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AP Calculus AB : Solving separable differential equations and using them in modeling

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Example Questions

Example Question #1 : Applications Of Antidifferentiation

Find (dy/dx).

sin(xy) = x + cos(y)

Possible Answers:

dy/dx = (1 – ycos(xy))/(xcos(xy) + sin(y))

dy/dx = (xcos(xy) + sin(y))/(1 – ycos(xy))

dy/dx = (1 – cos(xy))/(cos(xy) + sin(y))

dy/dx = (cos(xy) + sin(y))/(1 – cos(xy))

None of the above

Correct answer:

dy/dx = (1 – ycos(xy))/(xcos(xy) + sin(y))

Explanation:

The first step of the problem is to differentiate with respect to (dy/dx):

cos(xy)[(x)(dy/dx) + y(1)] = 1 – sin(y)(dy/dx)

*Note: When differentiating cos(xy) remember to use the product rule. (xy' + x'y)

Step 2: Clean the differentiated problem up

cos(xy)(x)(dy/dx) + cos(xy)y = 1 – sin(y)(dy/dx)

cos(xy)(x)(dy/dx) + sin(y)(dy/dx) = 1 – cos(xy)y

Step 3: Solve for (dy/dx)

dy/dx = (1 – ycos(xy))/(xcos(xy) + sin(y))

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Example Question #1 : Applications Of Antidifferentiation

Find the equation of the normal line at (2,0) on the graph $y=x^{3}-6x+4$ .

Possible Answers:

$y=\frac{-1}{6}x+2$

y=6x-12

$y'=3x^{2}-6$

$y=\frac{-1}{6}x+\frac{1}{3}$

y=6x+2

Correct answer:

$y=\frac{-1}{6}x+\frac{1}{3}$

Explanation:

The answer is $y=\frac{-1}{6}x+\frac{1}{3}$ .

$y=x^{3}-6x+4$

$y'=3x^{2}-6$

Now plug in x=2 .

$y'=3(2)^{2}-6 = 6$ now we know 6 is the slope for the tangent line. However, we aren't looking for the slope of the tangent line. The slope of the normal line is the negative reciprocal of the tangent's slope; meaning the slope of the normal is $\frac{-1}{6}$ . Now find the equation of the normal line.

$y-0=\frac{-1}{6}(x-2)$

$y=\frac{-1}{6}x+\frac{1}{3}$

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Example Question #2 : Applications Of Antidifferentiation

$f(x) = \frac{x^3}{1-x^2}$

What is the derivative of f(x) ?

Possible Answers:

$\frac{x^2(3-5x^2)}{1-x^2}$

$\frac{-2x^4-3 x^2(1-x^2)}{(1-x^2)^2}$

$-x$

Correct answer:

Explanation:

Use the quotient rule.

${y}' = \frac{(1-x^2)(3x^2)-x^3(-2x)}{(1-x^2)^2} = \frac{x^2(3-3x^2+ 2x^2)}{(1-x^2)^2} = \frac{x^2(3-x^2)}{(1-x^2)^2}$

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Example Question #3 : Applications Of Antidifferentiation

Find $y^{'}$ if

$y=\frac{ln(x)}{x^{3}}$

Possible Answers:

$\frac{1}{x^{4}}$

$\frac{ln(x)-1}{x^{4}}$

$\frac{1+3ln(x)}{x^{4}}$

$\frac{3}{x^{3}}$

$y'=\frac{1-3ln(x)}{x^{4}}$

Correct answer:

$y'=\frac{1-3ln(x)}{x^{4}}$

Explanation:

The answer is

$y'=\frac{1-3ln(x)}{x^{4}}$

$y=\frac{ln(x)}{x^{3}}$

$y'=\frac{(\frac{1}{x})x^{3}-ln(x)(3x^{3})}{x^{6}}$

$y'=\frac{x^{2}(1-3ln(x))}{x^{6}}$

$y'=\frac{1-3ln(x)}{x^{4}}$

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Example Question #4 : Applications Of Antidifferentiation

