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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)

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    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

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

    I have solved it with integrating factor method with [tex] I=\exp^{\int \frac{1}{D} \alpha^2 dt}[/tex] as integrating factor and [tex]\frac{K}{S_s} = \frac{1}{D}[/tex]

    I have tried to solve it with following steps

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

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

    [tex]\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}[/tex]

    [tex]\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}[/tex]

    [tex]\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[/tex]

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

    [tex]\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[/tex]

    [tex]\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[/tex]

    [tex]\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[/tex]

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

    Can someone please review whether I have solved it correctly or not?
    Last edited by stapel; 03-03-2018 at 09:55 PM. Reason: Fixing LaTeX.

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