Differential Equations

Question

Find a general solution.

1. u''+7u=0

 

2. y''-2y'+26y=0

 

Solve the given initial value problem.
3. y''+9y=0

 

 

Would someone mind explaining these problems?

Answer 1

Jessica,

 

A few tutors have already posted solutions to (1) and (3).  I will help you with (2).

 

y'' - 2y' + 26y = 0

 

Let y = e^ax

 

(a² - 2a + 26)e^ax = 0

 

The characteristic equation is 

 

a² - 2a + 26 = 0

 

a = (1/2) [2 ± √(4 - 4·26)] = (1/2)(2 ± 10i) = 1 ± 5i

 

The general solution will then be

 

y(x) = e^x [A sin(5x) + B cos(5x)]

 

where A and B are determined by 2 independent boundary conditions.