Solve the following ODE by using the method of undetermined coefficients in which Euler's formula needs to be utilized: y′′−2y′+y=sin(t)

Violet Woodward 2022-07-22 Answered
Solve the following ODE by using the method of undetermined coefficients in which Euler's formula needs to be utilized:
y 2 y + y = sin ( t )
The way that I solved this doesn't involve Euler's formula, and I was wondering how I might use the formula here.

My approach:
The formula can be written as y ( t ) = y h ( t ) + y p ( t ) where y h ( t ) is the "homogeneous version" of the ODE and y p ( t ) is the particular solution that we'll obtain via the basic rule of the method of undetermined coefficients.

y h ( t ):
Putting r ( t ) = sin ( t ) = 0 in the original equation, the ODE we need to solve is:
y 2 y + y = 0
where we can set the general solution as y = e λ t and obtain the characteristic equation:
λ 2 2 λ + 1 = 0
which has a real double root, hence giving us the solution:
y h ( t ) = ( c 1 + c 2 t ) e t

y p ( t ):
Judging by the fact that r ( t ) is shape k sin ( ω t ) and we know that ω = 1 we can set the general solution to be of form:
y p ( t ) = K cos ( t ) + M sin ( t ) y p ( t ) = K sin ( t ) + M cos ( t ) y p ( t ) = K cos ( t ) M sin ( t )
substituting these equations into the original equation and then simplifying gives us:
y p ( t ) = 1 2 cos ( t )
And in conclusion, we can write that the solution to the given ODE is:
y ( t ) = y h ( t ) + y p ( t ) = ( c 1 + c 2 t ) e t + 1 2 cos ( t )
How would we be able to derive this conclusion via Euler's formula? Thanks in advance.
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Answers (2)

repotasonwf
Answered 2022-07-23 Author has 12 answers
So, here Euler's formula means using sin t == e i t e i t 2 i The particular integral will be found as
f ( D ) y = e a t y p ( t ) = e a t f ( a )
Where we have
( D 1 ) 2 y = e i t e i t 2 i y p ( t ) = ( 2 i ) 1 ( e i t ( i 1 ) 2 e i t ( i 1 ) 2 ) = 1 4 [ e i t + e i t ]
= 1 2 cos t .
y h ( t ) remains the same as you found.
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Luz Stokes
Answered 2022-07-24 Author has 3 answers
Consider a new differential equation
(1) d 2 y d t 2 2 d y d t + 1 = ( d 2 d t 2 d d t + 1 ) y = e i t .
Let y = e i t g ( t ). By the product rule
d y d t = d d t ( e i t g ( t ) ) = e i t d d t g ( t ) + ( d d t e i t ) g ( t ) = e i t ( d d t + i ) g ( t ) .
Then the differential equation (1) becomes
e i t [ ( d d t + i ) 2 2 ( d d t + i ) + 1 ] g ( t ) = e i t
so that after cancelling e i t from both sides and expanding the expression in parenthesis on the LHS
[ ( d d t + i ) 2 2 ( d d t + i ) + 1 ] g ( t ) = ( d 2 d t 2 + ( 2 i 2 ) d d t 2 i ) g ( t ) = 1
so that g ( t ) = c for some constant c C . Therefore, g ( t ) = i / 2. A particular solution of y would then be
y = i 2 e i t = i 2 ( cos t + i sin t ) .
Since sin t = I m ( e i t ), then a particular solution to
d 2 y d t 2 2 d y d t + 1 = ( d 2 d t 2 d d t + 1 ) y p = sin t
is y p = I m ( y ) = I m ( i 2 ( cos t + i sin t ) ) = 1 2 cos t
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