SOLUTION: i need a solution to the ordinary differential equation x"+2x'+2x=0

Question 212136: i need a solution to the ordinary differential equation
x"+2x'+2x=0
Found 2 solutions by Fombitz, drj:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
Let x=e^(rt), x'=r*x, x''=r^2*x
x"+2x'+2x=0
r^2*x+2r*x+2x=0
r^2+2r+2=0
Using the quadratic formula,
r=-1+/-i
x=e^((-1+/i)t)
x=e^(-t)*(C1*cos(t)+C2*sin(t)), where C1 and C2 are constants.
Answer by drj(1380) (Show Source): You can put this solution on YOUR website!
I need a solution to the ordinary differential equation
x"+2x'+2x=0

Step 1. Assume the solution is

If this is true, then we need to solve for m.

Step 2. Take the derivatives of x:

x" =

x' =

Step 3. Substitute above derivatives and x into the given equation

x"+2x'+2x=

Step 4. Factor out

to get

Step 5. The only way to get 0 is that the quadratic expression is zero. That is,

Now, we can solve for m using the quadratic formula below.

where

a=1, b=2 and c=2

Step 6. See standard procedure of solving quadratic equation below:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

The discriminant -4 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.

In the field of imaginary numbers, the square root of -4 is + or - .

The solution is

Here's your graph:

Step 7. Note the roots are complex. So your solution will consists of complex exponentials. Also, please ignore the graph since it's only applicable for real roots.

Using the above roots m1 and m2, the solution is

where the c1 and c2 are arbitrary constants. We need initial values to get values of c1 and c2.

I believe the above problem is above the skill level of algebra. However,
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