Variation of parameters method

In summary: They are used to find the particular solution by setting the coefficients of the sines and cosines equal to 0 in the equation y_p=v_{1}\cos(x)+v_{2}\sin(x). This is a common method used to solve differential equations with non-constant coefficients.
  • #1
djeitnstine
Gold Member
614
0

Homework Statement



[tex]y''+y=tan(x)+e^{3x}-1[/tex]

Homework Equations



homogeneous solution:
[tex]y_{hom..}=C_{1}cos(x)+C_{2}sin(x)[/tex]

particular solution:
[tex]y_{parti..}=v_{1}' cos(x)+v_{2}' sin(x)[/tex]

The Attempt at a Solution



[tex]v_{1}' cos(x)+v_{2}' sin(x)=0[/tex] (1)

[tex]-v_{1}' sin(x)+v_{2}' cos(x) = tan(x)+e^{3x}-1 [/tex] (2)

(1)* by sin (2)* by cos

and solve for v2

[tex]sin(x)+e^{3x}cos(x)-cos(x)=v_{2}'[/tex]

[tex]-cos(x)+ \frac{3}{10}e^{3x}cos(x)-\frac{1}{10}sin(x)-sin(x)=v_{2}[/tex]

Now finding v1 is the problem here. if I reverse the multiplication and do (1)* by cos (2)* by sin it seems even worse =(. I saw the complete answer and if I remember correctly its [tex]y_{p}=\frac{1}{2}cos(x)ln(sin(x)+1)-\frac{1}{2}cos(x)ln(sin(x)-1)-\frac{1}{10}e^{3x}+1[/tex]

I really just need to know what to multiply by to get going
 
Physics news on Phys.org
  • #2
djeitnstine said:
particular solution:
[tex]y_{parti..}=v_{1}' cos(x)+v_{2}' sin(x)[/tex]

Why the primes? Why not just use [tex]y_p=v_{1}\cos(x)+v_{2}\sin(x)[/tex] as your ansatz (assumed form)?

[tex]v_{1}' cos(x)+v_{2}' sin(x)=0[/tex] (1)

[tex]-v_{1}' sin(x)+v_{2}' cos(x) = tan(x)+e^{3x}-1 [/tex] (2)

Where are these equations coming from?:confused:

Your particular solution should satisfy [itex]y_p''+y_p=\tan(x)+e^{3x}-1[/itex] not [itex]y_p=0[/itex] or (2)
 
  • #3
As gabbagabbahey said, drop the primes from you solution- they are confusing.
y(x)= v1(x)cos(x)+ v2(x)sin(x)
Differenitate:
y'= v1' cos(x)- v1 sin(x)+ v2' sin(x)+ v2 cos(x)
Now, we can simplify this by "restricting" our search. Out of the infinitely many v1 and v2 that would satisfy the equation, we only look for those such that
v1' cos(x)+ v2' sin(x)= 0

So now
y'= -v1 sin(x)+ v2 cos(x)
Differentiate again: y"= -v1' sin(x)- v1 cos(x)+ v2' cos(x)- v2 sin(x).

Put that and the equation for y' into the original differential equation and it gives one equation for v1' and v2' (all of the v1 and v2 terms without ' will cancel). That, together with v1' cos(x)+ v2' sin(x)= 0 gives to equations to solve (algebraically) for v1' and v2' and then you integrate to find v1 and v2.
 
  • #4
Ok thanks I finally got the answer.
 
  • #5
gabbagabbahey said:
Why the primes? Why not just use [tex]y_p=v_{1}\cos(x)+v_{2}\sin(x)[/tex] as your ansatz (assumed form)?



Where are these equations coming from?:confused:

Your particular solution should satisfy [itex]y_p''+y_p=\tan(x)+e^{3x}-1[/itex] not [itex]y_p=0[/itex] or (2)

These equations come from a proof shown by various texts including my professor. One example is at http://tutorial.math.lamar.edu/ under the Diff Eq. pdf if you'd like to see.
 

1. What is the Variation of Parameters method?

The Variation of Parameters method is a technique used in solving non-homogeneous linear differential equations. It involves finding a particular solution by varying the parameters of a general solution to the corresponding homogeneous equation.

2. When is the Variation of Parameters method used?

This method is used when the non-homogeneous term in a linear differential equation is a function that can be expressed as a linear combination of known functions. It is also used when the coefficients in the differential equation are constants.

3. How does the Variation of Parameters method work?

The method involves finding the general solution to the corresponding homogeneous equation, then using it to find the particular solution by varying the parameters. The parameters are chosen such that the particular solution satisfies the non-homogeneous equation.

4. What are the advantages of using the Variation of Parameters method?

One advantage is that it can be used to find the particular solution to a wide range of non-homogeneous linear differential equations. It also allows for the use of known functions in finding the solution, making it easier to solve complicated equations.

5. Are there any limitations to the Variation of Parameters method?

Yes, this method can only be used for linear differential equations with constant coefficients and non-homogeneous terms that can be expressed as a linear combination of known functions. It also requires some algebraic manipulation and can be time-consuming for more complex equations.

Similar threads

  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
233
  • Calculus and Beyond Homework Help
Replies
3
Views
268
  • Calculus and Beyond Homework Help
Replies
2
Views
857
  • Calculus and Beyond Homework Help
Replies
21
Views
754
  • Calculus and Beyond Homework Help
Replies
3
Views
487
  • Calculus and Beyond Homework Help
Replies
6
Views
677
  • Calculus and Beyond Homework Help
Replies
3
Views
768
  • Calculus and Beyond Homework Help
Replies
8
Views
115
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
Back
Top