Problem using variation of parameters

In summary, the conversation is about finding the general solution of a differential equation using variation of parameters. The solution involves finding the values of u1 and u2 and plugging them back into the original equation. The final answer is given in terms of the exact solution, y(x), which includes several terms and constants. The person asking for help confirms that the answer provided by the expert is correct.
  • #1
Tarhead
7
0
the problem is fin the general solution of the differential eq :

y''+y=2sect + 3 (-pi/2 < t < pi/2)

using variation of parameters.

I just needed a check to make sure my answer was correct.

r^2+1 = 0
r= -i
r= i
y1= cost
y2= sint
g(t)= 2sect+ 3

y(t) = c1cost + c2sint + Y(t)
Y(t) = u1y1 + u2y2
u1 = -(integal) (y2*g(t))/W in which W = 1
= -(int) sint(2sect+ 3)
= -(int) sint(2/cost+3)
= -(2 (int) tant + 3 (int) sint)
is this correct, where do i go from here

u2= (integral) y1*g(t)/ W
= (int) cost(2sect + 3)/ W
= (int) (2*(cost/cost) + 3cost)
= (int) 2 + (int)3 cost
= 2+3(sint)
is this correct

and then I plug these back into the Y(t) eq and add this to y(t)?
 
Physics news on Phys.org
  • #2
Yup. And your answer is correct.
 
  • #3
I don't know what u did,but here's where you have to get...

[tex] \frac{d^2y}{dx^2}+y=2\sec x+3 [/tex]

Exact solution is :

[tex]y\left( x\right) =2\ln \left( 1-\cos x-\sin x\right) \cos x+3\cos x+2\ln \left( 1-\cos x+\sin x\right) \cos x-4\ln \left( \sin x\right) \cos x[/tex]
[tex]\allowbreak +2\ln \left( \cos x+1\right) \cos x-2\ln 2\cos x+4\sin x\arctan \left(\frac{\sin x}{\cos x+1}\right)+3+C_1\cos x+\allowbreak C_2\sin x [/tex]

Daniel.
 
Last edited:

Related to Problem using variation of parameters

1. What is the variation of parameters method?

The variation of parameters method is a technique used in solving ordinary differential equations (ODEs) by expressing the solution as a linear combination of known functions. This method is particularly useful for solving non-homogeneous linear ODEs.

2. When is the variation of parameters method used?

The variation of parameters method is used when solving non-homogeneous linear ODEs, where the non-homogeneous term is a sum of known functions. It can also be used for certain types of non-linear ODEs, although the process may be more complicated.

3. What are the steps involved in using the variation of parameters method?

The first step is to find the complementary solution, which is the solution to the corresponding homogeneous equation. Then, the particular solution is found by assuming a linear combination of known functions and solving for the coefficients using a system of equations. Finally, the general solution is obtained by combining the complementary and particular solutions.

4. How does the variation of parameters method differ from other methods of solving ODEs?

The variation of parameters method is different from other methods such as the method of undetermined coefficients and the method of annihilators because it does not rely on guessing a particular form of the solution. Instead, the solution is expressed as a linear combination of known functions, making it a more general and versatile technique.

5. Are there any limitations or drawbacks to using the variation of parameters method?

One limitation of the variation of parameters method is that it can only be used for linear ODEs. Additionally, the process of finding the particular solution can be more complicated and time-consuming compared to other methods, especially for higher-order ODEs. It is also not always possible to find a closed-form solution using this method.

Similar threads

  • Calculus and Beyond Homework Help
Replies
7
Views
552
  • Introductory Physics Homework Help
Replies
6
Views
515
Replies
1
Views
1K
Replies
33
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Differential Equations
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Differential Equations
Replies
1
Views
1K
Replies
1
Views
673
  • Introductory Physics Homework Help
Replies
7
Views
1K
Back
Top