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

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Example Questions

Example Question #11 : Applications Of Antidifferentiation

Solve the separable differential equation

$\frac{\mathrm{d} y}{\mathrm{d} x}=x^2y$

given the condition y(0)=3

Possible Answers:

y=0

None of the other answers

$y=Ce^{\frac{x^3}{3}}$

y=C

$y=3e^{\frac{x^3}{3}}$

Correct answer:

$y=3e^{\frac{x^3}{3}}$

Explanation:

To solve this equation, we must separate the variables such that terms containing x and y are on the same side as dx and dy, respectively:

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

Integrating both sides of the equation, we get

$\ln \left | y\right |+C_1=\frac{x^3}{3}+C_2$

The integrals were found using the following rules:

$\int \frac{dx}{x}=\ln \left | x\right |+C$ , $\int x^n dx=\frac{x^{n+1}}{n+1}+C$

After combining the constants of integration into a single C, exponentiating both sides, and using the properties of exponents to simplify, we get

$y=e^{\frac{x^3}{3}+C}=e^{\frac{x^3}{3}}\cdot e^C=Ce^{\frac{x^3}{3}}$

To solve for C, we use the condition given:

3=Ce^0

3=C

Our final answer is

$y=3e^{\frac{x^3}{3}}$

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Example Question #12 : Applications Of Antidifferentiation

is a function of . Solve for in this differential equation:

$\frac{\mathrm{d} y}{\mathrm{d} x} = \sqrt{\frac{ e^{y}}{x}}$

Possible Answers:

$y =-2 \ln ( \sqrt{x}+ C)$

$y =- \ln (- \sqrt{x}+ C)$

$y =\ln ( \sqrt{x}+ C)$

$y =-2 \ln (- \sqrt{x}+ C)$

$y =2 \ln (- \sqrt{x}+ C)$

Correct answer:

$y =-2 \ln (- \sqrt{x}+ C)$

Explanation:

First, rewrite the expression on the right as the $\frac{1}{2}$ power of the radicand:

$\frac{\mathrm{d} y}{\mathrm{d} x} = \sqrt{\frac{ e^{y}}{x}}$

$\frac{\mathrm{d} y}{\mathrm{d} x} =\left ( \frac{ e^{y}}{x} \right )^{\frac{1}{2}}$

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{ (e^{y})^{\frac{1}{2}}}{x^{\frac{1}{2}}}$

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{ e^ {\frac{y}{2}}}{x^{\frac{1}{2}}}$

The expressions with can be separated from those with by multiplying both sides by $e^ {-\frac{y}{2} }\mathrm{\; d} x$ :

$\frac{\mathrm{d} y}{\mathrm{d} x} \cdot e^ {-\frac{y}{2} }\mathrm{\; d} x = \frac{ e^ {\frac{y}{2}}}{x^{\frac{1}{2}}} \cdot e^ {-\frac{y}{2} }\mathrm{\; d} x$

$e^ {-\frac{y}{2} } \mathrm{d} y= \frac{ 1}{x^{\frac{1}{2}}} \mathrm{\; d} x$

$e^ {-\frac{1}{2} y} \mathrm{d} y= x^{-\frac{1}{2}} \mathrm{\; d} x$

Find the indefinite integral of both sides:

$\int e^ {-\frac{1}{2} y} \mathrm{d} y= \int x^{-\frac{1}{2}} \mathrm{\; d} x$

The expression on the right can be integrated using the Power Rule. On the right, use some -substitution, setting $u = -\frac{1}{2} y$ ; this makes $\mathrm{d} u = -\frac{1}{2} \mathrm{d}y$ and $\mathrm{d}y = -2 \mathrm{\; d} u$ :

$-2 \int e^ {u} \mathrm{d} u= \frac{x\frac{1}{2}}{\frac{1}{2}}+ C$

$-2 e^ {u} =2 \sqrt{x}+ C$

Apply some algebra to solve for :

$e^ {u} =- \sqrt{x}+ C$

$u =\ln (- \sqrt{x}+ C)$

Substitute $-\frac{1}{2} y$ back for , and apply some algebra:

