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Q23E

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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 391
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

solve the given initial value problem using the method of Laplace transforms. Sketch the graph of the solution.y''+y=t--(t--4)u(t--2);y(0)=0,y'(0)=1

On solving the given initial value problem using the method of Laplace transforms the solution is y(t)=t-(t-4+2cos(t-2)-sin(t-2))u(t-2)and the graph is

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Definition

The Laplace transform, is an integral transform that converts a function of a real variable usually t, the time domain to a function of a complex variable s.

Step 2: Applying Laplace transform and using its linearity  

Ly''+y=L{t-(t-4)u(t-2)}Ly''+L{y}=L{t}-L{(t-4)u(t-2)}

s2Y(s)-sy(0)-y(0)+Y(s)=1s2-L{(t-4)u(t-2)}s2Y(s)-1+Y(s)=1s2-L{(t-4)u(t-2)}

Since,

L{(t-4)u(t-2)}=e-2sL{t-2}=e-2s(L{t}-2L{1})=e-2s1s2-2s

We have that,

s2+1Y(s)=1s2-e-2s1s2-2s+1s2+1Y(s)=1+s2s2-e-2s1-2ss2Y(s)=1s2-e-2s1-2ss2s2+1

Using partial fractions we get,

1-2ss2s2+1=1s2-2s+2s-1s2+1

And inverse Laplace transform gives,

L-1e-2ss2=(t-2)u(t-2)L-12e-2ss=2u(t-2)

2L-12e-2sss2+1=2cos(t-2)u(t-2)L-1e-2ss2+1=sin(t-2)u(t-2)

Step 3: The solution of given IVP

y(t)=L-11s2-e-2s1-2ss2s2+1=L-11s2-L-1e-2ss2-2e-2ss+2e-2sss2+1-e-2ss2+1=t-((t-2)u(t-2)-2u(t-2)+2cos(t-2)u(t-2)-sin(t-2)u(t-2))=t-(t-4+2cos(t-2)-sin(t-2))u(t-2)

Hence, y(t)=t-(t-4+2cos(t-2)-sin(t-2))u(t-2)

and the graph is

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