Example of an ODE with initial conditions 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. 
 A: 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. 
A: Now, I am not sure what your textbook is doing, but I am going to show you how I would solve this equation using the methods I have learned. First, as you said, we have:
$$(-\lambda^2+p\lambda+q)u=0$$
Thus, we need to divide by $u$ and find the solutions to $\lambda$. There are then three possibilities, assuming $p$ and $q$ are real numbers.


*

*There are two, distinct real solutions.

*There is only one solution with multiplicity of two.

*There are two, distinct complex solutions.


In order to make this simpler, I am going to focus on Possibility 1 and I will refer to the two possible values of $\lambda$ as $\lambda_1$ and $\lambda_2$. Now, we have two possible solutions:
$$u=e^{\lambda_1(t-t_0)}, u=e^{\lambda_2(t-t_0)}$$
Now, you will learn the reasons for why this is later, but these two solutions are called a "fundamental set of solutions" because they generate all the possible solutions to this ODE. Thus, we write the general solution as a linear combination of these two solutions, like so:
$$u=c_1e^{\lambda_1(t-t_0)}+c_2e^{\lambda_2(t-t_0)}$$
Here, $c_1$ and $c_2$ are constants we have to solve for using the initial conditions. As you can see, the problem gives you two initial conditions and there are two constants to solve for, so it works out. First, let's get the equation from $u(t_0)=u_0$:
$$u(t_0)=c_1e^{\lambda_1(t_0-t_0)}+c_2e^{\lambda_2(t_0-t_0)}=c_1+c_2=u_0$$
Now, to get the equation from $u'(t_0)=v_0$, we first have to figure out what $u'$ is. Using chain rule, it is easy to take the derivative of $u$:
$$u'=\lambda_1c_1e^{\lambda_1(t-t_0)}+\lambda_2c_2e^{\lambda_2(t-t_0)}$$
Thus, from $u'(t_0)=v_0$, we find that:
$$u'(t_0)=\lambda_1c_1e^{\lambda_1(t_0-t_0)}+\lambda_2c_2e^{\lambda_2(t_0-t_0)}=\lambda_1c_1+\lambda_2c_2=v_0$$
This, combined with the $c_1+c_2=u_0$ equation, allows us to solve for $c_1$ and $c_2$ since we have a system of two linear equations with two unknowns. Using Cramer's Rule, we get:
$$c_1=\frac{-u_0\lambda_2+v_0}{\lambda_1-\lambda_2}, c_2=\frac{u_0\lambda_1-v_0}{\lambda_1-\lambda_2}$$
Thus, our final solution to this ODE with initial conditions is:
$$u=\frac{-u_0\lambda_2+v_0}{\lambda_1-\lambda_2}e^{\lambda_1(t-t_0)}+\frac{u_0\lambda_1-v_0}{\lambda_1-\lambda_2}e^{\lambda_2(t-t_0)}$$
Now, this is a pretty long solution and it only covers the case where $\lambda_1, \lambda_2$ are distinct, real numbers. However, that is the beautiful yet messy world of differential equations for you. I hope this example helps you understand initial condition problems more!
