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Calculus 2 : Derivatives

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Example Questions

Example Question #5 : Euler's Method

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=\frac{y}{x}$ at x=2 with the initial condition y(1)=1 and step size h=0.25 .

Possible Answers:

(2,1)

(2,3)

(2,2)

(2,1.75)

Correct answer:

(2,2)

Explanation:

Euler's method uses iterative equations to find a numerical solution to a differential equation. The following equations

$x_{n+1}=x_n+h$

$y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value. f(x_n,y_n) is the solution to the differential equation.

In this problem,

$f(x,y)=\frac{dy}{dx}=\frac{y}{x}$

h=0.25

Starting at the initial point (x_0,y_0)=(1,1)

$x_{1}=x_0+h=1+0.25=1.25$

$y_{1}=y_0+h*f(x_0,y_0)=1+0.25*(1/1)=1.25$

(x_1,y_1)=(1.25,1.25)

$x_{2}=x_1+h=1.25+0.25=1.5$

$y_{2}=y_1+h*f(x_1,y_1)=1.25+0.25*(1.25/1.25)=1.5$

(x_2,y_2)=(1.5,1.5)

We continue using Euler's method until x=2 . The results of Euler's method are in the table below.

Problem 3

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Example Question #6 : Euler's Method

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=\frac{y}{x}$ at x=2 with the initial condition y(1)=1 and step size h=0.1 .

Possible Answers:

(2,1)

(2,3)

(2,2)

(2,1.75)

Correct answer:

(2,2)

Explanation:

Euler's method uses iterative equations to find a numerical solution to a differential equation. The following equations

$x_{n+1}=x_n+h$

$y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value. f(x_n,y_n) is the solution to the differential equation.

In this problem,

$f(x,y)=\frac{dy}{dx}=\frac{y}{x}$

h=0.1

Starting at the initial point (x_0,y_0)=(1,1)

$x_{1}=x_0+h=1+0.1=1.1$

$y_{1}=y_0+h*f(x_0,y_0)=1+0.1*(1/1)=1.1$

(x_1,y_1)=(1.1,1.1)

$x_{2}=x_1+h=1.1+0.1=1.2$

$y_{2}=y_1+h*f(x_1,y_1)=1.1+0.1*(1.1/1.1)=1.2$

(x_2,y_2)=(1.2,1.2)

We continue using Euler's method until x=2 . The results of Euler's method are in the table below.

Problem 4

Note: Due to the simplicity of the differential equation, Euler's method finds the exact solution, even with a large step size, Using a smaller step size is unnecessary and more time consuming.

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Example Question #7 : Euler's Method

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=cos(x)$ at $x=\pi/2$ with the initial condition y(0)=0 and step size $h= \pi/10$ .

Possible Answers:

$(\pi/2,0.314)$

$(\pi/2,1.57)$

$(\pi/2,1.25)$

$(\pi/2,1.14)$

Correct answer:

$(\pi/2,1.14)$

Explanation:

Euler's method uses iterative equations to find a numerical solution to a differential equation. The following equations

$x_{n+1}=x_n+h$

$y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value. f(x_n,y_n) is the solution to the differential equation.

In this problem,

$f(x,y)=\frac{dy}{dx}=cos(x)$

$h= \pi/10$

Starting at the initial point (x_0,y_0)=(0,0)

$x_{1}=x_0+h=0+\pi/10=\pi/10$

$y_{1}=y_0+h*f(x_0,y_0)=0+(\pi/10)*cos(0)=\pi/10$

$(x_1,y_1)=(\pi/10,\pi/10)$

$x_{2}=x_1+h=\pi/10+\pi/10=\pi/5$

$y_{2}=y_1+h*f(x_1,y_1)=\pi/10+(\pi/10)*cos(\pi/10)=\pi/5$

$(x_2,y_2)=(\pi/5,\pi/5)$

We continue using Euler's method until $x=\pi/2$ . The results of Euler's method are in the table below.

Problem 7

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Example Question #2 : Euler's Method

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=cos(x)$ at $x=\pi/2$ with the initial condition y(0)=0 and step size $h= \pi/20$ .

Possible Answers:

$(\pi/2,2)$

$(\pi/2,1.25)$

$(\pi/2,0.314)$

$(\pi/2,1.07)$

Correct answer:

$(\pi/2,1.07)$

Explanation:

Euler's method uses iterative equations to find a numerical solution to a differential equation. The following equations

$x_{n+1}=x_n+h$

$y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value. f(x_n,y_n) is the solution to the differential equation.

