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

Study concepts, example questions & explanations for Calculus AB

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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:

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

A fancy wording for factoring

Integrating with two variables

Taking the derivative of two variables

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. \(\displaystyle dy\) or \(\displaystyle dx\)), then we can integrate each side with respect to each variable.

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

When would one need to use separation of variables?

Possible Answers:

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

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

When we need to find the derivative of a differential equation

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

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.

Example Question #2 : Find General And Particular Solutions Using Separation Of Variables

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

 

Possible Answers:

\(\displaystyle y = ce^{x^2}\)

\(\displaystyle \frac{-1}{2y^2} = x^2 + C\)

\(\displaystyle y = ln|x^2| + C\)

\(\displaystyle y +C = x^2 + D\)

Correct answer:

\(\displaystyle y = ce^{x^2}\)

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.

\(\displaystyle \frac{dy}{dx} = 2yx\)

 

 

\(\displaystyle \frac{dy}{y} = 2xdx\)                                   (divided by \(\displaystyle y\) and multiplied by \(\displaystyle dx\))

 

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

 

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

 

\(\displaystyle ln|y| + C= 2\frac{x^2}{2} + D\)

 

\(\displaystyle ln|y| = x^2 + a\)

 

\(\displaystyle e^{ln|y|} = e^{x^2 + a}\)

 

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

 

\(\displaystyle y = ce^{x^2}\)

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

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

Possible Answers:

\(\displaystyle y = 4x^2 + c\)

\(\displaystyle y = 8x^2 + c\)

\(\displaystyle y = \frac{1}{x^2} + c\)

\(\displaystyle y = \frac{x^2}{2} + c\)

Correct answer:

\(\displaystyle y = 4x^2 + c\)

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.

 

\(\displaystyle \frac{dy}{dx} = \frac{4x}{y}\)

 

\(\displaystyle ydy = 4xdx\)                                   (multiplied by \(\displaystyle y\) and multiplied by \(\displaystyle dx\))

 

\(\displaystyle \int y dy = \int 4x dx\)

 

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

 

\(\displaystyle \frac{y^2}{2} = \frac{4x^2}{2} + a\)

 

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

 

\(\displaystyle y = 4x^2 + c\)

Example Question #52 : Differential Equations

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

Possible Answers:

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

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

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

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

Correct answer:

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

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.

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

 

\(\displaystyle y dy = (2-3x^2)^2 dx\)                                                    (multiplied by \(\displaystyle y\) and \(\displaystyle dx\))

 

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

 

\(\displaystyle \int y dy= \int 9x^4 - 12x^2 + 4 dx\)

 

\(\displaystyle \int y dy= \int 9x^4 dx - \int 12x^2 dx + \int 4 dx\)

 

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

 

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

 

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

 

Example Question #2 : 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.

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

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

\(\displaystyle \frac{dy}{dx} = 2x\)

 

 

Possible Answers:

\(\displaystyle y = \sqrt{x}\)

\(\displaystyle y = x^2\)

\(\displaystyle y = x^2 + c\)

\(\displaystyle y = x\)

Correct answer:

\(\displaystyle y = x^2\)

Explanation:

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

\(\displaystyle \frac{dy}{dx} =2x\)

 

\(\displaystyle dy = 2 dx\)                             (multiplied by \(\displaystyle dx\))

 

\(\displaystyle \int dy = \int 2x dx\)

 

 

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

 

\(\displaystyle y = x^2 + c\)

 

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

 

\(\displaystyle 1 = 1^2 + c\)

 

\(\displaystyle 0 = c\)

 

And so our solution is

 

 

\(\displaystyle y = x^2\)

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

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

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

 

 

Possible Answers:

\(\displaystyle ln|y| = \sqrt{\frac{x^4}{4} + 1}\)

\(\displaystyle y = e^{\frac{x^4}{4} + 1}\)

\(\displaystyle y = \frac{x^4}{4} + 1\)

\(\displaystyle y = \frac{x^4}{4}\)

Correct answer:

\(\displaystyle ln|y| = \sqrt{\frac{x^4}{4} + 1}\)

Explanation:

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

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

\(\displaystyle \int ydy = 2x^3dx\)

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

\(\displaystyle \frac{y^2}{2} = \frac{x^4}{2} + a\)

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

\(\displaystyle y = \sqrt{\frac{x^4}{4} + c}\)

 

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

 

\(\displaystyle 1 = \sqrt{\frac{0^4}{4} +c}\)

\(\displaystyle 1 = \sqrt{c}\)

\(\displaystyle 1 = c\)

 

So our solution at this point is:

 

\(\displaystyle y = \sqrt{\frac{x^4}{4} + 1}\)

 

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

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

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

 

 

Possible Answers:

\(\displaystyle y = 1\)

\(\displaystyle y = \sqrt{1 + c}\)

\(\displaystyle y =0\)

\(\displaystyle y = \pm 1\)

Correct answer:

\(\displaystyle y = \pm 1\)

Explanation:

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

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

\(\displaystyle 2ydy = -sin(x)dx\)                                        (multiplying by \(\displaystyle 2y\) and \(\displaystyle dx\))

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

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

\(\displaystyle y^2 = cos(x) + c\)

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

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

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

\(\displaystyle 0 = \sqrt{c}\)

\(\displaystyle 0 = c\)

 

Now we will solve for \(\displaystyle y\) when  \(\displaystyle x = \pi\)

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

 

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

\(\displaystyle y = \sqrt{1}\)

\(\displaystyle y = \pm 1\)

Example Question #2 : 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

they have no special name

separable equations

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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