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Calculus AB : Find General and Particular Solutions Using Separation of Variables

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Example Questions

Example Question #1 :Find General And Particular Solutions Using Separation Of Variables

What is separation of variables?

Possible Answers:

Taking the derivative of two variables

A fancy wording for factoring

Integrating with two variables

Using algebra to rewrite a differential equation so that two different variables are on opposite sides

Correct answer:

Using algebra to rewrite a differential equation so that two different variables are on opposite sides

Explanation:

When we are trying to integrate a differential equation, sometimes we have to use a method called separation of variables. This is because we need to integrate with respect to each variable but are unable to do so when they are on the same side. When we are able to get each variable on different sides with their respective differential (i.e.or), then we can integrate each side with respect to each variable.

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Example Question #1 :Find General And Particular Solutions Using Separation Of Variables

When would one need to use separation of variables?

Possible Answers:

When we are solving a function with two different variables at a certain point

We never actually have to use separation of variables it is just a shortcut

When we need to find the derivative of a differential equation

When we need to integrate a differential equation with two different variables

Correct answer:

When we need to integrate a differential equation with two different variables

Explanation:

By using separation of variables we are able to integrate to solve the differential equation.

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Example Question #3 :Find General And Particular Solutions Using Separation Of Variables

Use separation of variables to solve the following differential equation: $\frac{dy}{dx} = 2yx$

Possible Answers:

y +C = x^2 + D

$y = ce^{x^2}$

$\frac{-1}{2y^2} = x^2 + C$

y = ln|x^2| + C

Correct answer:

$y = ce^{x^2}$

Explanation:

We must manipulate this differential equation to get each variable and its own side with its differential. Once we do that we must integrate each side accordingly.

$\frac{dy}{dx} = 2yx$

$\frac{dy}{y} = 2xdx$ (divided byand multiplied by)

$\frac{1}{y}dy = 2xdx$ (just rearranging)

$\int\frac{1}{y}dy = 2\int xdx$ (Let a = D-C )

$ln|y| + C= 2\frac{x^2}{2} + D$

ln|y| = x^2 + a

$e^{ln|y|} = e^{x^2 + a}$

$y = e^{x^2}e^a$ ( e^a is just a constant so we rename it)

$y = ce^{x^2}$

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Example Question #4 :Find General And Particular Solutions Using Separation Of Variables

Use separation of variable to solve the following differential equation: $\frac{dy}{dx} = \frac{4x}{y}$

Possible Answers:

$y = \frac{1}{x^2} + c$

y = 4x^2 + c

$y = \frac{x^2}{2} + c$

y = 8x^2 + c

Correct answer:

y = 4x^2 + c

Explanation:

We must manipulate this differential equation to get each variable and its own side with its differential. Once we do that we must integrate each side accordingly.

$\frac{dy}{dx} = \frac{4x}{y}$

ydy = 4xdx (multiplied byand multiplied by)

$\int y dy = \int 4x dx$

$\frac{y^2}{2} + C = \frac{4x^2}{2} + D$ (Let D-C = a )

$\frac{y^2}{2} = \frac{4x^2}{2} + a$

y = 4x^2 + 2a (is just a constant so we will rename it)

y = 4x^2 + c

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Example Question #5 :Find General And Particular Solutions Using Separation Of Variables

Use separation of variable to solve the following differential equations: $\frac{dy}{dx} = \frac{(2 - 3x^2)^2}{y}$

Possible Answers:

$y = \sqrt{\frac{18x^2}{5} - 8x^3 + 8x + c}$

$y = \frac{18x^2}{5} - 8x^3 + 8x + c$

$y = ln|\frac{18x^2}{5} - 8x^3 + 8x + c|$

$y = \frac{18x^2}{5} - 8x^3 + 8x + c$

Correct answer:

$y = \sqrt{\frac{18x^2}{5} - 8x^3 + 8x + c}$

Explanation:

We must manipulate this differential equation to get each variable and its own side with its differential. Once we do that we must integrate each side accordingly.

