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

Second order differential

  • Thread starter cemar.
  • Start date
41
0
y'' - 8y' + 16 = 0 ; y(0) = 4 ; y'(0) = 9


this should be not too bad but im stuck in the same place.
m^2 - 8m + 16
m1 = m2 = 4

y = C1e^(mx) + C2(e^(mx))
sub in y(0) = 4
4 = C1 + C2
C1 = 4-C2

y' = mC1e^(mx) + mC2e^(mx)
sub in y'(0) = 9
9 = mC1 + mC2
C1 = 4 - C2 (from above), m=4
9 = 4(4-C2) + 4C2
9 = 16 - 4C2 + 4C2
9 = 16
In this case would there just be no solution or am i missing something?!!?
Thank you!!!!
 

Answers and Replies

nicksauce
Science Advisor
Homework Helper
1,272
5
If I recall correctly, the general solution in the case of a repeated root for a second order constant coefficients ODE is
[tex]c_1e^{m_1x} + c_2xe^{m_1x}[/tex]
 
263
70
Hey
The formula you are using for y is only valid if you have to different solutions m, i.e.
[tex]m_{1}\neq{m_{2}} [/tex]

In your case y is given by

[tex]y=(C_{1}+C_{2}x)e^{mx}[/tex]
 
41
0
thanks!!!!!
 
HallsofIvy
Science Advisor
Homework Helper
41,738
898
Well done, guys.
 

Related Threads for: Second order differential

  • Last Post
Replies
18
Views
3K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
1
Views
524
  • Last Post
Replies
3
Views
827
  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
1
Views
1K
Top