Prove that $k(f+g)_{xx} = (f+g)_t$

Question

Prove that $k(f+g)_{xx} = (f+g)_t$, given that:

$kf_{xx} = f_t$ and $kg_{xx} = g_t$, for $f(x,t)$ and $g(x,t)$.

My understanding: I think we need to prove $(f+g)_{xx} = f_{xx}+g_{xx}$ and $(f+g)_t= f_t + g_t$. Stuck here.

Thanks for any help.

Answer 1

$(f+g)_{xx}=(f_{xx}+g_{xx})$,

and

$(f+g)_{t}=(f_{t}+g_{t})$,

since the differentiation is a linear operator.

Then:

$k(f+g)_{xx}=k(f_{xx}+g_{xx})=k f_{xx}+k g_{xx}=f_{t}+g_{t}=(f+g)_{t}$.