Determine if the set of all solutions to an equation is a vector space. Without solving, decide if the set of all solutions of the following equation is a vector space:$$x\frac{d^2y}{dx^2}-e^x\frac{dy}{dx}+ysin(x) = 0$$According to the book that this problem is from, the answer is that it is a vector space.  I am not sure how to figure this out on my own.  I assume that a solution to the equation is a pair $(x,y)$, and that to solve the problem, I need to check if the properties of a vector space hold with these pairs.  One such property is the existence of an additive inverse.  I assume the additive inverse in this case would be $(0,0)$, but when plugging this in to the above equation I get $-\frac{dy}{dx} = 0$ , which doesn't seem to be true.
Maybe the book has the wrong answer, or maybe I'm approaching this problem the wrong way.  In either case, can anyone offer some insight into how to solve this problem and problems of this sort in general?  Keep in mind that this book is an introductory linear algebra book and doesn't require advanced mathematical knowledge.  
 A: Hint:
Your equation is linear in $y$ and its derivatives, so it is a linear differential equation, and this means that a linear combination of any two solutions is also a solution.
Here the vector space is a space of functions. If $y_1$ and $y_2$ are solutions of the equations, than it is simple to verify that also $y_1+\lambda y_2$ is a solution:
$$x\frac{d^2(y_1+\lambda y_2)}{dx^2}-e^x\frac{d(y_1+\lambda y_2)}{dx}+(y_1+\lambda y_2)\sin(x) = $$
$$=x\left(\frac{d^2y_1}{dx^2}+\lambda\frac{d^2y_2}{dx^2}\right)-e^x\left(\frac{dy_1}{dx}+\lambda\frac{dy_2}{dx}\right)+\left(y_1+\lambda y_2 \right)\sin(x) =  $$
$$=x\left(\frac{d^2y_1}{dx^2}\right)+\lambda x\left(\frac{d^2y_2}{dx^2}\right)-e^x\left(\frac{dy_1}{dx}\right)-\lambda e^x\left(\frac{dy_2}{dx}\right)+y_1 \sin(x)+\lambda y_2 \sin(x) =  $$
$$=x\left(\frac{d^2y_1}{dx^2}\right)-e^x\left(\frac{dy_1}{dx}\right)+y_1\sin(x)+\lambda \left[ x\left(\frac{d^2y_2}{dx^2}\right)- e^x\left(\frac{dy_2}{dx}\right)+ y_2 \sin(x)\right] =$$$$= 0+\lambda \cdot 0 =0 $$
