Question

Determine the solution y(x) satisfying y'' + 2y' + 2y = 0 satisfying y(1)=2 and y'(1) =1.
What technique would I use?

Answer 1

That's a nice constant-coefficient, homogeneous ordinary differential equation.
We can get it's characteristic polynomial:
\[r^2+2r+2=0\]
Then solve for the polynomial's roots (r1 and r2). The fundamental set of solutions will then be of the form
\[y_1(t)= e ^{r_1*t} ; y_2(t) = e^{r_2*t} \]
If we get a single repeated root r1, the fundamental solution set will be
\[y_1(t)= e ^{r_1*t} ; y_2(t) = t*e^{r_1*t} \]
Then we write the general solution as
\[y(t)=c_1y_1(t) + c_2y_2(t)\]
And finally use the initial conditions to solve for c1 and c2.

Answer 2

\[r=\frac{ -2\pm \sqrt{2^{2}-4*1*2} }{2*1 }=-1 \pm \iota \]
solution is
\[y=e ^{-x}\left[ c1 \cos x+c2 \sin x \right]\]
find c1 and c2 from the initial conditions.