Can Variable Coefficients in an ODE Be Simplified for Easier Solution?

In summary, the conversation revolved around finding a solution for the differential equation y''(x) - A*y'(x) - B*exp(-C*A*x)*y(x) = M*exp(-N*x) and whether there are any other methods besides using series solutions. The possibility of using Laplace transforms and the limitations of this method were also discussed. Additionally, the use of substitution and the resulting solution containing Bessel functions was mentioned. The conversation also touched on the possibility of using the method of undetermined coefficients for finding a particular integral, but the complications involved with the Wronskian were noted. Finally, the idea of using Green's function as a solution was suggested, though it may still involve an integral form
  • #1
iwasthere
2
0
i want to solve the following differential equation:

y''(x) - A*y'(x) - B*exp(-C*A*x)*y(x) = M*exp(-N*x)
A,B,C,M,N are constants.

-is there any solution of the above equation (except series solution)?
-is there any proper substitution that can turn the variable coefficient into constant?
-can i use method of undetermined coefficient to obtain particular integral?
 
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  • #2
I'd try using laplace transforms to see if it has any analytical non-series solution. No idea if you can use undetermined coefficients, but I can't imagine why you'd want to. A transform looks much simpler, especially with the e term in there.
 
  • #3
thanks for the reply...

for the homogeneous equation:
y''(x) - A*y'(x) - B*exp(-C*A*x)*y(x) = 0

if the substitution z =exp(-C*A*x) is used,it becomes something like the following one( if i am not wrong):

z*y''(z) + (1 + 1/C)*y'(z) - B/(C*A)^2*y(z) = 0

which has a solution containing bessel function. But how can i obtain the particular integral (without containing any integral form) for :

y''(x) - A*y'(x) - B*exp(-C*A*x)*y(x) = M*exp(-N*x)

i can use method of variation of parameter for particular integral, but the wronskian becomes complicated (i have to leave the solution in integral form).
therefore i was thinking about method of undetermined coefficient to obtain the particular integral.

i need the general solution without keeping any integral form in the solution.
please help me...
many many thanks in advance.......
 
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  • #4
Angry Citizen said:
I'd try using laplace transforms to see if it has any analytical non-series solution. No idea if you can use undetermined coefficients, but I can't imagine why you'd want to. A transform looks much simpler, especially with the e term in there.

I don't think that will work well. I don't remember Laplace transforms being very good for ODE's with variable coefficients. In particular, the [itex]\exp(-c x)y(x)[/itex] term will give a [itex]Y(s+c)[/itex] term, which means the Laplace-domain equation for the Laplace transform variable [itex]Y(s)[/itex] is a functional equation instead of an algebraic equation.

iwasthere said:
thanks for the reply...

for the homogeneous equation:
y''(x) - A*y'(x) - B*exp(-C*A*x)*y(x) = 0

if the substitution z =exp(-C*A*x) is used,it becomes something like the following one( if i am not wrong):

z*y''(z) + (1 + 1/C)*y'(z) - B/(C*A)^2*y(z) = 0

which has a solution containing bessel function. But how can i obtain the particular integral (without containing any integral form) for :

y''(x) - A*y'(x) - B*exp(-C*A*x)*y(x) = M*exp(-N*x)

i can use method of variation of parameter for particular integral, but the wronskian becomes complicated (i have to leave the solution in integral form).
therefore i was thinking about method of undetermined coefficient to obtain the particular integral.

i need the general solution without keeping any integral form in the solution.
please help me...
many many thanks in advance.......

You may have to leave the solution in integral form. There's not necessarily a nice, closed form solution. Do you know what a Green's function is? Consider a modified form of your equation:

[tex]\mathcal L G(x;\xi) = \delta(x-\xi),[/tex]

where [itex]\delta(x-\xi)[/itex] is the dirac delta function and [itex]\mathcal L[/itex] is the differential operator,

[tex]\mathcal L = \frac{d^2}{dx^2} -A \frac{d}{dx} - B\exp(-ACx)[/tex]

If you solve that equation for [itex]G(x;\xi)[/itex], then to the solution for your equation with the [itex]M\exp(-Nx)[/itex] is just

[tex]y(x) = M\int d\xi G(x;\xi)\exp(-N\xi),[/tex]

where the integral is taken over your range of x. In a one-dimensional problem like yours, this is very similar to the variations of parameters method, I think, but this one generalizes to higher dimensions. Your solution may still be in terms of an integral, and there may be no way around that.
 
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  • #5


I am unable to provide specific solutions to equations without more context or information. However, I can provide some general information about this type of differential equation.

ODEs with variable coefficients are often more challenging to solve than those with constant coefficients. In this case, the variable coefficients are A, B, and C. It is possible that there may be a solution to this equation without using a series solution, but it would depend on the specific values of A, B, C, M, and N. The best approach would be to use numerical methods or a computer program to solve for a particular solution.

In terms of finding a proper substitution to turn the variable coefficients into constants, it is not always possible. However, one approach could be to use a change of variables, such as y(x) = z(x)*exp(-C*A*x), which can sometimes simplify the equation.

The method of undetermined coefficients can be used to obtain a particular solution for this type of equation, but it may not always be successful. It relies on guessing a particular form for the solution and then solving for the coefficients. This method works best when the inhomogeneous term (M*exp(-N*x) in this case) is a simple function.

In conclusion, while it may be possible to find a solution to this ODE without using a series solution, it would require more information or a numerical approach. There may also be ways to simplify the equation by using a change of variables, but it is not always possible. The method of undetermined coefficients can be used, but its success may depend on the form of the inhomogeneous term.
 

Related to Can Variable Coefficients in an ODE Be Simplified for Easier Solution?

1. What is an ODE with variable coefficient?

An ODE (ordinary differential equation) with variable coefficient is a differential equation that contains a coefficient that changes with respect to the independent variable. This means that the coefficient is not a constant, but rather a function of the independent variable.

2. How is an ODE with variable coefficient solved?

The solution to an ODE with variable coefficient involves finding a function that satisfies the equation. This can be done through various methods such as separation of variables, substitution, or using an integrating factor.

3. What are the applications of ODEs with variable coefficient?

ODEs with variable coefficient have many real-world applications, including modeling population growth, chemical reactions, and fluid dynamics. They are also used in engineering to describe systems with changing parameters.

4. What are the challenges in solving ODEs with variable coefficient?

One of the main challenges in solving ODEs with variable coefficient is that they do not have a general solution like ODEs with constant coefficients. This means that different methods may need to be used for different types of variable coefficients. Additionally, the solutions can become more complicated and difficult to interpret.

5. How does the behavior of ODEs with variable coefficient differ from ODEs with constant coefficient?

The behavior of ODEs with variable coefficient can be more complex and unpredictable compared to ODEs with constant coefficient. This is because the changing coefficient can significantly affect the solution, leading to behaviors such as instability or oscillations. In contrast, ODEs with constant coefficient have more regular and predictable solutions.

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