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\frac{d \bar h}{dt} + \frac{K}{S_s} \alpha^2 \bar h = -\frac{K}{S_s} \alpha H h_b(t)

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I have to solve this ODE which I have done using integrating factor using following steps

taking integrating factor [itex] I=\exp^{\int \frac{1}{D} \alpha^2 dt} [/itex] and [itex]\frac{K}{S_s} = \frac{1}{D}[/itex]

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I \frac{d \bar h}{dt} + I \frac{1}{D} \alpha^2 \bar h = -I \frac{1}{D} \alpha H h_b(t)

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I \frac{d \bar h}{dt} + I \frac{1}{D} \alpha^2 \bar h= -I \frac{1}{D} \alpha H h_b(t)

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\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}

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\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}

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\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

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\bar h I = - \frac{1}{D} \alpha H \int_0^t h_b(t) \exp^{\int \frac{1}{D} \alpha^2 dt} dt

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\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

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\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

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\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

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\bar h = - \frac{1}{D} \alpha H \int_0^t h_b(t) \exp^{\frac{1}{D} \alpha^2 ( \tau - t)} dt\\

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This solution is to be used in following equation

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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

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which results in following equation

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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

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I need to simplify this above equation. As a hint I know following integral will be employed during its simplification.

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\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}}

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