• Support PF! Buy your school textbooks, materials and every day products Here!

Problem using variation of parameters

  • Thread starter Tarhead
  • Start date
  • #1
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)?
 

Answers and Replies

  • #2
321
0
Yup. And your answer is correct.
 
  • #3
dextercioby
Science Advisor
Homework Helper
Insights Author
12,977
540
I dunno 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 Threads for: Problem using variation of parameters

  • Last Post
Replies
0
Views
1K
Replies
1
Views
834
Replies
0
Views
2K
Replies
1
Views
4K
  • Last Post
Replies
1
Views
1K
Replies
3
Views
497
  • Last Post
Replies
4
Views
2K
Top