Solving $u_{\alpha} + u_{t} = -\mu t u, t>0$ .

An age structured population with distribution $$u(a,t)$$ over age $$a$$ has a death rate increasing linearly with time and constant birth rate $$b$$, $$u(\alpha,0) = u_{0}(\alpha)$$.

Model is $$u_{\alpha} + u_{t} = -\mu t u, t>0$$ $$u(a,0) = u_{0}(a), a \geq 0$$ $$u(0,t) = F(t) = b\int_{0}^{\infty} u(a,t) da$$(which is the non-local boundary condition), $$\mu,b$$ are constants.

We need to find the solution $$u(\alpha,t)$$ for $$\alpha > t > 0$$ and for $$0.

Seeing the non-local boundary condition I thought of $$F(t) = b\int_{0}^{t} u(a,t)da +b\int_{t}^{\infty} u(a,t) da$$ (in case it may be useful)

I am clueless about how to obtain $$u(a,t)$$.

Thinking of method of characteristics, we have $$\frac{d\alpha}{1} = \frac{dt}{1} = \frac{du}{-\mu t}$$ and taking last equality we have $$u = -\mu t^2 +c$$

Since the equation is linear, let's define $$u(\alpha,t)=f(\alpha)g(t)$$then by substituting in the equation we obtain$$f'(\alpha)g(t)+f(\alpha)g'(t)=-\mu t f(\alpha)g(t)$$after dividing by $$u(\alpha ,t)$$ and solving the outcome ODEs we finally conclude that:$$u(\alpha,t)=Ae^{k(\alpha-t)}e^{-{1\over 2}\mu t^2}$$where $$k$$ and $$A$$ are complex constants. Since the equation is linear, then any linear combination of the above answers is also itself an answer, therefore$$u(\alpha,t)=\int_{0}^{\infty}A(\omega)e^{k(\omega)\cdot (\alpha-t)}e^{-{1\over 2}\mu t^2}d\omega$$Now if we decide to have a real answer with the assumption that $$u(a,0)$$ has a Fourier Transform, then only the values of $$k(\omega)$$ that are pure imaginary are needed. In that case, $$\Re\{A(-\omega)\}=\Re\{A(\omega)\}\\\Im\{A(-\omega)\}=-\Im\{A(\omega)\}$$Also without loss of generality, we assume $$k(\omega)=i\omega$$. These constraints assure that our final answer has the following real-valued form:$$u(\alpha,t)=e^{-{1\over 2}\mu t^2}\int_0^\infty C(\omega)\cos \omega (\alpha-t)+D(\omega)\sin \omega (\alpha-t)d\omega$$now if we expand $$u(a,0)$$ evenly around zero we obtain$$D(\omega)=0\\C(\omega)={2\over \pi}\int_0^\infty u(a,0)\cos \omega ada$$and the final answer becomes:$$u(\alpha,t)=e^{-{1\over 2}\mu t^2}\int_0^\infty C(\omega)\cos \omega (\alpha-t)d\omega$$