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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 1
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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Short Answer

Use Euler’s method with step size h = 0.1 to approximate the solution to the initial value problem

y'=x-y2, y (1) = 0 at the points x=1.1, 1.2, 1.3, 1.4 and 1.5 .

xn1.11.21.31.41.5
yn0.10.2090.3250.4440.564
See the step by step solution

Step by Step Solution

Write the recursive formula

For the solution use the Euler’s formula yn+1=yn+h.fxn,yn

Apply recursive formula

We have, fx,y=x-y2,x0=1,y0=0,h=0.1

Then, yn+1=yn+h.fxn,yn=yn+0.1x-y2

Put  n=0 to find  y1

Now, find the value of y1

y1=y0+0.1x0-y02=0+0.11=0.1

Hence, the value of y1=0.1 when x1=1.1

Put  n=1 to find  y2

The value of y2 is

y2=y1+0.1x1-y12=0.1+0.11.1-0.01=0.1+0.109=0.209

Consequently, the value is y2=0.209 when x2=1.2

Put  n=2 to find  y3

Now the value of y3

y3=y2+0.1x2-y22=0.209+0.11.2-0.043=0.209+0.116=0.325

So, the value of y3=0.325 when x3=1.3

Put  n=3 to find  y4

The value of y4 is

y4=y3+0.1x3-y32=0.325+0.11.3-0.106=0.325+0.119=0.444

Thus, the value is y4=0.444 when x4=1.4

Put  n=4 to find  y5

The value of y5 is

y5=y4+0.1x4-y42=0.444+0.11.4-0.197=0.444+0.12=0.564

Therefore, the value is y5=0.564 when x5=1.5

Therefore the solution is

role="math" localid="1663939345323" xn1.11.21.31.41.5
yn0.10.2090.3250.4440.564

Most popular questions for Math Textbooks

Variation of Parameters. Here is another procedure for solving linear equations that is particularly useful for higher-order linear equations. This method is called variation of parameters. It is based on the idea that just by knowing the form of the solution, we can substitute into the given equation and solve for any unknowns. Here we illustrate the method for first-order equations (see Sections 4.6 and 6.4 for the generalization to higher-order equations).

(a) Show that the general solution to (20)dydx+P(x)y=Q(x) has the formy(x)=Cyh(x)+yp(x), where yh ( 0is a solution to equation (20) when Q(x)=0 ,

C is a constant, andyp(x)=v(x)yh(x) for a suitable function v(x). [Hint: Show that we can take yh=μ-1(x) and then use equation (8).] We can in fact determine the unknown function yhby solving a separable equation. Then direct substitution of vyh in the original equation will give a simple equation that can be solved for v.

Use this procedure to find the general solution to (21) localid="1663920708127" dydx+3xy=x2 , x > 0 by completing the following steps:

(b) Find a nontrivial solution yh to the separable equation (22) localid="1663920724944" dydx+3xy=0 , localid="1663920736626" x>0 .

(c) Assuming (21) has a solution of the formlocalid="1663920777078" yp(x)=v(x)yh(x) , substitute this into equation (21), and simplify to obtain localid="1663920789271" v'(x)=x2yh(x).

d) Now integrate to get localid="1663920800433" vx

(e) Verify that localid="1663920811828" y(x)=Cyh(x)+v(x)yh(x) is a general solution to (21).

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