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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 413
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 the method of Laplace transforms to solve the given initial value problem. Here x′, y′, etc., denotes differentiation with respect to t; so does the symbol D.

x'-2y=2x(1)=1

x'+x-y'=t2+2t-1y(1)=0

The solution is x(t)=t2,y(t)=t-1.

See the step by step solution

Step by Step Solution

Step 1: Given information

Since the initial conditions are given at t=1, we need to shift the argument to the zero, so let

x1(t)=x(t+1)y1(t)=y(t+1)

Then x1(0)=x(1)=1 and y1(0)=y(1)=0 and

x1'(t)=x'(t+1)(t+1)=x'(t+1)y1'(t)=y'(t+1)(t+1)=y'(t+1)

Then the given system becomes

{x1'-2y1=2x1'+x1-y1'=(t+1)2-2(t+1)-1{x1'-2y1=2,   x1(0)=1x1'+x1-y1'=t2+4t+2,   y1(0)=0

Step 2: Take Laplace transform

Applying the Laplace transform gives

{L{x1'-2y1}(s)=L{2}(s)L{x1'+x1-y1'}(s)=L{t2+4t+2}(s)

Simplify equation , further as:

{sX1(s)-x1(0)-2Y1(s)=2ssX1(s)-x1(0)+X1(s)-sY1(s)-y1(0)=2s3+4s2+2s

{sX1(s)-1+2Y1(s)=2ssX1(s)-1+X1(s)-sY1(s)=2s3+4s2+2s

{X1(s)=2s2+2sY1(s)+1s(2+2s-s)Y1(s)=2s3-1s+2s2

Simplify equation(2+2s-s)Y1(s)=2s3-1s+2s2 , further as:

Y1(s)=2-s2+2ss3=1s2

Step 3: Finding the functions

Substituting 1s2 for Y1(s) in equation (1) gives

X1(s)=2s2+2s3+1s

Applying inverse Laplace we get

L-1{Y1(s)}(t)=L-1{1s2}(t)Þy1(t)=tL-1{X1(s)}(t)=L-1{2s2+2s3+1s}(t)

Take inverse Laplace transform we get;

x1(t)=t2+2t+1=t+12

Now we find

x(t)=x1(t-1)=t2y(t)=y1(t-1)=t-1

Therefore

x(t)=t2,y(t)=t-1

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