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Calculus 1 : Differential Equations

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Example Questions

Example Question #41 : Differential Equations

Find the equilibrium values for the following differential equation: y^2 + y'=y

Possible Answers:

0,1

-1,1

Correct answer:

0,1

Explanation:

To find the equilibrium values, substitute y'=0 and solve for . The equilibrium values are the solutions of the differential equation in constant form.

y^2 + y'=y

y^2 + 0=y

y^2-y=0

y(y-1)=0

y=0,1

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Example Question #31 : How To Find Solutions To Differential Equations

Suppose is greater than zero. Solve the differential equation: $y'= \frac{x^2}{y^\frac{1}{2}}$

Possible Answers:

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

$y=\left(\frac{1}{2}x^3+C\right)^\frac{3}{2}$

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

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

$y=\left(\frac{1}{2}x^3+C\right)^\frac{2}{3}$

Correct answer:

$y=\left(\frac{1}{2}x^3+C\right)^\frac{2}{3}$

Explanation:

Rewrite $y'= \frac{x^2}{y^\frac{1}{2}}$ so that the same variables $\textup{x, dx, y, dy}$ are aligned correctly on the left and right of the equal sign.

$y^\frac{1}{2}\cdot y'= x^2$

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

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

Integrate both sides and solve for .

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

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

$y^\frac{3}{2}=\left(\frac{3}{2}\right)\left(\frac{1}{3}x^3+C\right)$

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

$y=\left(\frac{1}{2}x^3+C\right)^\frac{2}{3}$

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Example Question #1336 : Functions

Find the function whose slope at the point (x,y) is -4x^3+5x and passes through the point (4,5) .

Possible Answers:

$y= -x^4+\frac{5}{2}x^2$

y=181

y=0

$y= -x^4+\frac{5}{2}x^2+221$

$y= -x^4+\frac{5}{2}x^2+c$

Correct answer:

$y= -x^4+\frac{5}{2}x^2+221$

Explanation:

This question requires us to use differential equations. Begin as follows:

$\frac{dy}{dx}=-4x^3+5x$

dy}=-4x^3+5xdx

$y=\int -4x^3+5xdx$

$y= -x^4+\frac{5}{2}x^2+c$

Now, we know that y must pass through the point (4,5), so we can use this point to find c

$5= -(4)^4+\frac{5}{2}4^2+c$

5= -256+40+c

c=221

So our function is as follows:

$y= -x^4+\frac{5}{2}x^2+221$

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Example Question #1337 : Functions

Solve:

$\frac{dy}{dx}=10y$

Possible Answers:

y=10e^x

y=Ce^0^.^1^x

y=Ce^1^0^x

y=Ce^x

$y=\frac{1}{10}e^x$

Correct answer:

y=Ce^1^0^x

Explanation:

Multiply and divide on both sides of the equation $\frac{dy}{dx}=10y$ .

$\frac{1}{y}dy=10 \: dx$

Integrate both sides.

$\int \frac{1}{y}dy=\int 10\: dx$

$ln\left |y \right |=10x+C$

Use base to eliminate the natural log.

$e^{ln(y)}=e^1^0^x^+^C$

y=Ce^1^0^x

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Example Question #41 : Differential Equations

$y=e^{2x}$ .

Calculate y'(1)-y'(0).

Possible Answers:

7.39

10.2

12.78

Correct answer:

12.78

Explanation:

Remember that the derivative of $e^u = e^u \cdot u'$ .

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

Now plug in the x=0,1 to find the corresponding values.

y'(1)=14.78

y'(0)=2.

Substitute these into the desired formula.

y'(1)-y'(0).

14.78-2=12.78

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Example Question #42 : Differential Equations

Determine the general solution to the following differential equation:

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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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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Calculus 1 : Differential Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #41 : Differential Equations

Example Question #31 : How To Find Solutions To Differential Equations

Example Question #1336 : Functions

Example Question #1337 : Functions

Example Question #41 : Differential Equations

Example Question #42 : Differential Equations

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

All Calculus 1 Resources

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