In this problem,

$f(x,y)=\frac{dy}{dx}=cos(x)$

$h= \pi/20$

Starting at the initial point (x_0,y_0)=(0,0)

$x_{1}=x_0+h=0+\pi/20=\pi/20$

$y_{1}=y_0+h*f(x_0,y_0)=0+(\pi/20)*cos(0)=\pi/20$

$(x_1,y_1)=(\pi/20,\pi/20)$

$x_{2}=x_1+h=\pi/20+\pi/20=\pi/10$

$y_{2}=y_1+h*f(x_1,y_1)=\pi/20+(\pi/20)*cos(\pi/20)=\pi/10$

$(x_2,y_2)=(\pi/10,\pi/10)$

We continue using Euler's method until $x=\pi/2$ . The results of Euler's method are in the table below.

Problem 8

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Example Question #9 : Euler's Method

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=y^2e^{x}$ at x=6 with the initial condition y(0)=0.01 and step size h= 1 .

Possible Answers:

(6,1.0157)

(6,0.0239)

(6,0.1089)

(6,0.1013)

Correct answer:

(6,0.1089)

Explanation:

Euler's method uses iterative equations to find a numerical solution to a differential equation. The following equations

$x_{n+1}=x_n+h$

$y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value. f(x_n,y_n) is the solution to the differential equation.

In this problem,

$f(x,y)=\frac{dy}{dx}=y^2e^{x}$

h= 1

Starting at the initial point (x_0,y_0)=(0,0.01)

$x_{1}=x_0+h=0+1=1$

$y_{1}=y_0+h*f(x_0,y_0)=0.01+1*(0.01)^2*e^{0}=0.0101$

(x_1,y_1)=(1,0.0101)

$x_{2}=x_1+h=1+1=2$

$y_{2}=y_1+h*f(x_1,y_1)=0.0101+1*(0.0101)*e^{1}=0.01038$

(x_2,y_2)=(2,0.01038)

We continue using Euler's method until x=6 . The results of Euler's method are in the table below.

Problem 11

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Example Question #10 : Euler's Method

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=y^2e^{x}$ at x=6 with the initial condition y(0)=0.01 and step size h= 0.5 .

Possible Answers:

(6,0.013)

(6,0.012)

(6,5.113)

(6,0.1089)

Correct answer:

(6,5.113)

Explanation:

Euler's method uses iterative equations to find a numerical solution to a differential equation. The following equations

$x_{n+1}=x_n+h$

$y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value. f(x_n,y_n) is the solution to the differential equation.

In this problem,

$f(x,y)=\frac{dy}{dx}=y^2e^{x}$

h= 0.5

Starting at the initial point (x_0,y_0)=(0,0.01)

$x_{1}=x_0+h=0+0.5=0.5$

$y_{1}=y_0+h*f(x_0,y_0)=0.01+0.5*(0.01)^2*e^{0}=0.01005$

(x_1,y_1)=(0.5,0.01005)

$x_{2}=x_1+h=0.5+0.5=1$

$y_{2}=y_1+h*f(x_1,y_1)=0.01005+0.5*(0.01005)*e^{1}=0.01013$

(x_2,y_2)=(1,0.01013)

We continue using Euler's method until x=6 . The results of Euler's method are in the table below.

Problem 12

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Example Question #11 : Euler's Method

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=ln\left (\frac{y}{x} \right )$ at x=3 with the initial condition y(1)=1 and step size h=0.5 .

Possible Answers:

(3,3)

(3,1)

(3,2.75)

(3,2)

Correct answer:

(3,2)

Explanation:

Euler's method uses iterative equations to find a numerical solution to a differential equation. The following equations

$x_{n+1}=x_n+h$

$y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value. f(x_n,y_n) is the solution to the differential equation.

In this problem,

$f(x)=\frac{dy}{dx}=ln\left (\frac{y}{x} \right )$

h=0.5

Starting at the initial point (x_0,y_0)=(1,1)

$x_{1}=x_0+h=1+0.5=1.5$

$y_{1}=y_0+h*f(x_0,y_0)=1+0.5*ln(1/1)=1$

(x_1,y_1)=(1.5,1)

$x_{2}=x_1+h=1.5+0.5=2$

$y_{2}=y_1+h*f(x_1,y_1)=1+0.5*ln(1/1.5)=1.333$

(x_2,y_2)=(2,1.333)

We continue using Euler's method until x=3 . The results of Euler's method are in the table below.

