How to find a solution to this linear ODE?

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Discussion Overview

The discussion revolves around finding a solution to a linear ordinary differential equation (ODE) of the form $$ \frac{d \bar h}{dt} + \frac{K}{S_s} \alpha^2 \bar h = -\frac{K}{S_s} \alpha H h_b(t) $$ using the integrating factor method. Participants explore the steps taken to solve the ODE, including the use of an integrating factor and the implications of constants involved.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a detailed solution using the integrating factor method, specifying the integrating factor as $$ I=\exp^{\int \frac{1}{D} \alpha^2 dt} $$ and relating $$\frac{K}{S_s}$$ to $$\frac{1}{D}$$.
  • Another participant questions whether the constants K, S, and $$\alpha$$ are time-dependent or constant, seeking clarification on their nature.
  • A later reply confirms that K, S, and H are constants, while $$\alpha$$ is derived from a Fourier transform.
  • One participant identifies a potential error in the third equation of the original solution, suggesting that the first term should be $$\frac{d \bar h}{dt} \exp^{\int \frac{1}{D} \alpha^2 dt}$$ instead of the presented form.
  • Another participant expresses that the overall solution appears correct but criticizes the clarity of the notation used, particularly the lack of parentheses, and notes the absence of an initial condition unless $$h = 0$$ at $$t = 0$$.

Areas of Agreement / Disagreement

Participants have not reached a consensus on the correctness of the solution presented. While some express confidence in the approach, others highlight potential errors and ambiguities in the notation and assumptions.

Contextual Notes

There are unresolved questions regarding the dependence of certain parameters on time, as well as the clarity and completeness of the solution steps, particularly concerning initial conditions.

Atr cheema
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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?
 
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Are K, S and ##\alpha## constant or do they depend on t?

Thanks.
 
Gene Naden said:
Are K, S and ##\alpha## constant or do they depend on t?

Thanks.
A, K and H are constand and ##\alpha## comes from Fourier transform.
 
The third equation seems wrong; you wrote:
##\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}
\\##
The first term should be, I believe:
##\frac{d \bar h}{dt} \exp^{\int \frac{1}{D} \alpha^2 dt}##
 
Atr cheema said:
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?
Your writing of these equations leaves a lot to be desired, especially the lack of proper use of parentheses. However, it seems to me your result is correct, except for not including the initial condition (unless h = 0 at t = 0).
 

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