I was given the following example as an introduction towards ODEs. But I have an issue with it:
Let $$-u'' + pu' + qu = 0,\ p,q \in \mathbb{R} \\
u(t_0) = u_0 \\
u'(t_0) = v_0, \\ u_0,v_0\in \mathbb{R}
$$
If we say, that
$$u = e^{\lambda(t-t_0)} u_0
$$
We fulfill the first initial condition. Then we can write the equation as
$$(-\lambda^2 +p\lambda + q )u = 0$$
If we solve $-\lambda^2 +p\lambda + q = 0$ for $\lambda$, we might have either a single or a two-fold zero. Which is way too restricted to let us fulfill the second initial conditon $u'(t_0)=v_0$, because
$$u'(t) = \lambda e^{\lambda (t-t_0)}u_0$$ and so $$u'(t_0) = \lambda u_0 \stackrel{!}{=} v_0 \Rightarrow \lambda = v_0 / u_0$$
But we are already forced to choose $\lambda$ from the at most two solutions to the 2nd degree polynomial.
Best Answer
First of all you find the general solution to your differential equation.
You solve the characteristic polynomial and find your eignvalues and form the general solution which is a linear combination, either $$y= c_1 e^{\lambda _1 t} + c_2 e^{\lambda _2 t}$$ in case of distinct eigenvalues or
$$y= c_1 e^{\lambda t} + c_2 te^{\lambda t}$$ in case of an eigenvalue with multiplicity two.
Then you find the parameters $c_1$ and $c_2$ from your initial conditions.