How to simplify the solution of the following linear homogeneous ODE?

[Atr cheema]

During solution of a PDE I came across following ODE
$\frac{d \bar h}{dt} + \frac{K}{S_s} \alpha^2 \bar h = -\frac{K}{S_s} \alpha H h_b(t)$
I have to solve this ODE which I have done using integrating factor using following steps
taking integrating factor $I=\exp^{\int \frac{1}{D} \alpha^2 dt}$ and $\frac{K}{S_s} = \frac{1}{D}$
$I \frac{d \bar h}{dt} + I \frac{1}{D} \alpha^2 \bar h = -I \frac{1}{D} \alpha H h_b(t) \\ I \frac{d \bar h}{dt} + I \frac{1}{D} \alpha^2 \bar h= -I \frac{1}{D} \alpha H h_b(t) \\ \frac{d \bar h}{dt} \exp^{\frac{1}{D} \alpha^2 dt} + \frac{1}{D} \alpha^2 \bar h \exp^{\int \frac{1}{D} \alpha^2 dt} = - \frac{1}{D} \alpha H h_b(t) \exp^{\int \frac{1}{D} \alpha^2 dt} \\ \frac{d \bar h}{dt} \exp^{\int \frac{1}{D} \alpha^2 dt} = - \frac{1}{D} \alpha H h_b(t) \exp^{\int \frac{1}{D} \alpha^2 dt} \\ \int_0^t \frac{d \bar h}{dt} \exp^{\int \frac{1}{D} \alpha^2 dt} = \int_0^t - \frac{1}{D} \alpha H h_b(t) \exp^{\int \frac{1}{D} \alpha^2 dt} dt \\ \bar h I = - \frac{1}{D} \alpha H \int_0^t h_b(t) \exp^{\int \frac{1}{D} \alpha^2 dt} dt \\ \bar h = - \frac{1}{D} \alpha H \int_0^t h_b(t) \exp^{\int \frac{1}{D} \alpha^2 d \tau} \exp^{- \int \frac{1}{D} \alpha^2 dt} dt \\ \bar h = - \frac{1}{D} \alpha H \int_0^t h_b(t) \exp^{\int \frac{1}{D} \alpha^2 d \tau - \int \frac{1}{D} \alpha^2 dt} dt \\ \bar h = - \frac{1}{D} \alpha H \int_0^t h_b(t) \exp^{\frac{1}{D} \alpha^2 \int d \tau - \int dt} dt \\ \bar h = - \frac{1}{D} \alpha H \int_0^t h_b(t) \exp^{\frac{1}{D} \alpha^2 ( \tau - t)} dt\\$
This solution is to be used in following equation
$h(x,t) = \frac{2}{\pi} \int_0^{\infty} \bar h \frac{\alpha \cos (\alpha x) + H \sin (\alpha x)}{H^2 + \alpha^2} d\alpha$
which results in following equation
$h(x,t) = \frac{2}{\pi} \int_0^{\infty} [\frac{-1}{D} \alpha H \int_0^t h_b(t) \exp^{\frac{1}{D} \alpha^2 (\tau - t)} dt] \frac{\alpha \cos (\alpha x) + H \sin (\alpha x)}{H^2 + \alpha^2} d \alpha$

I need to simplify this above equation. As a hint I know following integral will be employed during its simplification.
$\int_0^{\infty} \exp^{- {\alpha}^2 {\alpha}^2}. \frac{{\alpha}^2 \cos (\alpha x) + \alpha c \sin (\alpha x)}{c^2 + \alpha^2} d\alpha = \frac{\pi^{0.5}}{2 \alpha} \exp^{\frac{-x^2}{4 {\alpha}^2}} - \frac{c \pi}{2} \exp^{{\alpha}^2 c^2 +cx} .erfc{\alpha c + \frac{x}{2 \alpha}}$

Can somebody help me how can second last equation be simplified using the integral in last equation?

 

[jvicens]

Thank you for sharing your solution to the ODE. It looks like you have made some good progress in solving it using the integrating factor method.

To simplify the final equation, you can use the following steps:

1. First, let's rewrite the integral in the last equation as follows:
$\int_0^{\infty} \exp^{- {\alpha}^2 {\alpha}^2}. \frac{{\alpha}^2 \cos (\alpha x) + \alpha c \sin (\alpha x)}{c^2 + \alpha^2} d\alpha = \int_0^{\infty} \exp^{- {\alpha}^2 {\alpha}^2} \frac{\alpha (\alpha \cos (\alpha x) + c \sin (\alpha x))}{c^2 + \alpha^2} d\alpha$

2. Next, we can use the identity $\exp^{-{\alpha}^2} = \frac{\pi^{0.5}}{2 \alpha} \exp^{\frac{-x^2}{4 {\alpha}^2}}$ to rewrite the integral as follows:
$\int_0^{\infty} \frac{\pi^{0.5}}{2 \alpha} \exp^{\frac{-x^2}{4 {\alpha}^2}} \frac{\alpha (\alpha \cos (\alpha x) + c \sin (\alpha x))}{c^2 + \alpha^2} d\alpha$

3. Now, we can use the substitution $u = \frac{x}{2 \alpha}$ to rewrite the integral as follows:
$\frac{\pi^{0.5}}{2} \int_0^{\infty} \exp^{-u^2} \frac{2u \cos (2u) + c \sin (2u)}{c^2 + (2u)^2} du$

4. Next, we can use the fact that $\sin (2u) = 2 \cos (2u) - 1$ to rewrite the integral as follows:
##
\frac{\pi^{0.5}}{2} \int_0^{\infty} \exp^{-u^2} \frac{2u \cos (2u) + c (2 \cos (2u) - 1)}{