Particular Solutions of Undetermined Coefficients

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SUMMARY

The discussion focuses on finding particular solutions to the complex differential equation y'' + 2y' + 3y = -1 exp(2it). The initial solution y_c is derived using the method of undetermined coefficients, leading to the expression y(t) = e^(-t)(cos(sqrt(2)t) + sin(sqrt(2)t)) - 1/(4i-1)e^(2it). The correct particular solution for the equation involving sin(2t) is proposed as yp = (1/17)(1 + 4i)e^(2it), achieved by rationalizing the denominator of the earlier derived term. The discussion also highlights the importance of notation clarity in mathematical expressions.

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Homework Statement



Find a particular solution y_c to the complex differential equation
y'' + 2 y' + 3 y = -1 exp(2 i t)

y_c=

Using the solution y_c, construct a particular solution y_{1p} to the following differential equation
y'' + 2 y' + 3 y = -1 sin(2 t)
y_{1p} =

Again using the solution y_c, construct a particular solution y_{2p} to the following differential equation
y'' + 2 y' + 3 y = -1 cos(2 t)
y_{2p} =

Homework Equations



y = Ae^(2it)

The Attempt at a Solution



I let y'' + 2y' + 3y = 0 and then solve getting
e^(-t)(cos(sqrt2t) + sin(sqrt2t)) + g(h)

then solve for the g(h) by first finding A

y = Ae^(2it), y' = A2ie^(2it), y'' = -A4e^(2it)

substitute and solve for A getting -1/(4i - 1)


→ y(t) = e^(-t)(cos(sqrt2t) + sin(sqrt2t)) -1/(4i-1)e^(2it)
the Constants of integration can be anything for this problem so i just let them equal 1


but this is still not the right answer

please help
 
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osheari1 said:

Homework Statement



Find a particular solution y_c to the complex differential equation
y'' + 2 y' + 3 y = -1 exp(2 i t)

y_c=

Using the solution y_c, construct a particular solution y_{1p} to the following differential equation
y'' + 2 y' + 3 y = -1 sin(2 t)
y_{1p} =

Again using the solution y_c, construct a particular solution y_{2p} to the following differential equation
y'' + 2 y' + 3 y = -1 cos(2 t)
y_{2p} =

Homework Equations



y = Ae^(2it)

The Attempt at a Solution



I let y'' + 2y' + 3y = 0 and then solve getting
e^(-t)(cos(sqrt2t) + sin(sqrt2t)) + g(h)

then solve for the g(h) by first finding A

y = Ae^(2it), y' = A2ie^(2it), y'' = -A4e^(2it)

substitute and solve for A getting -1/(4i - 1)


→ y(t) = e^(-t)(cos(sqrt2t) + sin(sqrt2t)) -1/(4i-1)e^(2it)
the Constants of integration can be anything for this problem so i just let them equal 1


but this is still not the right answer

please help

Is the right answer yp = (1/17)(1 + 4i)e2it? If so, all I did was to convert -1/(4i - 1) by multiplying by the complex conjugate of the denominator over itself.

BTW, when you write cos(sqrt2t) + sin(sqrt2t), I can't tell if you mean
[itex]\sqrt{2t}[/itex] or [itex]\sqrt{2}t[/itex].

If you don't use LaTeX, write it like this: sqrt(2)t or t*sqrt(2).
 

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