$-\frac{1}{2} y =\ln (- \sqrt{x}+ C)$

$y =-2 \ln (- \sqrt{x}+ C)$

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Example Question #13 : Solving Separable Differential Equations And Using Them In Modeling

Solve the separable differential equation

$\frac{\mathrm{d} y}{\mathrm{d} x}=y(x^2+5x+2)$

where y(0)=7

Possible Answers:

$y=7e^{\frac{x^3}{3}+\frac{5x^2}{2}+2x}$

y=0

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

y=7

Correct answer:

$y=7e^{\frac{x^3}{3}+\frac{5x^2}{2}+2x}$

Explanation:

To solve the separable differential equation, we must separate x and y to the same sides as their respective derivatives:

$\frac{dy}{y}=(x^2+5x+2)dx$

Next, we integrate both sides:

$\ln \left | y\right |=\frac{x^3}{3}+\frac{5x^2}{2}+2x+C$

The integrals were solved using the following rules:

$\int \frac{dx}{x}=\ln \left | x\right |+C$ , $\int x^n dx=\frac{x^{n+1}}{n+1}+C$

The two constants of integration were combined to make a single one.

Now, we exponentiate both sides to solve for y, keeping in mind rules for exponents which allow us to move the integration constant to the front:

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

To solve for the constant of integration, we use the condition given:

$7=Ce^{0}=C(1)=C$

Our final answer is

$y=7e^{\frac{x^3}{3}+\frac{5x^2}{2}+2x}$

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Example Question #14 : Solving Separable Differential Equations And Using Them In Modeling

Solve the following separable differential equation:

$\frac{\mathrm{d} y}{\mathrm{d} x}= y\sec^2(x)$

given the condition that at $x=\pi, y=-2$

Possible Answers:

$y=Ce^{\tan(x)}$

$y=-2e^{\tan(x)}$

y=0

$y=2e^{\tan(x)}$

y=-2

Correct answer:

$y=-2e^{\tan(x)}$

Explanation:

To solve the separable differential equation, we must separate x and y and their respective derivatives to either side of the equal sign:

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

Now, we integrate both sides of the equation:

$\ln \left | y\right |=\tan(x)+C$

The integrals were found using their identical rules.

Exponentiating both sides of the equation to solve for y - and keeping in mind the rules of exponents - we get

$y=e^{\tan(x)+C}=e^C\cdot e^{\tan(x)}=Ce^{\tan(x)}$

Now, we solve for the integration constant by using the condition given:

$-2=Ce^{\tan(\pi)}=Ce^0=C$

Our final answer is

$y=-2e^{\tan(x)}$

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Example Question #15 : Solving Separable Differential Equations And Using Them In Modeling

Solve the following separable differential equation:

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{x^2}{y\cos(y^2)}$

Possible Answers:

$y=\sqrt{\arcsin(\frac{2x^3}{3})+C}$

$y=\arcsin(\frac{2x^3}{3})+C$

$y=\sqrt{\arcsin(\frac{2x^3}{3}+C)}$

$y=\sqrt{\arcsin(\frac{2y^3}{3})}+C$

Correct answer:

$y=\sqrt{\arcsin(\frac{2x^3}{3}+C)}$

Explanation:

To solve the separable differential equation, we must separate x and y to the same sides as their respective derivatives:

$y\cos(y^2)dy=x^2dx$

Next, we integrate both sides, where on the lefthand side, the following substitution is made:

u=y^2, du=2ydy

$\frac{1}{2}\int\cos(u)du=\frac{1}{2}\sin(y^2)=\frac{x^3}{3}+C$

The integrals were solved using the following rules:

$\int \cos(x)dx=\sin(x)+C, \int x^ndx=\frac{x^{n+1}}{n+1}+C$

The two constants of integration were combined to make a single one.

