ODE for the Fourier transform

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Consider the PDE $$ u_t=u_{xx}+u_x+u. $$ Fourier transforming both sides gives $$ \hat{u}_t=(ik)^2+ik+\hat{u},~~~(*) $$ where $\hat{u}$ is the Fourier transform of $u$.

Now, for fixed $k$, $(*)$ is an ODE in $t$.

It is said that solutions of $(*)$ are of the form $$ \hat{u}=\exp(\lambda(k)t)\cdot v_0(k). $$

Why?