Find the solution of the given initial value problem. y'+(2/t)y=(cos t)t^2 , y(3.14)=0, t>0

Deromediqm 2022-07-29 Answered
Find the solution of the given initial value problem.
y + ( 2 / t ) y = ( cos t ) t 2 , y ( 3.14 ) = 0 , t > 0
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Answers (1)

repotasonwf
Answered 2022-07-30 Author has 12 answers
Given equation is
y 1 + 2 t y = cos t t 2 ( 1 )
This is a linear equation of first order in y
where P ( t ) = 2 t a n d Q ( t ) = cos t t 2
P ( t ) d t = 2 t d t = 2 log t = log t 2
Integrating Factor = e P ( t ) d t = e log t 2 = t 2
Multiplying (1) with t 2 , we get
t 2 y 1 + 2 t y = cos t ( 2 )
(2) can be written as
d d t [ y t 2 ] = cos t
d d t [ y t 2 ] d t = cos t d t
y t 2 = sin t + C
y ( t ) = sin t t 2 + C t 2
we know that y ( π ) = 0
so sin π π 2 + C π 2 = 0 C = 0
Therefore the general solution of (1) is y = sin t t 2
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