   Chapter 6, Problem 14RE

Chapter
Section
Textbook Problem

Euler's Method In Exercises 13 and 14, use Euler’s Method to make a table of values for the approximate solution of the differential equation with the specified initial value. Use n steps of size h. y ' = 5 x − 2 y , y ( 0 ) = 2 , n = 10 , h = 0.01

To determine

To calculate: The table of values for the approximate solution of the differential equation y=5x2y  , y(0)=2 , n=10, h=0.1 by Euler’ method.

Explanation

Given:

y=5x2y  , y(0)=2 , n=10, h=0.1

Formula used:

If y=f(x,y) and it passes through point (x0,y0). Then proceed in the direction indicated by slope with a step size h. Then, move along the tangent line, to get another point (x1,y1),

x1=x0+h  and  y1=y0+hf(x0,y0).

Continue the above process to get:

x1=x0+h  and    y1=y0+hf(x0,y0)x2=x1+h  and    y2=y1+hf(x1,y1)                               xn=xn1+h  and  yn=yn1+hf(xn1,yn1)

Calculation:

The differential equation y=5x2y passes through point (0,2).

So,

x0=0,y0=2,h=0.1 and f(x,y)=5x2y

x1 can be calculated as:

x1=x0+h=0+0.1=0.1

The first approximation is:

y1=y0+hf(x0,y0)=2+(0.1)(5(0)2(2))=2+(0.1)(4)=1.6

Thus, the point (x1,y1) is equal to (0.1,1.6).

x2 can be calculated as:

x2=x1+h=0.1+0.1=0.2

The second approximation is:

y2=y1+hf(x1,y1)=1.6+(0.1)(5(0.1)2(1.6))=1.33

Thus, the point (x2,y2) is equal to (0.2,1.33).

x3 can be calculated as:

x3=x2+h=0.2+0.1=0.3

The third approximation is:

y3=y2+hf(x2,y2)=1.33+(0.1)(5(0.2)2(1.33))=1.164

Thus, the point (x3,y3) is equal to (0.3,1.164).

x4 can be calculated as:

x4=x3+h=0.3+0.1=0.4

The fourth approximation is:

y4=y3+hf(x3,y3)=1.164+(0.1)(5(0.3)2(1.164))=1.0812

Thus, the point (x4,y4) is equal to (0.4,1.081).

x5 can be calculated as:

x5=x4+h=0

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