Solve Particular Solution Differential Equation: {\theta}''(t)-{\theta}(t)=tsint

In summary: I don't actually use complex solutions, I just plug my particular solution into the homogeneous equation and differentiate. But I understand what you're saying, it's just that I'm used to it and it's easier for me.
  • #1
Precursor
222
0
Homework Statement
Find a particular solution to the differential equation: [tex]{\theta}''(t)-{\theta}(t)=tsint[/tex]

The attempt at a solution
So I started by using the particular solution [tex]{\theta}_{p}=(At+B)(Csint+Dcost)[/tex]
Before I continue with the rest of the solution, is this correct so far?
 
Last edited:
Physics news on Phys.org
  • #2
Looks right, you should have 4 coefficients to determine for a particular solution of that form.
 
  • #3
So I continue by finding the first and second derivatives of that particular solution:
[tex]{\theta}'_{p}=(At+B)(Ccost-Dsint)+A(Csint+Dcost)[/tex]
[tex]{\theta}''_{p}=(At+B)(-Csint-Dcost)+A(Ccost-Dsint)+A(Ccost-Dsint)[/tex]
Now I substitute back into the differential equation and get
[tex]-2(At+B)(Csint+Dcost)+2A(Ccost-Dsint)=tsint[/tex]
From here I'm having a bit of difficulty. How do I solve for the coefficients?
 
  • #4
Well I would've personally multiplied everything out in your trial solution before differentiating, so you'd have Atcost + Btsint + Ccost + Dsint. Remember your coefficients arbitrary, but in your case you must try 4.

Then differentiate for Y'p and Y''p, plug into your homogenous equation, regroup/factor everything in terms of sint, cost, tsint, tcost, then set equal to tsint and solve the system of equations.

These problems stress one's neatness and eyesight, be careful haha.
 
  • #5
Did it myself(tempted as my final is coming up lol), and this is the correct solution:
http://www.wolframalpha.com/input/?i=y''+-y+=+xsinx

(y and x are your theta and t, wolfram is a lifesaver!)

Also always be sure to check that your particular solutions are linearly independent with the solutions to the homogeneous part. Clear sails on this one as you've got real-valued exponentials for your homogeneous part.
 
  • #6
The coefficients are ultimately determined using the initial conditions, where y(t) or y'(t) are given for some value t. If no initial conditions are given, the coefficients remain unknown.
 
  • #7
You mean the coefficients of the homogeneous part of the solution yes?
 
  • #8
So I've worked it through and I get

AC = -1/2
AD = 0
AD = -BC
AC = BD

This doesn't seem right though.
 
  • #9
I have no idea how you reached that. Here's how I did it:
http://img685.imageshack.us/img685/1977/p1000999sx.jpg [Broken]
 
Last edited by a moderator:
  • #10
So this particular solution is not correct: [tex]{\theta}_{p}=(At+B)(Csint+Dcost)[/tex]

Because when I multiply through I get: [tex]ACtsint+ADtcost+BCsint+BDcost[/tex]


Oh, do I replace the coefficients AC, AD, BC, BD to A, B, C, D?
 
  • #11
Yep, that's what I meant by "your coefficients are arbitrary".
 
  • #12
Alright I've finally got the correct solution. But you managed to do it showing a lot less work. Did you skip some simplification steps?

But thanks for the help!
 
  • #13
You're welcome. I don't expand anything out unless I have to, avoids a possible source of mistakes in my experience. When I plug my particular and its derivatives back in I just factor the x terms out of everything(identifying them term by term) so I can see my coefficients clearly and what each equation is equal to: (0,1,0,0) in this case. Very "linear", some people do the same thing but with columns.
 
  • #14
If you've already learned complex calculus, I strongly recommend you to use complex solutions, it makes it much easier
 
  • #15
Do you mean trying a trial solution with complex exponentials, so you'd only have 2 coefficients to solve? Never thought of that, but it sounds like a clever idea.
 

1. What is a particular solution differential equation?

A particular solution differential equation is an equation that represents the relationship between a variable and its rate of change. It involves finding a specific solution that satisfies the equation among an infinite number of possible solutions.

2. How do you solve a particular solution differential equation?

To solve a particular solution differential equation, you can use various methods such as separation of variables, integrating factors, or using a substitution. In this case, you can use the method of undetermined coefficients to find a particular solution that satisfies the given equation.

3. What does {\theta}''(t) and {\theta}(t) represent in this equation?

In this equation, {\theta}''(t) represents the second derivative of the function {\theta}(t) with respect to time. This represents the rate of change of the rate of change or the acceleration of the function {\theta}(t). {\theta}(t) represents the function itself, which could represent various physical quantities such as position, velocity, or angle depending on the context.

4. What does the term "sint" represent in this equation?

The term "sint" represents the sine function with the input being the variable t. This means that the function {\theta}(t) is being multiplied by the sine of t in the given equation.

5. What are some real-life applications of solving particular solution differential equations?

Solving particular solution differential equations can be applied in many scientific fields such as physics, engineering, biology, and economics. For example, in physics, it can be used to model the motion of objects under the influence of forces. In engineering, it can be used to design control systems for various machines. In biology, it can be used to study population growth or chemical reactions. In economics, it can be used to model the changes in market demand or supply over time.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
226
  • Calculus and Beyond Homework Help
Replies
1
Views
646
  • Calculus and Beyond Homework Help
Replies
2
Views
180
  • Calculus and Beyond Homework Help
Replies
1
Views
771
  • Calculus and Beyond Homework Help
Replies
7
Views
489
Replies
7
Views
439
  • Calculus and Beyond Homework Help
Replies
3
Views
474
  • Calculus and Beyond Homework Help
Replies
4
Views
864
  • Calculus and Beyond Homework Help
Replies
5
Views
834
  • Calculus and Beyond Homework Help
Replies
1
Views
879
Back
Top