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.