$\frac{dy}{dx} = \frac{(2 - 3x^2)^2}{y}$

y dy = (2-3x^2)^2 dx (multiplied byand)

ydy =9x^4 -12x^2+4 dx (expansion)

$\int y dy= \int 9x^4 - 12x^2 + 4 dx$

$\int y dy= \int 9x^4 dx - \int 12x^2 dx + \int 4 dx$

$\frac{y^2}{2} + C = \frac{9x^5} - 4x^3 + 4x + D$ (Let D-C = a )

$y^2 = \frac{18x^2}{5} - 8x^3 + 8x + 2a$ (is a constant so call it)

$y = \sqrt{\frac{18x^2}{5} - 8x^3 + 8x + c}$

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Example Question #6 :Find General And Particular Solutions Using Separation Of Variables

True or False: We can use separation of variables to solve a differential equation at a particular solution.

Possible Answers:

True

False

Correct answer:

True

Explanation:

Just like integrating for a function of a single variable at a particular solution. We are also able to solve using separation of variables and integrating for our particular solution.

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Example Question #7 :Find General And Particular Solutions Using Separation Of Variables

Find the general solution of the following differential equation at the point (1,1) .

$\frac{dy}{dx} = 2x$

Possible Answers:

y = x^2

$y = \sqrt{x}$

y = x^2 + c

y = x

Correct answer:

y = x^2

Explanation:

We first must use separation of variables to solve the general equation, then we will be able to find the general solution.

$\frac{dy}{dx} =2x$

dy = 2 dx (multiplied by)

$\int dy = \int 2x dx$

y + C = x^2 + D (Let D-C = c )

y = x^2 + c

Now we plug in our particular solution (1,1) to solve for our constant

1 = 1^2 + c

0 = c

And so our solution is

y = x^2

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Example Question #8 :Find General And Particular Solutions Using Separation Of Variables

Find the general solution of the following differential equation at the point (1,0) .

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

Possible Answers:

$y = e^{\frac{x^4}{4} + 1}$

$ln|y| = \sqrt{\frac{x^4}{4} + 1}$

$y = \frac{x^4}{4} + 1$

$y = \frac{x^4}{4}$

Correct answer:

$ln|y| = \sqrt{\frac{x^4}{4} + 1}$

Explanation:

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

ydy = 2x^3dx (multiplying by y^2 and multiplying by)

$\int ydy = 2x^3dx$

$\frac{y^2}{2} + C = \frac{2x^4}{4} + D$ (Let D-C=a )

$\frac{y^2}{2} = \frac{x^4}{2} + a$

$y^2 = \frac{x^4}{4} + \frac{a}{2}$ ( $\frac{a}{2}$ is just a constant so rename it)

$y = \sqrt{\frac{x^4}{4} + c}$

Now we plug in our point (1,0) to solve for.

$1 = \sqrt{\frac{0^4}{4} +c}$

$1 = \sqrt{c}$

1 = c

So our solution at this point is:

$y = \sqrt{\frac{x^4}{4} + 1}$

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Example Question #9 :Find General And Particular Solutions Using Separation Of Variables

Find the particular solution for $x ={\pi}$ using the point $(\frac{pi}{2},0)$ of the following differential equation.

$\frac{dy}{dx} = \frac{-sin(x)}{2y}$

Possible Answers:

y =0

$y = \sqrt{1 + c}$

y = 1

$y = \pm 1$

Correct answer:

$y = \pm 1$

Explanation:

We first must use separation of variables to solve the general equation, then we will be able to find the particular solution.

$\frac{dy}{dx} = \frac{-sin(x)}{2y}$

2ydy = -sin(x)dx (multiplying byand)

$\int 2ydy = \int -sin(x)dx$

$\frac{2y^2}{2} + C = cos(x) + D$ (Let D-C = c )

y^2 = cos(x) + c

$y = \sqrt(cos(x) + c)$

Now we plug in our initial condition that we were given $(\frac{pi}{2},0)$

$0 = \sqrt{cos(\frac{\pi}{2}) + c}$

$0 = \sqrt{c}$

0 = c

Now we will solve forwhen $x = \pi$

$y = \sqrt{cos(x)}$

$y = \sqrt{cos(\pi)}$

$y = \sqrt{1}$

$y = \pm 1$

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Example Question #1 :Find General And Particular Solutions Using Separation Of Variables

What do we call a differential equation that can be solved by using separation of variables.

Possible Answers:

separable partials

separable equations

they have no special name

separable derivatives

Correct answer:

separable equations

Explanation:

Separable equations are what we call differential equations that we are able to solve by using separation of variables.

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Calculus AB : Find General and Particular Solutions Using Separation of Variables

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #1 :Find General And Particular Solutions Using Separation Of Variables

Example Question #1 :Find General And Particular Solutions Using Separation Of Variables

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