# Thread: How to solve this linear hom. ODE? (d bar{h})/dt+(K/S_s)a^2 bar{h}=-(K/S_s)aH(h_b)(t)

1. ## How to solve this linear hom. ODE? (d bar{h})/dt+(K/S_s)a^2 bar{h}=-(K/S_s)aH(h_b)(t)

I want to find solution to 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 solved it with integrating factor method with $I=\exp^{\int \frac{1}{D} \alpha^2 dt}$ as integrating factor and $\frac{K}{S_s} = \frac{1}{D}$

I have tried to solve it with following steps

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

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

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

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

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

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

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

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

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

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

Can someone please review whether I have solved it correctly or not?