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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 47
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

Question: Consider the initial value problem dydx+1+sin2xy=x,y(0)=2.

(a) Using definite integration, show that the integrating factor for the differential equation can be written as μ(x)=(0x1+sin2tdt) and that the solution to the initial value problem is y(x)=1μ(x)0xμ(s)sds+2μ(x)

(b) Obtain an approximation to the solution at x = 1 by using numerical integration (such as Simpson’s rule, Appendix C) in a nested loop to estimate values of μ(x)and, thereby, the value of 01μ(s)ds.

[Hint: First, use Simpson’s rule to approximate μ(x) at x = 0.1, 0.2, . . . , 1. Then use these values and apply Simpson’s rule again to approximate 01μ(s)ds]

(c) Use Euler’s method (Section 1.4) to approximate the solution at x = 1, with step sizes h = 0.1 and 0.05. [A direct comparison of the merits of the two numerical schemes in parts (b) and (c) is very complicated, since it should take into account the number of functional evaluations in each algorithm as well as the inherent accuracies.]

  1. 0.9960
  2. 0.9486
  3. 0.9729
See the step by step solution

Step by Step Solution

Step 1(a): Find the value of y(x)

Since Pt=1+sin2t is a continuous real valued function on an open interval which contains points 0 we can use fundamental theorem of calculus to obtain


We can write integrating factor as μ(x)=exp0x1+sin2tdt)

Multiplying by μx




Step 2: Find the values of μ(x)

Here ys=1μs0sxμxdx+2μs

Put x=1 then y1=1μ10sxμsdx+2μ1

We find the value of01μssdx.

Using Simpson’s rule

abfxdx=x3k=1nfx2k-2+4fx2k-2+fx2kx=b-a2n,xk=a+kx, k=0.1,0.2,...,1

So, μ(x)=exp0x1+sin2tdt) at x=0.1,0.2, ……1

The values are

x = 0, =1

x = 0.1, =1.105354

x = 0.2, =1.223010

x = 0 .3, =1.355761

x = 0 .4, =1.506975

x = 0.5, =1.680635

x = 0 .6, =1.881401

x = 0 .7, =2.114679

x = 0 .8, =2.386713

x = 0 .9, =2.70670

x = 1, =3.076723

Now, using the previous conclusions



role="math" localid="1663932388162" 01sμsds=0.13[0·μ0+40.1·μ0.1+20.2·μ0.2+40.3·μ0.3+20.4·μ0.4+40.5·μ0.5+20.6·μ0.6 +40.7·μ0.7+20.8·μ0.8+40.9·μ0.9+1·μ1]=1.064539role="math" localid="1663932537871" y1=1μ10sxμsdx+2μ1=0.9960

Therefore, y(1)=0.9960

 Step 3(b): Find the value of  y(1) at h=0.1

The differential equation is y'=x-1+sin2xy

Use the recursive formula

localid="1663933001609" xn+1=xn+hyn+1=yn+hfxnyn=yn+hfxn-1+sin2xnyn


x0=0 and y0=0, h=0.1, N=10 steps at x=1

The values are

x1=0.1, y1=1.8x2=0.2, y2=1.6921x3=0.3, y3=1.4830x4=0.4, y4=1.3584x5=0.5, y5=1.2526x6=0.6, y6=1.1637x7=0.7, y7=1.09x8=0.8, y8=1.0304x9=0.9, y9=0.9836x10=1, y10=0.9586

Therefore, y=0.9586

 Step 4 (c): Determine the value of y(1) at h=0.05

x0=0andy0=0, h=0.1, N=20 steps at x=1

x1=0.05, y1=1.9x2=0.1, y2=1.8074x3=0.15, y3=1.4830x4=0.2, y4=1.642x5=0.25, y5=1.5683x6=0.3, y6=1.5x7=0.35, y7=1.4368x8=0.4, y8=1.3784x9=0.45, y9=1.3244x10=0.5, y10=1.2747x11=0.55, y11=1.229x12=0.6, y12=1.187x13=0.65, y13=1.149x14=0.7, y14=1.1144x15=0.75, y15=1.0831x16=0.8, y16=1.0551x17=0.85, y17=1.0301x18=0.9, y18=1.0082x19=0.95, y19=0.9892x20=1, y20=0.9729

Therefore, y(1)=0.9729 y(1)=0.9729

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