How do I factorise:?

Question

x'' - 3x' + 2x = 0
where x(0)= x and x'(0)= 6

Solution is to be given in the form of a function, using t as variable:
x(t)= ...

Answer 1

integrating once we have : x' + 3x + x^2+ c1 =0
again integrating we have : x + 3/2 x^2 + x^3/3 +c1x+c2 =0
c2=x and c1 = -6
x^3 /3 +(3/2)x^2-4x =0 put t in place of x

Answer 2

The auxilliary equation is u^2 - 3 u + 2 = (u - 2)(u - 1) = 0, with solutions u = 1 and u = 2. Therefore, the general solution is x(t) = Ae^t + Be^(2t). From x'(0) = 6 we see that A + 2B = 6. You messed up the initial condition at x(0). (Clearly, x(0) = x is not correct. Use the correct initial condition to find another equation in A and B, and then you can solve for A and B to get the particular solution.)

Answer 3

Characteristic equation:

r^2 - 3r + 2 = 0
r^2 - r - 2r + 2 = 0
r(r-1)-2(r-1) = 0
(r-1)(r-2) =0
r = {1, 2}

General solution
x(t) = Ae^t + Be^2t
x(0) = ?
Assuming you mean x(0) = 0
A + B = 0
B = -A

x'(t) = Ae^t + 2Be^(2t)
x'(0) = A + 2B

A + 2B = 6
A - 2A = 6
A = -6
B = 6

x(t) = 6e^(2t) - 6e^(t)

Answer 4

m^2-3m+2=0
(m-2)(m-1)=0
m=2,1

x=c1e^2t+c2e^t
x'=2c1e^2t+c2e^t


applying the initial conditions
x=c1+c2
6=2c1+c2
subtract the two equations
x-6=-c1
c1=6-x
c2=x-c1=x-(6-x)=6+2x

x=(6-x)e^2t+(6+2x)e^t