Particular Solutions of Undetermined Coefficients

In summary, the conversation was about finding particular solutions to complex differential equations using a given solution. The solution for y_c was found to be e^(-t)(cos(sqrt2t) + sin(sqrt2t)) -1/(4i-1)e^(2it). However, the solution for y_{1p} and y_{2p} were not correct and needed to be adjusted by converting -1/(4i-1) using the complex conjugate of the denominator. The correct solution for y_{1p} was found to be (1/17)(1 + 4i)e2it.
  • #1
osheari1
5
0

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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  • #2
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).
 

1. How do you determine the particular solution of undetermined coefficients?

The particular solution of undetermined coefficients is determined by first finding the general solution of the homogeneous equation. Then, a particular solution is found by using the method of undetermined coefficients, where the coefficients are chosen based on the form of the non-homogeneous term in the original equation.

2. What is the method of undetermined coefficients?

The method of undetermined coefficients is a technique used to find a particular solution to a non-homogeneous linear differential equation. It involves choosing a particular form for the solution and then determining the coefficients of that form by plugging it into the original equation.

3. When is the method of undetermined coefficients applicable?

The method of undetermined coefficients is applicable when the non-homogeneous term in the original equation is a polynomial, exponential, sine, cosine, or a combination of these functions. It is not applicable for non-linear differential equations.

4. What are the limitations of the method of undetermined coefficients?

The method of undetermined coefficients has limitations when the non-homogeneous term is a more complicated function, such as a product of trigonometric functions or a polynomial multiplied by an exponential. In these cases, other methods such as variation of parameters or the annihilator method may be used.

5. How do you know if the particular solution is correct?

The particular solution can be checked by plugging it back into the original equation and verifying that it satisfies the equation. Additionally, the general solution should also satisfy the equation and the particular solution should be unique.

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