Problem 5

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Example Question #12 : Euler's Method

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=ln\left (\frac{y}{x} \right )$ at x=3 with the initial condition y(1)=1 and step size h=0.2 .

Possible Answers:

(3,2.5)

(3,2)

(3,1.25)

(3,3)

Correct answer:

(3,2.5)

Explanation:

Euler's method uses iterative equations to find a numerical solution to a differential equation. The following equations

$x_{n+1}=x_n+h$

$y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value. f(x_n,y_n) is the solution to the differential equation.

In this problem,

$f(x)=\frac{dy}{dx}=ln\left (\frac{y}{x} \right )$

h=0.2

Starting at the initial point (x_0,y_0)=(1,1)

$x_{1}=x_0+h=1+0.2=1.2$

$y_{1}=y_0+h*f(x_0,y_0)=1+0.2*ln(1/1)=1$

(x_1,y_1)=(1.2,1)

$x_{2}=x_1+h=1.2+0.2=1.4$

$y_{2}=y_1+h*f(x_1,y_1)=1+0.2*ln(1/1.2)=1.167$

(x_2,y_2)=(1.4,1.167)

We continue using Euler's method until x=3 . The results of Euler's method are in the table below.

Problem 6

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Example Question #13 : Euler's Method

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=e^{x}$ at x=5 with the initial condition y(0)=1 and step size h= 1 .

Possible Answers:

(5,32.19)

(5,86.79)

(5,65.49)

(5,94.77)

Correct answer:

(5,86.79)

Explanation:

Euler's method uses iterative equations to find a numerical solution to a differential equation. The following equations

$x_{n+1}=x_n+h$

$y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value. f(x_n,y_n) is the solution to the differential equation.

In this problem,

$f(x,y)=\frac{dy}{dx}=e^{x}$

h= 1

Starting at the initial point (x_0,y_0)=(0,1)

$x_{1}=x_0+h=0+1=1$

$y_{1}=y_0+h*f(x_0,y_0)=1+1*e^{0}=2$

(x_1,y_1)=(1,2)

$x_{2}=x_1+h=1+1=2$

$y_{2}=y_1+h*f(x_1,y_1)=2+1*e^1=4.718$

(x_2,y_2)=(2,4.718)

We continue using Euler's method until x=5 . The results of Euler's method are in the table below.

Problem 9

Note: Using a step size of gives an answer that is not close to the correct answer, y(5)=148.413 . This is because the step size is too small.

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Example Question #14 : Euler's Method

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=e^{x}$ at x=5 with the initial condition y(0)=1 and step size h= 0.5 .

Possible Answers:

(5,129.26)

(5,114.62)

(5,132.77)

(5,69.61)

Correct answer:

(5,114.62)

Explanation:

Euler's method uses iterative equations to find a numerical solution to a differential equation. The following equations

$x_{n+1}=x_n+h$

$y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value. f(x_n,y_n) is the solution to the differential equation.

In this problem,

$f(x,y)=\frac{dy}{dx}=e^{x}$

h= 1

Starting at the initial point (x_0,y_0)=(0,1)

$x_{1}=x_0+h=0+0.5=0.5$

$y_{1}=y_0+h*f(x_0,y_0)=1+0.5*e^{0}=1.5$

(x_1,y_1)=(0.5,1.5)

$x_{2}=x_1+h=0.5+0.5=1$

$y_{2}=y_1+h*f(x_1,y_1)=1.5+0.5*e^{0.5}=2.324$

(x_2,y_2)=(1,2.324)

We continue using Euler's method until x=5 . The results of Euler's method are in the table below.

Problem 10

Note: Using a larger step size got us closer the correct answer, y(5)=148.413 .

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Calculus 2 : Derivatives

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #5 : Euler's Method

Example Question #6 : Euler's Method

Example Question #7 : Euler's Method

Example Question #2 : Euler's Method

Example Question #9 : Euler's Method

Example Question #10 : Euler's Method

Example Question #11 : Euler's Method

Example Question #12 : Euler's Method

Example Question #13 : Euler's Method

Example Question #14 : Euler's Method

All Calculus 2 Resources

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