Find the derivative:

f(x)=x^3-3x^2

Possible Answers:

f'(x)=3x^3-6x^2

f'(x)=3x-6

f'(x)=x^2-6x

f'(x)=3x^2-6x

Correct answer:

f'(x)=3x^2-6x

Explanation:

To find the derivative, multiply the exponent by the coefficent in front of the x term and then decrease the exponent by 1:

f'(x)=3x^2-6x

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Example Question #6 : Solving Separable Differential Equations And Using Them In Modeling

Find the solution to the equation $y'=y$ at $x=2$ with initial condition $y(0)=2$ .

Possible Answers:

$2e^2$

$1$

$e$

$2e$

$e^2$

Correct answer:

$2e^2$

Explanation:

First, we need to solve the differential equation of $y'=y$ .

y'=y

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

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

$\int \frac{dy}{y}=\int dx$

ln(y)=x+c , where is a constant

y=e^xe^c

y=Ce^x , where is a constant

To find , use the initial condition, y(0)=2 , and solve:

C=2

Therefore, $y=2e^x$ .

Finally, at x=2 , $y(2)=2e^2$ .

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Example Question #7 : Solving Separable Differential Equations And Using Them In Modeling

Solve the differential equation: $\frac{dy}{dx}=-2xy$

Note that (-1,2) is on the curve.

Possible Answers:

y=-2

y=ln(2)+1

y=-x^2+ln(2)+1

$y=2e^{1-x^2}$

Correct answer:

$y=2e^{1-x^2}$

Explanation:

In order to solve differential equations, you must separate the variables first.

$\frac{dy}{y}=-2xdx$

$\int \frac{dy}{y}= \int -2xdx$

ln(y)=-x^2+C

Since point (-1,2) is on the curve, ln(2)=-1+C .

ln(2)+1=C

ln(y)=-x^2+ln(2)+1

To get rid of the log, raise every term to the power of e:

$e^{ln(y)}=e^{-x^2}*e^{ln(2)}*e^1$

$y=2e^{1-x^2}$

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Example Question #8 : Solving Separable Differential Equations And Using Them In Modeling

Find the solution to the differential equation

$\frac{dy}{dx} = \frac{x^{2}+ 4x}{y^{2}}$ when f(0) = 3 .

Possible Answers:

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

$y^{3} = x^{3} + 2x^{2} + 9$

$\frac{y^{3}}{3} = \frac{x^{3}}{3} + 2x^{2} - 27$

$\frac{1}{y} = \frac{1}{x} + 2x^{2} + 9$

$\frac{y^{3}}{3} = \frac{x^{3}}{3} + 2x^{2} + 9$

Correct answer:

$\frac{y^{3}}{3} = \frac{x^{3}}{3} + 2x^{2} + 9$

Explanation:

First, separate the variables of the original differential equation:

$\frac{dy}{dx} = \frac{x^{2}+ 4x}{y^{2}} \Rightarrow y^{2}dy = (x^{2} + 4x)dx$ .

Then, take the antiderivative of both sides, which gives

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

Use the given condition f(0) = 3 , plugging in

x = 0 and y=3 , to solve for . This gives c=9 , so the correct answer is

$\frac{y^{3}}{3} = \frac{x^{3}}{3} + 2x^{2} + 9$ .

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Example Question #9 : Solving Separable Differential Equations And Using Them In Modeling

Solve the following separable differential equation y' = ye^x with initial condition y(0) = e^2 .

Possible Answers:

$y=e^{e^x}$

$y=ee^{e^x}$

$y=e^{x^2}$

y=e^2e^x

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

Correct answer:

$y=ee^{e^x}$

Explanation:

We proceed as follows

y'=ye^x . Start

$\frac{dy}{dx} = ye^x$ . Rewrite as $\frac{dy}{dx}$ .

$\frac{1}{y}dy=e^xdx$ . Multiply both sides by , and divide both sides by .

