For the Klein Gordon Field, the equations of motion are described by the equation
$$(\partial_{\mu}\partial^{\mu} + m^2)\phi(\vec{x},t)=0$$
Which when the field is expressed as a Fourier transform of the momentum we can get that $$(\partial_{\mu}\partial^{\mu} + m^2)\int \frac{d^3k}{(2\pi)^3}\tilde{\phi}(k,t) e^{i \vec{k} \cdot \vec{x}}=0$$
Which is the same as
$$\int \frac{d^3k}{(2\pi)^3}(\partial_{\mu}\partial^{\mu} + m^2)\tilde{\phi}(k,t) e^{i \vec{k} \cdot \vec{x}}=0$$
From this we get a differential equation for $\tilde{\phi}$ and we get that $$\phi(\vec{x},t)=\int \frac{d^3 p}{(2\pi)^3 }(a(\vec{p})e^{-ip \cdot x} + a^{\dagger}(\vec{p})e^{-ip \cdot x})$$
For the Dirac field which equations of motion are given by
$$(i\gamma^\mu\partial_\mu - m)\psi(x,t) = 0$$
Can we arrive at the form:
$$\psi(\vec{x},t)= \sum_{s} \int \frac{d^3 p}{(2\pi)^3}(a_s(\vec{p})u_s(\vec{p})e^{-ip \cdot x} + b_s^{\dagger}(\vec{p}) v_s(\vec{p})e^{ip \cdot x})$$
Using the same technique that we used for the Klein Gordon equation where $$(i\gamma^\mu\partial_\mu - m)\int \frac{d^3 k}{(2\pi)^3}\tilde{\psi}(k,t)e^{-ik \cdot x}=0$$ Any help would be greatly appreciated, Thanks.