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Problem 3.09PP
Solve the following ODEs using Laplace transforms;
(a) 5 ( 0 + 5 ( 0 + 3y (0 = 0; y (0 ) = 1, y (0 ) = 2
(b) 5 (0 - 25 (0 + 4 y (0 = 0 ; y (0 ) = 1. 5 ((0 = 2
(a) 5 (0 + j ( 0 = sin) = 1,5ii ( y ( 4 ) - 4 + i ] + y ( 4 ) = l 
(4*+l)y(4) = ̂ - l + 4 
y ( 4 ) = - r y i ---- r + - ^
^ ' 4^ (4^ + !) 4 * + l
4_____ l _
■ 4^(4 ’ + 1) 4’ + l 4 '+ l
__1_ 1 4 1
4’ 4’ + l * 4 ’ + l 4’ + !
= - L - 2 - ! - + - J -
4’ 4’ + l 4^+1
Apply inverse Laplace transform. 
Use the following formula;
Step 19 of 19
r'
c ' | = sini(U+4^
r' ( * |̂ = cos4fU + i’;
Therefore,
y(»)=/-2sin»+cosf
Therefore, the expression for y ( ' ) is, 
|y ( t ) = / - 2 s in f+ c o s f |.