How to find a solution to this linear ODE?

In summary: Also, I don't know why you are making the problem so complicated with the use of the integrating factor and the substitution ##\tau = D t##. I would have done it this way: Divide both sides by ##K/S_s## and then multiply both sides by ##dt## to put it into standard form:$$-\alpha h + \frac{d \bar h}{dt} = \alpha H h_b(t)$$Now, integrate both sides with respect to t:$$\int \left( -\alpha h + \frac{d \bar h}{dt}\right)\,dt = \alpha H \int h_b(t)\,dt$$$$-\alpha \int h\,
  • #1
Atr cheema
69
0
I want to find solution to following ODE
$$ \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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  • #2
Are K, S and ##\alpha## constant or do they depend on t?

Thanks.
 
  • #3
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.
 
  • #4
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}##
 
  • #5
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).
 

1. What is a linear ODE?

A linear ODE (ordinary differential equation) is a mathematical equation that involves a function and its derivatives. It is called linear because the function and its derivatives only appear in a linear form, meaning they are not multiplied together or raised to a power.

2. How do I solve a linear ODE?

To solve a linear ODE, you need to first determine its order (the highest derivative present) and then apply specific techniques such as substitution, integration, or power series to find the solution. It is also important to have initial or boundary conditions to fully determine the solution.

3. What are initial and boundary conditions?

Initial conditions are given values for the function and its derivatives at a specific point, usually denoted by t = 0. Boundary conditions, on the other hand, are given values for the function and its derivatives at different points in the domain of the equation. These conditions are necessary to fully determine the solution to a linear ODE.

4. Can all linear ODEs be solved analytically?

No, not all linear ODEs can be solved analytically. Some equations may have solutions that cannot be expressed in terms of known functions. In these cases, numerical methods such as Euler's method or the Runge-Kutta method can be used to approximate the solution.

5. Are there any tips for solving linear ODEs more efficiently?

Yes, there are a few tips that can help make the process of solving linear ODEs more efficient. These include identifying symmetry or special properties of the equation, using appropriate substitution or transformation techniques, and practicing with various types of equations to improve problem-solving skills.

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