Question 212136
I need a solution to the ordinary differential equation 
x"+2x'+2x=0


Step 1.  Assume the solution is 


{{{x=e^mx}}}


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


Step 2. Take the derivatives of x:


x" =  {{{m^2e^mx}}}


x' = {{{me^mx }}}


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


x"+2x'+2x= {{{0=m^2e^mx+2me^mx+2e^mx}}}


Step 4.  Factor out


{{{e^mx}}}


to get


{{{(e^mx)(m^2+2m+2)=0}}}


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


{{{m^2+2m+2=0}}}


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


 {{{m = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


where 


a=1, b=2 and c=2


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


*[invoke quadratic "m", 1, 2, 2 ]



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


{{{y=c1e^m1+c2e^m2}}}


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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