differential Equations

Study concepts, example questions & explanations for Differential Equations

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Example Questions

Example Question #1 : Initial Value Problems

Solve the separable differential equation

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

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

Possible Answers:

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

none of these answers

y=arcsin(cos(x))

$y=arcsin(-cos(x)+\frac{\sqrt{2}}{4})$

$y=arcsin(-cos(x)-\frac{\sqrt{2}}{4})$

Correct answer:

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

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.

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.

cos(y)dy=sin(x)dx Integrating, we get:

sin(y)=-cos(x)+C 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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Example Question #1 : Initial Value Problems

Solve the differential equation

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

Subject to: y(0)=0

Possible Answers:

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

$y=\sqrt{\frac{x^3}{3}+\frac{2x^{5}}{4}}$

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

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

none of these answers

Correct answer:

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

Explanation:

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 x^2 and pulling it outside of the square root to get:

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

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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,

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

Report an Error

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Differential Equations : Introduction to Differential Equations

Study concepts, example questions & explanations for Differential Equations

All Differential Equations Resources

Example Questions

Example Question #1 : Initial Value Problems

Example Question #1 : Initial Value Problems

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

All Differential Equations Resources

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