Now, we solve for y:

$y^2=\arcsin(\frac{2x^3}{3}+C)$

$y=\sqrt{\arcsin(\frac{2x^3}{3}+C)}$

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Example Question #16 : Solving Separable Differential Equations And Using Them In Modeling

Solve the separable differential equation

$\frac{\mathrm{d} y}{\mathrm{d} x} = y(2x+x^2)$

given the initial condition

f(0)=14

Possible Answers:

$y=14e^{\frac{x^2}{2}+\frac{x^3}{3}}$

$y=14e^{2x+\frac{x^3}{3}}$

$y=14e^{x^2+\frac{x^2}{3}}$

$y=14e^{x^2+\frac{x^3}{3}}$

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

Correct answer:

$y=14e^{x^2+\frac{x^3}{3}}$

Explanation:

To solve the separable differential equation, we must separate x and y to the same sides as their respective derivatives:

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

Next, we integrate both sides:

$\ln \left | y\right |= x^2+\frac{x^3}{3}+C$

The integrals were solved using the following rules:

$\int \frac{dx}{x}=\ln \left | x\right |+C$ , $\int x^n dx = \frac{x^{n+1}}{n+1}+C$

The two constants of integration were combined to make a single constant.

Now, exponentiate both sides to isolate y, and use the properties of exponents to rearrange the integration constant:

$e^{\ln\left | y\right |}= e^{x^2+\frac{x^3}{3}+C}$

$y=e^{x^2+\frac{x^3}{3}}\cdot e^C=Ce^{x^2+\frac{x^3}{3}}$

(The exponential of the constant is another constant.)

Finally, we solve for the integration constant using the initial condition:

14=Ce^0=C

Our final answer is

$y=14e^{x^2+\frac{x^3}{3}}$

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Example Question #16 : Solving Separable Differential Equations And Using Them In Modeling

Solve the separable differential equation:

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{x^4+e^x}{2y}, y< 0$

Possible Answers:

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

$y=-\sqrt{\frac{x^5}{5}+e^x+C}$

$y=-\sqrt{{x^5}+e^x+C}$

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

Correct answer:

$y=-\sqrt{\frac{x^5}{5}+e^x+C}$

Explanation:

To solve the separable differential equation, we must separate x and y to the same sides as their respective derivatives:

2y dy=(x^4+x)dx

Next, we integrate both sides:

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

The integrals were solved using the following rules:

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$ , $\int e^x dx=e^x+C$

The two constants of integration were combined to make a single one.

Now, we solve for y:

$y=\pm \sqrt{\frac{x^5}{5}+e^x+C}$

Because the problem statement said that y is negative - and y cannot be zero - our final answer is

$y=-\sqrt{\frac{x^5}{5}+e^x+C}$

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Example Question #17 : Solving Separable Differential Equations And Using Them In Modeling

Solve the separable differential equation:

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{4y}{x^2}, x>0$

and at x=1, y=2

Possible Answers:

$y=2e^{-\frac{4}{x}}$

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

$y=2e^{-\frac{4}{x}}+C$

$y=2e^{4-\frac{4}{x}}$

y=2

Correct answer:

$y=2e^{4-\frac{4}{x}}$

Explanation:

To solve the separable differential equation, we must separate x and y to the same sides as their respective derivatives:

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

Next, we integrate both sides:

$\ln \left | y\right | = -\frac{1}{x}+C$

The integrals were solved using the following rules:

$\int \frac{dx}{x}= \ln \left | x\right |+C$ , $\int x^n dx = \frac{x^{n+1}}{n+1}+C$

The two constants of integration were combined to make a single one.

Now, we exponentiate both sides to solve for y:

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

Using the properties of exponents, we can rearrange the integration constant:

$y=e^{-\frac{1}{x}+C} = e^{-\frac{1}{x}}\cdot e^C = Ce^{-\frac{1}{x}}$

(The exponential of the constant is itself a constant.)

Using the given condition, we can solve for C:

$2=Ce^{-4}$

C=2e^4

Our final answer is

$y=2e^{4-\frac{4}{x}}$

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Example Question #17 : Solving Separable Differential Equations And Using Them In Modeling

The rate of a chemical reaction is given by the following differential equation:

$\frac{-d[A]}{dt} = k[A]^{2}$ ,

where $\left [ A \right ]$ is the concentration of compound at a given time, . Which one of the following equations describes $\frac{1}{\left [ A \right ]}$ as a function of time? Let $\left [ A \right ]_{0}$ be the concentration of compound when t=0 .

