Finding basis for the differential equation So the question is to find a basis for the solution of following differential equation that and when $y(0)=3,y'(0)=2$ what is the solution
$y'' + 2y' + y = 0$
So my attempt was to put it as auxiliary polynomial,which
$p(t) = t^2 +2t +1$
$=(t+1)^2$
Hence zeros are -1 and -1, so that $\{e^{-t},te^{-t}\}$
So the solution is that
$y(t)= b_1e^{-t} + b_2te^{-t}$
$y(0)= b_1e^{-0} + b_2(0)e^{-0}$
$=b_1 = 3$
$y'(t)= -b_1e^{-t} + b_2e^{-t} - b_2te^{-t}$
$=-b_1e^{-t} + b_2(1-t)e^{-t}$
$y'(0)= -b_1e^{-0} + b_2(1-0)e^{-0}$
$=b_2-b_1=2$, which $b_2 = 5$
Hence, the solution $y(t)= 3e^{-t} + 5te^{-t}$
Am I doing it right? New to linear Algebra
 A: TL;DR: yes, your work is correct.
To solve a linear homogeneous ODE with constant coefficients, one tries solutions of the form $y(t) = {\rm e}^{rt}$. The resulting characteristic equation $r^2+2r+1=0$ says that the values of $r$ solving such equation are the values of $r$ for which $y(t) = {\rm e}^{rt}$ is actually a solution of $y''+2y'+y=0$. In this case, $r=-1$ is a double root. This tells us that $y_1(t) = {\rm e}^{-t}$ is a solution, but we need a second solution $y_2$ such that $\{y_1,y_2\}$ is linearly independent. Dealing with roots with multiplicity, we obtain more solutions by multiplying by powers of $t$. Verify that $y_2(t) = t{\rm e}^{-t}$ is also a solution of $y''+2y'+y=0$. Since the equation is of order $2$, the space of solutions has dimension $2$, so that any solution is a linear combination of $y_1$ and $y_2$. The general solution of $y''+2y'+y=0$ is of the form $$y(t) = c_1{\rm e}^{-t} + c_2t{\rm e}^{-t}, \qquad c_1,c_2\in \Bbb R.$$You may think of that as $y(t) = (c_1+c_2t){\rm e}^{-t}$: a degree $1$ polynomial times ${\rm e}^{-t}$. In any case, prescribing $y(0)$ and $y'(0)$ allows us to find $c_1$ and $c_2$, which is to say, we may find the unique solution among all the infinitely many ones listed which, in addition, satisfies the given initial conditions. Then $3=y(0) = c_1$ and $y'(t) = -3{\rm e}^{-t}+c_2{\rm e}^{-t}-c_2t{\rm e}^{-t}$, so that $2=y'(0) = -3+c_2$ leads to $c_2 = 5$, and so the desired solution to the initial value problem is $$y(t) = 3{\rm e}^{-t}+5t{\rm e}^{-t}.$$
