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Calculus 1 : Equations

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Example Questions

Example Question #33 : How To Find Solutions To Differential Equations

Consider the chemical reaction $A\rightarrow B$ .

Initially there are moles of . The change in concentration of occurs at the following rate.

$\frac{dC_A}{dt}=-kC_A^2$

Where $k=0.0572 min^{-1}$ .

Find C_A at minutes.

Possible Answers:

2.73 mols

7.21 mols

0.45 mols

3.36 mols

1.04 mols

Correct answer:

1.04 mols

Explanation:

Rearrange the equation to have all the C_A terms on one side and the rest of the terms on the other side.

$\frac{dC_A}{dt}=-kC_A^2$

$\frac{1}{C_A^2}dC_A=-kdt$

From here we need to take the integral of the function.

$\int \frac{1}{C_A^2}dC_A=\int -kdt$

$-\frac{1}{C_A}=-kt+C$ Where C is some constant.

We know that the initial concentration of A is 10 mols. In otherwords, at

t=0, C_A=10

$-\frac{1}{10}=-k(0)+C$

$C=-\frac{1}{10} mols$

Solving for C_A gives

$C_A(t)=\frac{1}{kt+\frac{1}{10}}$ .

Plugging everything in gives

C_A(15)=1.04 mols .

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Example Question #34 : How To Find Solutions To Differential Equations

Find the particular solution for the following initial value problem:

$x\frac{dy}{dx}-y\sqrt{x}=0$

y(0)=1

Possible Answers:

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

$y=2\sqrt{x}+1$

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

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

$y=e^{2\sqrt{x}}+1$

Correct answer:

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

Explanation:

To find the particular solution, we start by finding the general solution. First we rearrange the differential equation such that is on one side with any terms and is on the other side with any terms. We can then integrate each side with respect to the appropriate variable and solve for to find the general solution for the differential equation. Finally, we plug in the given initial condition to determine the value of the constant, which gives us the particular solution:

$x\frac{dy}{dx}-y\sqrt{x}=0$

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

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

$\frac{1}{y}dy=\frac{1}{\sqrt{x}}dx$

$\int \frac{1}{y}dy=\int \frac{1}{\sqrt{x}}dx$

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

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

$y=e^{2\sqrt{x}+C}=e^Ce^{2\sqrt{x}}=Ce^{2\sqrt{x}}$

$y(0)=1\rightarrow Ce^{2\sqrt{0}}=1\rightarrow C=1$

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

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Example Question #1342 : Functions

Find the particular solution for the following differential equation:

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

y(1)=-1

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To find the particular solution, we start by finding the general solution. First we rearrange the differential equation such that is on one side with any terms and is on the other side with any terms. We can then integrate each side with respect to the appropriate variable and solve for to find the general solution for the differential equation. Finally, we plug in the and values of the given point to determine the value of the constant, which gives us the particular solution:

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

$\frac{1}{y}dy=\frac{1}{x^2}dx$

$\int \frac{1}{y}dy=\int \frac{1}{x^2}dx$

$\ln(y)=-x^{-1}+C$

$e^{\ln(y)}=e^{-x^{-1}+C}$

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

$y(1)=-1\rightarrow Ce^{-\frac{1}{1}}=-1\rightarrow C=-e$

$y=(-e)e^{-\frac{1}{x}}=-e^{1-\frac{1}{x}}$

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Example Question #35 : How To Find Solutions To Differential Equations

Determine the general solution to the following differential equation:

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

Possible Answers:

y=x^3-x^2+C

y=6x-2+C

y=6x+C

y=3x^2-2x+C

y=3x^3-2x^2+C

Correct answer:

y=x^3-x^2+C

Explanation:

This is a separable differential equation, which means we are able to separate and , placing them on opposite sides of the equation with their corresponding variables. There are no terms, so we place alone on one side, and on the other side with the terms containing . We can then integrate each side with respect to the appropriate variable, which gives an equation for that is the general solution to the differential equation:

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

dy=(3x^2-2x)dx

$\int dy=\int(3x^2-2x)dx$

Remember when we integrate, we increase the exponent by one and then divide the term by the value of the new exponent. We will need to integrate each term that contains a .

y=x^3-x^2+C

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Example Question #41 : How To Find Solutions To Differential Equations

Find the general solution for the following differential equation:

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

Possible Answers:

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

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

$y=-\frac{1}{3}x^3+\frac{2}{3}x+C$

$y=\frac{1}{3}x^3-\frac{2}{3}x+C$

$y=-\frac{1}{2}x^2+\frac{3}{2}x+C$

Correct answer:

$y=\frac{1}{3}x^3-\frac{2}{3}x+C$

Explanation:

First we must rearrange this separable differential equation so that we can place alone on one side and on the other side with the terms involving and any constants. We then integrate each side with respect to the appropriate variable and solve the result for to find the general solution of the differential equation:

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

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

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

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

$\int dy=\int \left(x^2-\frac{2}{3}\right)dx$

Remember when integrating, we increase the exponent by one and then divide the whole term by the value of the new exponent. Will we need to integrate each term that contains in this fashion.