$\ln|y| =e^x +C$ . Integrate both sides. Do not forget the on one of the sides.

Substitute the initial condition y(0)=e^2 .

$\ln|e^2| = e^0+C$ .

1=C . Solve for .

$\ln|y| = e^x+1$

$y = e^{e^x+1}$ . Exponentiate both sides .

$y = ee^{e^x}$ . Rule of exponents.

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Example Question #10 : Solving Separable Differential Equations And Using Them In Modeling

Solve the separable, first-order differential equation for :

$(y+1)\frac{dy}{dx}=2x-8+\frac{dy}{dx}$

y(0)=8

Possible Answers:

$y =-\sqrt{2x^2 -16x+8 }$

y = x^2 - 8x+8

$y =\sqrt{2x^2 -16x+16 }$

$y =\sqrt{x^2 -4x+8 }$

y = 8x-x^2 -8

Correct answer:

$y =\sqrt{2x^2 -16x+16 }$

Explanation:

Solve the separable, first-order differential equation for :

$(y+1)\frac{dy}{dx}=2x-8+\frac{dy}{dx}$

First collect all the terms with the derivative to one side of the equation.

$(y+1)\frac{dy}{dx}-\frac{dy}{dx}=2x-8$

$\frac{dy}{dx}[(y+1)-1]=2x-8$

$y\frac{dy}{dx}=2x-8$

Important Conceptual Note: often in texts on differential equations differentials often appear to have been rearranged algebraically as if $\frac{dy}{dx}$ is a "fraction," making it appear as if we "multiplied both sides" by to get: . This is not the case. The derivative is a limit by definition and, when the limit exists, can take on any real number which includes irrational numbers i.e. numbers which cannot be written as a ratio of two integers.

For instance, we cannot represent $\pi$ as a ratio, but some functions may have a derivative at a point such that the derivative is equal to $\pi$ , or a funciton may simply have an irrational number like $\pi$ as a derivative. For instance, if $y=\pi x$ we write the derivative $\frac{dy}{dx}=\pi$ . Claiming that and are representative of a "numerator" and "denominator" respectively, we would essentially be claiming to have found a way to write an irrational number, such as $\pi$ as a ratio, which is preposterous. The expression $\frac{dy}{dx}$ is simply notation.

Here is what we are really doing.

$\int y\frac{dy}{dx}dx=\int (2x-8)dx$

$\int y dy=\int (2x-8)dx$

$\frac{1}{2}y^2 +C_1 =x^2 -8x +C_2$

Note that the constants of integration can just be combined into one constant by defining .

Solve for :

y^2 =2(x^2 -8x +C)

$y =\pm \sqrt{2x^2 -16x+C }$

Applying the initial condition:

$y(0)=\pm \sqrt{C}=8$

C = 16

$y =\pm \sqrt{2x^2 -16x+16 }$

Here we have two possible solutions. However, because of the initial condition, we can easily rule out the negative solution. y(0) must be equal to positive .

$y(0)=8\: \: \: \: \: \: \Rightarrow\: \: \: \: \: \:\sqrt{16}=8$

$y =\sqrt{2x^2 -16x+16 }$

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AP Calculus AB : Solving separable differential equations and using them in modeling

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #1 : Applications Of Antidifferentiation

Example Question #1 : Applications Of Antidifferentiation

Example Question #2 : Applications Of Antidifferentiation

Example Question #3 : Applications Of Antidifferentiation

Example Question #4 : Applications Of Antidifferentiation

Example Question #6 : Solving Separable Differential Equations And Using Them In Modeling

Example Question #7 : Solving Separable Differential Equations And Using Them In Modeling

Example Question #8 : Solving Separable Differential Equations And Using Them In Modeling

Example Question #9 : Solving Separable Differential Equations And Using Them In Modeling

Example Question #10 : Solving Separable Differential Equations And Using Them In Modeling

All AP Calculus AB Resources

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