Show that the solution set $S$ to the differential equation $y''+4y'=x^2$ is not a vector space

Question

Show $S$ is not closed to scalar multiplication by assuming $y_1$ is a solution(so $y_1''+4y_1=x^2$) and showing the scalar multiple $y=2y_1$ is not a solution.

Given that $y_1$ is a solution then multiple $y_1$ by any scalar multiple should be a solution also however $2(y_1)$

$(2y_1)''+4(2y_1)=x^2$

Then $(2y_1)''+8(y_1)=x^2$ So is it now fair to say that thus $2y_1$ is not a solution?

Also its not closed under addition since given $y_1$ and $y_2$ as solutions then

$y_1''+4y_1'+y_2''+4y_2'=(y_1+y_2)''+8(y_1+y_2)$ which is showing its solutions are not closed under addition?

Answer 1

If $y_1$ and $y_2$ are solutions and $y=2y_1$ and $z=y_1+y_2$, then

$y''+4y=2x^2$ and $z''+4z=2x^2$.

Hence $y$ and $z$ are not solutions of the diff. equation.

Easier: $y=0$ is not a solutions of the diff. equation.

So: $S$ is not a vector space.