$y=\frac{1}{3}x^3-\frac{2}{3}x+C$

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Example Question #42 : How To Find Solutions To Differential Equations

Find the general solution for the following differential equation:

$y\frac{dy}{dx}=x^2-1$

Possible Answers:

$y=\pm\sqrt{\frac{1}{2}x^3+\frac{3}{2}x+C}$

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

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

$y=\pm\sqrt{\frac{1}{3}x^3+x+C}$

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

Correct answer:

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

Explanation:

First we must multiply to the right side of the equation so we have the terms with and the terms with . We can then integrate each side with respect to the appropriate variable and then solve the result for to find the general solution for the differential equation:

$y\frac{dy}{dx}=x^2-1$

(y)dy=(x^2-1)dx

$\int (y)dy=\int (x^2-1)dx$

Remember when integrating, we increase the exponent by one and then divide the term by that new exponent value. We will do this with each term on both sides.

$\frac{1}{2}y^2=\frac{1}{3}x^3-x+C$

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

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

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Example Question #43 : How To Find Solutions To Differential Equations

Determine the general solution to the following differential equation:

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

Possible Answers:

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

$y=Ce^{2x}$

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

y=2x+C

$y=Ce^{x^2}$

Correct answer:

$y=Ce^{x^2}$

Explanation:

In order to separate the differential equation such that is with the terms and is with the terms, we must divide both sides of the equation by and then multiply both sides by . We can then integrate each side with respect to the appropriate variable, and then solve for to find the general solution to the differential equation:

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

$\left(\frac{1}{y}\right)dy=(2x)dx$

$\int \left(\frac{1}{y}\right)dy=\int (2x)dx$

$\ln (y)=x^2+C$

$e^{\ln (y)}=e^{x^2+C}$

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

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

$y=Ce^{x^2}$

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Example Question #44 : How To Find Solutions To Differential Equations

Determine the general solution to the following differential equation:

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

Possible Answers:

$y=\pm \sqrt{\frac{1}{2}x^4+C}$

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

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

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

$y=\pm \sqrt{\frac{1}{3}x^3+C}$

Correct answer:

$y=\pm \sqrt{\frac{1}{2}x^4+C}$

Explanation:

For this separable differential equation, we can see that if we cross multiply we will have the term with , and the term with . After getting the equation into this form, we can integrate each side with respect to the appropriate variable, and then solve for to find the general solution to the differential equation:

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

(y)dy=(x^3)dx

$\int (y)dy=\int (x^3)dx$

Remember when integrating, we increase the exponent by one and then divide the term by that new exponent value. We will do this with each term on both sides.

$\frac{1}{2}y^2=\frac{1}{4}x^4+C$

$y^2=\frac{1}{2}x^4+C$

$y=\pm \sqrt{\frac{1}{2}x^4+C}$

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Example Question #45 : How To Find Solutions To Differential Equations

Find the particular solution for the following initial value problem:

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

y(0)=2

Possible Answers:

y=3x^3-2x^2+2

y=x^3-x^2

y=3x^2-2x

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

y=x^3-x^2+2

Correct answer:

y=x^3-x^2+2

Explanation:

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

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

dy=(3x^2-2x)dx

$\int dy=\int (3x^2-2x)dx$

Remember when integrating, we can use the power rule. This means to increase the exponent by one and then divide the term by that new exponent value. We will do this with each term on both sides.

y=x^3-x^2+C

Once we have found the general solution we plug in the initial conditions to solve for C.

$y(0)=2\rightarrow (0)^3-(0)^2+C=2\rightarrow C=2$

y=x^3-x^2+2

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Example Question #46 : How To Find Solutions To Differential Equations

Find the particular solution for the following differential equation:

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

y(0)=0

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To find the particular solution, we start by finding the general solution. First we rearrange the differential equation such that is on one side with any terms and is on the other side with any terms. For this problem we can see that the desired arrangement is achieved by simply cross multiplying the differential equation. We can then integrate each side with respect to the appropriate variable and solve for to find the general solution for the differential equation. Finally, we plug in the and values of the given point to determine the value of the constant, which gives us the particular solution:

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

(y)dy=(2x+1)dx

$\int (y)dy=\int (2x+1)dx$

Remember when integrating, we can use the power rule. The power rule states to increase the exponent by one and then divide the term by that new exponent value. We will do this with each term on both sides.

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

y^2=2x^2+2x+C

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

Now that we have the general solution we can plug in the initial values and solve for C.

$y(0)=0\rightarrow \pm \sqrt{2(0)^2+2(0)+C}=0\rightarrow C=0$

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

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Calculus 1 : Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #33 : How To Find Solutions To Differential Equations

Example Question #34 : How To Find Solutions To Differential Equations

Example Question #1342 : Functions

Example Question #35 : How To Find Solutions To Differential Equations

Example Question #41 : How To Find Solutions To Differential Equations

Example Question #42 : How To Find Solutions To Differential Equations

Example Question #43 : How To Find Solutions To Differential Equations

Example Question #44 : How To Find Solutions To Differential Equations

Example Question #45 : How To Find Solutions To Differential Equations

Example Question #46 : How To Find Solutions To Differential Equations

All Calculus 1 Resources

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