Differential equations problem

y'' + 2y' + 5y = 0

y(t) = ?

y(t) = c(1)exp(-t)cos(2t) + c(2)exp(-t)sin(-2t)

You really should do your own homework Anonymous.

>>2
Opps forgot to mention that c(1) and c(2) are arbitary constants.

42

c(1)exp(-1+i2t) + c(2)exp(-1-i2t) == c(1)exp(-t)cos(2t) + c(2)exp(-t)sin(-2t)

how do you make the left hand side the same as the rhs?

What the fuck exactly do you mean by " and '??

probably the first and second derivatives of y(t)

>>5
LHS =/= RHS.  Both sides are incorrect solutions to the problem.

You have found the correct roots to the characteristic equation.  This gives the two complex solutions to the differential equation:
1. y1 = exp(-1 + 2i)t = exp(-t)(cos(2t) + isin(2t))
2. y2 = exp(-1 - 2i)t =  exp(-t)(cos(-2t) + isin(-2t)) = exp(-t)(cos(2t) - isin(2t))
Now y(t) = c(1)y1 + c(2)y2 is a solution.

Using linear combinations of equations 1 and 2 to get the most general real valued solution:
y2 + y1 =  2exp(-t)cos(2t)
y1 – y2 = 2iexp(-t)sin(2t)
The constants in front do not affect the end solution and can be arbitrarily chosen so we can get rid of them to get the solution:
y(t) = c(1)exp(-t)cos(2t) + c(2)exp(-t)sin(2t)

It perhaps easier just to note that given a complex root a + bi to the characteristic equation of a 2nd order differential equation then the most general solution is:
y(t) = c(1)exp(at)cos(bt) + c(2)exp(at)cos(bt)

>>6

gb2/school/

how to solve this equation?

f(x)=(2x+3)^2(2x^2+1)^1/2  ,find f'(-2)

>>10

you mean f(x)=(2x+3)^2(2x^2+1)^(1/2) ?

so we have something like this:
   f(x) = g(x)*h(x)
where:
   g(x)=(2x+3)^2
   h(x)=(2x^2+1)^(1/2)
as we know:
   f'(x) = g'(x)h(x)+g(x)h'(x)
first step:
   g'(x)=2(2x+3)2=4(2x+3)
second step:
   h'(x)=(1/2)(2x^2+1)^(-1/2)(2*2x)=2x/(2x^2+1)^(1/2)
final:
   f'(x)=4(2x+3)*(2x^2+1)^(1/2)+((2x+3)^2)*(2x/(2x^2+1)^(1/2))=

2(2x+3)(6x^2+3x+2)
------------------
  (2x^2+1)^(1/2)

substitution:
   f'(-2)= -13 and some