The system is as follows: i_1=i_2+i_3, 50sin t=6i_1+i'_2+5i'_2, 50sin t=6i_1+i'_3, Find i_2,i_3.

Tara Mayer

Tara Mayer

Answered question

2022-10-14

The system is as follows:
i 1 = i 2 + i 3 , 50 sin t = 6 i 1 + i 2 + 5 i 2 , 50 sin t = 6 i 1 + i 3 ,
Find i 2 , i 3 .

Answer & Explanation

scranna0o

scranna0o

Beginner2022-10-15Added 16 answers

Notice that the equations could be re-written as
i 1 = i 2 + i 3 i 2 = 50 sin t 6 i 1 , H 5 i 2 , H i 3 = 50 sin t 6 i 1 , H
now replacing from the second and third equations into the first one, after simplification, we could write
d d t [ i 1 i 2 i 3 ] = [ 12 5 0 6 5 0 6 0 0 ] [ i 1 i 2 i 3 ] + [ 100 sin t 50 sin t 50 sin t ]
now for the homogeneous part
d d t i H = [ 12 5 0 6 5 0 6 0 0 ] i H
Decomposition of the above matrix yields
[ 12 5 0 6 5 0 6 0 0 ] = [ 5 1 0 3 2 0 2 3 1 ] [ 15 0 0 0 2 0 0 0 0 ] [ 5 1 0 3 2 0 2 3 1 ] 1
so that by defining
j H = [ 5 1 0 3 2 0 2 3 1 ] 1 i H
we get
d d t j H = [ 15 0 0 0 2 0 0 0 0 ] j H
which are three uncoupled first order ODEs which yield
j H = [ c 1 e 15 t c 2 e 2 t c 3 ]
Also, for the particular part, according to the inhomogeneous part, we should have
i P = sin t [ A 1 A 2 A 3 ] + cos t [ B 1 B 2 B 3 ]
Replacing the particular solution form into the original equation gives
sin t ( [ 12 5 0 6 5 0 6 0 0 ] [ A 1 A 2 A 3 ] + [ B 1 B 2 B 3 ] + [ 100 50 50 ] ) + cos t ( [ A 1 A 2 A 3 ] [ 12 5 0 6 5 0 6 0 0 ] [ B 1 B 2 B 3 ] ) = [ 0 0 0 ]
Now you should only summarize these steps to get the final result.
omgespit9q

omgespit9q

Beginner2022-10-16Added 3 answers

Now an alternative (non-matrix based) solution. After eliminating 𝑖1 from the first equation we get
i 2 = 11 i 2 6 i 3 + 50 sin t i 3 = 6 i 2 6 i 3 + 50 sin t
Decomposing into homogeneous and particular parts for the unknowns, for the first part we have
i 2 , H = 11 i 2 , H 6 i 3 , H i 3 , H = 6 i 2 , H 6 i 3 , H
Now we shall assume the solutions to have an exponential behavior. That is
i 2 , H = A 2 e r t i 3 , H = A 3 e r t
Replacing into the equations, we get
( 11 + r ) A 2 + 6 A 3 = 0 6 A 2 + ( 6 + r ) A 3 = 0

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