Differential Equation Help, I'm Stuck and Need Help Figuring out the Next Step The problem is: Suppose $\phi_1(x)$ and $\phi_2(x)$ are solutions of the differential equation (1):
 $$ y''(x)+p(x)y'(x)+q(x)y(x)=0 $$
on the interval $I$. Show that $\lambda(x)=c_1\phi_1(x)+c_2\phi_2(x)$, where $c_1$ and $c_2$ denote arbitrary constants, is also a solution of (1) on the interval $I$.
What I did was to take the first and second derivative of $\lambda$ and to plug those in for $y''$, $y'$, and $y$ respectively: $\lambda'= c_1\phi_1'(x)+c_2\phi_2'(x)$ and 
$\lambda''= c_1\phi_1''(x)+c_2\phi_2''(x)$. Then I got the equation
 $$ c_1(\phi_1''(x)+p(x)\phi_1'(x)+q(x)\phi_1(x)) = c_2(\phi_2''(x)+p(x)\phi_2'(x)+q(x)\phi_2(x)). $$
I am having trouble proving that these two are equal since $\phi_1$ and $\phi_2$ could be different. I am looking for some direction as for where to go from here or if I went wrong on one of my steps. Thanks.
 A: Put $\rm\displaystyle\: D = \frac{d}{dx}\:.\: $ Since $\rm\ L = D^2 + p\ D + q\ $ is $\rm\:\mathbb C$-linear, $\rm\ L\:(c_1 \phi_1 + c_2\: \phi_2)\: =\: c_1\ L(\phi_1) + c_2\ L(\phi_2) = 0\:.$  
That $\rm\: L\:$ is $\mathbb C$-linear follows simply from the fact that it is a sum of compositions of $\mathbb C$-linear maps, namely the derivative $\rm\ f \to D(f) = f\:'\:,\ $ and the linear maps $\rm f \to p\ f,\ \ f\to q\ f\:.$ 
A: Katie: What does it mean that $\phi$ is a solution of an equation? It means that whatever the equation says, it is true if instead of the variable you use $\phi$. In this case the "variable" is the function $y(x)$ in (1), so to say that $\phi$ is a solution means that the following is true:
 $$ \phi''(x)+p(x)\phi'(x)+q(x)\phi(x)=0. $$
You are told that this is true when $\phi$ is $\phi_1$ and also when $\phi$ is $\phi_2$. 
What you need to prove is that it is true when $\phi$ is $\lambda$. 
What you have done is correct so far, but not enough. I am not sure why you think the equation at the end is what you need. It happens to be true, but it seems strange to focus on it. A big hint towards what needs to be done is that $0=c_10+c_20$.
A: You are (almost) done, but seem to have dropped a minus sign.  You should be looking at the sum of the two sides of your equal sign.  As you are told that $\phi_1$ and $\phi_2$ are solutions to the equation, the expressions in parentheses are zero.
