The solution depends on the value of \(k\). To make things simple, assume \(k>0\). The homogeneous part of the equation is
\(\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0\)
and has characteristic equation
\(r^2-16k=0\implies r=\pm4\sqrt k\)
which admits the characteristic solution \(y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}\).
For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be \(y_p=ae^{4x}+be^x\). Then
\(\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x\)
So you have
\(16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x\)
\((16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x\)
This means
\(16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}\)
\(b(1-16k)=30\implies b=\dfrac{30}{1-16k}\)
and so the general solution would be
\(y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x\)
