Question about differential equations?(nonhomogeneous)

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In summary, The conversation discusses finding the particular solution for a problem involving the differential equation y''+9y'=162sin(9t)+1296cos(9t). The speaker proposes a setup but the expert summarizer points out that it is incorrect and provides the correct particular solution. The speaker also mentions the roots of the complementary function, but the expert notes that it is not relevant in finding the particular solution. The expert advises the speaker to simply plug in the particular solution and solve for the constants.
  • #1
hard_assteel
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For this problem I want to find the particular solution;
but i am not sure if it is right setup.


y''+9y'=162sin(9t)+1296cos(9t)

the roots for complementary function are 0,-9
so it has distinct roots.

my setup is;
[Asin(9t)+Bcos(9t)]+[Asin(9t)+Bcos(9t)]
Axsin(9t)+Bxcos(9t)

am I going about it the right way?

Thank You
 
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  • #2
hard_assteel said:
For this problem I want to find the particular solution;
but i am not sure if it is right setup.


y''+9y'=162sin(9t)+1296cos(9t)

the roots for complementary function are 0,-9
so it has distinct roots.

my setup is;
[Asin(9t)+Bcos(9t)]+[Asin(9t)+Bcos(9t)]
Axsin(9t)+Bxcos(9t)

am I going about it the right way?

Thank You
No. Here's your particular solution: yp = Asin(9t)+Bcos(9t)
This would be your particular solution for a right hand side of any of the following:
c1sin(9t)
c2cos(9t)
c1sin(9t) + c2cos(9t)
 
  • #3
hard_assteel said:
y''+9y'=162sin(9t)+1296cos(9t)

the roots for complementary function are 0,-9
so it has distinct roots.

That's true, but it's of no use in finding the particular solution.

my setup is;
[Asin(9t)+Bcos(9t)]+[Asin(9t)+Bcos(9t)]
Axsin(9t)+Bxcos(9t)

am I going about it the right way?

To be honest, I can't figure out what you're thinking. But no, what you have written is not right. All you need is [itex]y_P=A\sin(9t)+B\cos(9t)[/itex]. Plug it into your differential equation and solve for [itex]A[/itex] and [itex]B[/itex].
 

1. What is the difference between homogeneous and nonhomogeneous differential equations?

Homogeneous differential equations are equations in which all terms contain the dependent variable or its derivatives. Nonhomogeneous differential equations, on the other hand, have at least one term that does not contain the dependent variable or its derivatives.

2. How do you solve nonhomogeneous differential equations?

To solve a nonhomogeneous differential equation, you need to first solve the corresponding homogeneous equation. Then, using the method of variation of parameters, you can find a particular solution to the nonhomogeneous equation.

3. What is the role of initial conditions in solving nonhomogeneous differential equations?

Initial conditions are used to find the particular solution to a nonhomogeneous differential equation. They help to determine the values of the arbitrary constants that are needed to fully solve the equation.

4. Can nonhomogeneous differential equations have multiple solutions?

Yes, a nonhomogeneous differential equation can have multiple solutions. This is because there can be an infinite number of particular solutions that satisfy the nonhomogeneous equation. However, the general solution will still contain a single arbitrary constant.

5. Are there real-world applications for nonhomogeneous differential equations?

Yes, nonhomogeneous differential equations have many real-world applications in fields such as physics, engineering, and economics. They are often used to model systems that involve external forces or inputs, such as electrical circuits, chemical reactions, and population growth.

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