Proving a differential equation is linear Prove that the following differential equation is linear:
$$y(t)\frac{df(t)}{dt} - 3f(t)x(t)= 0.$$
I thought it was linear looking at it. However is there any way I can prove it? Any help would be much appreciated.
 A: If we assume that $f(t)$ is the dependent variable, then a differential equation, when expressed in the form $L(f) = 0$ is said to be linear if $L$ is a linear function in $f$ and in its derivatives. Thus, if $y(t)$ and $x(t)$ are known functions of $t$:
$$L(f) = y f' - 3 x f =0, $$ so in order to prove linearity of $L(f)$ we should have:


*

*For $\lambda \in \mathbb{R},$ $L(\lambda f ) = \lambda \, L(f)$. In your case: 


$$ y(\lambda f)' - 3x \lambda f =  y \lambda f' - 3x \lambda f = \lambda( y f' - 3 x f ) = \lambda L(f) = 0.$$ 


*

*On the other hand, for the differential operator to be linear, if $f_1$ and $f_2$ are given functions of $t$, then $L(f_1 + f_2) = L(f_1) + L(f_2).$ Indeed:


$$L(f_1 + f_2 ) = y (f_1+f_2)' - 3x (f_1 + f_2) = y (f_1' + f_2') - 3x (f_1 + f_2) = f_1' + y f_2' - 3x f_1 - 3x f_2 = y f_1' - 3x f_1 + y f_2' - 3x f_2 = L(f_1) + L(f_2).$$
Hence, your ODE is linear. You could have proved linearity by checking that, for any constants $\alpha$ and $\beta$ and functions $f_1$ and $f_2$: 
$$L(\alpha f_1 + \beta f_2 ) = \alpha L(f_1) + \beta L(f_2).$$
Hope this helps!
Cheers.
