Can I solve $\partial_t(\partial_t +a ) W(x,t) = (\partial_t + b)\hat{L} W(x,t)$ using the solution $\partial_t W_0 = \hat{L}W_0$?

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I have a challenging partial differential equation which is second order in time: $$\partial_t(\partial_t +a ) W(x,t) = (\partial_t + b)\hat{L} W(x,t)$$ Here $L$ is some linear operator which involves $x$ only, not $t$.

I know the solution $W_0$ to a simpler problem with only one derivative: $$\partial_t W_0 = \hat{L}W_0$$

Is there any way I can use $W_0$ to obtain $W$? So far I attempted the convolution ansatz $$W(x,t) = \int_0^t ds K(t-s)W_0(x,s).$$ Using the resulting identity $$\partial_t W = K(0)W_0(x,t) + \hat{L} W$$ Does lead to an interesting integral equation that gives the kernel function $K$ in terms of $W_0$, although this violates the assumption that $K$ does not depend on $x$.

Is there any means by which I can use the solution to the simpler problem to understand the more challenging problem?