Possible Answers:

$\frac{1}{[A_{0}]} + kt = \frac{1}{[A]}$

$[A_{0}]e^{kt}= \frac{1}{[A]}$

$[A_{0}]kt= \frac{1}{[A]}$

$[A_{0}]e^{-kt}= \frac{1}{[A]}$

$\frac{1}{[A_{0}]} - kt = \frac{1}{[A]}$

Correct answer:

$\frac{1}{[A_{0}]} + kt = \frac{1}{[A]}$

Explanation:

To begin with, the differential equation needs to be rearranged so that each variable is one side of the equation:

$\frac{-d[A]}{dt} = k[A]^{2} \Rightarrow \frac{d[A]}{[A]^{2}} = -kdt$ .

Then, integrate each side of the rate law, bearing in mind that $\left [ A \right ]$ will range from $\left [ A \right ]_{0}$ to $\left [ A \right ]$ , and time will range from to :

$\int_{[A]_{0}}^{[A]}\frac{d[A]}{[A]^{2}} = \int_{0}^{t} -kdt$

After integrating each side, the equation becomes:

$\frac{-1}{[A]} = -kt$ .

The left side has to be evaluated from $\left [ A \right ]_{0}$ to $\left [ A \right ]$ , and the right side is evaluated from to :

$\frac{-1}{[A]} - ( \frac{-1}{[A_{0}]}) = -kt - (-k(0))$ . This becomes:

$\frac{1}{[A_{0}]} - \frac{1}{[A]} = -kt$ .

Finally, rearranging gives:

$\frac{1}{[A_{0}]} + kt = \frac{1}{[A]}$

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Example Question #21 : Applications Of Antidifferentiation

Given that y' = (2y + 2)(x -2) and y(0) = 1 , solve for y(t) . What is the value of ?

Possible Answers:

2e^2 - 1

2e - 1

Correct answer:

Explanation:

This is a separable differential equation. The simplest way to approach this is to turn into $\frac{\mathrm{d} y}{\mathrm{d} x}$ , and then by abusing the notation, "multiplying by dx" on both sides.

dy = (2y+2)(x - 2)dx

We then group all the y terms with dy, and all the x terms with dx.

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

Integrating both sides, we find

$\int \frac{1}{2y+2}dy = \frac{\ln|2y+2|}{2} =\int (x - 2)dx = \frac{x^2}{2} - 2x + C$

Here, the first integral is found by using substitution of variables, setting u = 2y +2 . In addition, we have chosen to only put a +C on the second integral, as if we put it on both, we would just combine them in any case.

To solve for y, we multiply both sides by two and raise e to both sides to get rid of the natural logarithm.

$|2y+2| = e^{x^2 - 4x + C}$

(Note, C was multiplied by two, but it's still just an arbitrary constant. If you prefer, you may call the new C value C_1 .)

Now we drop our absolute value signs, and note that we can take out a factor of e^C and stick in front of the right hand side.

$2y+2 =\pm e^Ce^{x^2 - 4x}$

As $\pm e^C$ is just another arbitrary constant, we can relabel this as C, or C_2 if you prefer. Solving for y gets us

$y = \frac{Ce^{x^2 - 4x} - 2}{2}$

Next, we plug in our initial condition to solve for C.

$1 =\frac{C - 2}{2}$ ; C = 4

Leaving us with a final equation of

$y = \frac{4e^{x^2 - 4x} - 2}{2}$

Plugging in x = 4, we have a final answer,

$y = \frac{4e^{16 - 16} - 2}{2} = \frac{4 - 2}{2} = 1$

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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 #11 : Applications Of Antidifferentiation

Example Question #12 : Applications Of Antidifferentiation

Example Question #13 : Solving Separable Differential Equations And Using Them In Modeling

Example Question #14 : Solving Separable Differential Equations And Using Them In Modeling

Example Question #15 : Solving Separable Differential Equations And Using Them In Modeling

Example Question #16 : Solving Separable Differential Equations And Using Them In Modeling

Example Question #16 : Solving Separable Differential Equations And Using Them In Modeling

Example Question #17 : Solving Separable Differential Equations And Using Them In Modeling

Example Question #17 : Solving Separable Differential Equations And Using Them In Modeling

Example Question #21 : Applications Of Antidifferentiation

All AP Calculus AB Resources

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