First order differential equation problem

Question

Suppose we have $$ \frac{dy}{dx} +f(x)y = r(x) $$ and it has two solutions $y_1(x)$ and $y_2(x)$ then how to prove that solution of differential equation $$ \frac{dy}{dx} +f(x)y = 2r(x) $$ Will be $y_1(x)+y_2(x)$ ? I think given differential equations is linear first order equation so its solution will be $$y.e^{\int f(x)dx} = \int r. e^{\int f(x)dx}dx $$ now do I establish two solution as $y_1$ and $y_2$ out of this equation?

Answer 1

I don't know if this is not just placing the solution in and determining that it satisfies..however I will proceed!

$$ y_1'' + f(x)y_1' = r(x)\\ y_2'' + f(x)y_2' = r(x). $$ adding the two equations together we obtain $$ y_1'' + y_2'' + f(x)y_1' + f(x)y_2' = 2r(x)\\ \left(y_1 + y_2\right)'' + f(x)\left(y_1'+y_2'\right) = \\ \left(y_1 + y_2\right)'' + f(x)\left(y_1+y_2\right)' = 2r(x). $$ so we find (as not surprising) the solution $y_1 + y_2$ satisfies the equation.

If this is not the kind of process you would like to follow then I can try to improve.