Solving and Elasticity Problem: Differential Equation

In summary: A + \frac{A}{r} = 0Solving for A, we get A = 0. Therefore, the particular solution is:f_{1p} = BSimilarly, for the second equation, we can assume a particular solution of the form:f_{2p} = Cr^2 + Dr^4Substituting this into the equation, we get:(\frac{d^2}{dr^2}+\frac{1}{r}\frac{d}{dr} -\frac{4}{r^2})(\frac{d^2f_{2p}}{dr^2}+\frac{1}{r}\frac{df_{2
  • #1
TheClincher
8
0

Homework Statement



Show that the general solutions to the equations:
[tex](\frac{d^2}{dr^2}+\frac{1}{r})(\frac{d^2f_1}{dr^2} + \frac{1}{r}\frac{df_1}{dr}) = 0[/tex]

[tex](\frac{d^2}{dr^2}+\frac{1}{r}\frac{d}{dr} -\frac{4}{r^2})(\frac{d^2f_2}{dr^2}+\frac{1}{r}\frac{df_2}{dr}-\frac{4f_2}{r^2})=0[/tex]

are:

[tex]f_1=c_1r^2\ln r + c_2r^2 + c_3\ln r + c_4[/tex]
[tex]f_2=c_5r^2+c_6r^4+\frac{c_7}{r^2}+c_8[/tex]

Homework Equations


-

The Attempt at a Solution



I'm not familiar with this kind of differential equation. I've got homogeneous linear higher order differential equations down (thanks to http://tutorial.math.lamar.edu/Classes/DE/HOHomogeneousDE.aspx ), but I'm not sure how to approach this one. A lead into what this is called or some characteristic equation and the general solution would be nice.

Cheers,
Justin
 
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  • #2


Dear Justin,

Thank you for your post. The equations you have provided are known as second-order differential equations with variable coefficients. These types of equations can be solved using the method of variation of parameters. This method involves finding a particular solution and then adding it to the general solution of the corresponding homogeneous equation.

To solve these equations, we first need to find the general solutions of the corresponding homogeneous equations:

(\frac{d^2}{dr^2}+\frac{1}{r})(\frac{d^2f_1}{dr^2} + \frac{1}{r}\frac{df_1}{dr}) = 0

The corresponding characteristic equation for this equation is:

r^2 + 1 = 0

Solving for r, we get two complex roots: r = ±i. Therefore, the general solution to this equation is:

f_1 = c_1r^2 + c_2r^2 + c_3\ln r + c_4

Similarly, for the second equation:

(\frac{d^2}{dr^2}+\frac{1}{r}\frac{d}{dr} -\frac{4}{r^2})(\frac{d^2f_2}{dr^2}+\frac{1}{r}\frac{df_2}{dr}-\frac{4f_2}{r^2})=0

The corresponding characteristic equation is:

r^2 + \frac{1}{r} - \frac{4}{r^2} = 0

Solving for r, we get two real roots: r = 2, -\frac{1}{2}. Therefore, the general solution to this equation is:

f_2 = c_5r^2 + c_6r^4 + \frac{c_7}{r^2} + c_8

Now, to find the particular solutions for the original equations, we use the method of variation of parameters. For the first equation, we can assume a particular solution of the form:

f_{1p} = A\ln r + B

Substituting this into the equation, we get:

(\frac{d^2}{dr^2}+\frac{1}{r})(\frac{d^2f_{1p}}{dr^2} + \frac{1}{r}\frac{df_{1p}}
 
  • #3


I would like to provide some guidance on how to approach this problem. The first step would be to recognize that the given equations are second-order linear differential equations with variable coefficients. This means that the equations involve the second derivative of a function and the coefficients of the equation are not constant, but instead depend on the independent variable r.

To solve these types of differential equations, we can use the method of Frobenius, which involves assuming a solution of the form f(r) = r^m, where m is a constant. Plugging this into the equations and solving for m will give us the characteristic equation, which will help us find the general solution.

In the first equation, we can see that m=0 is a repeated root of the characteristic equation, meaning that the general solution will involve a term of the form c_1r^2\ln r + c_2r^2. The second equation has a more complicated characteristic equation, but we can still use the same method to find the general solution.

It is important to note that the given solutions are just one possible form of the general solution, as there may be other combinations of constants that satisfy the equations. Therefore, it is always a good idea to check if the given solutions actually satisfy the equations by plugging them in and solving for the constants.

I hope this helps in understanding how to approach and solve this type of differential equation. it is important to have a strong understanding of mathematical tools and techniques, such as differential equations, to effectively solve problems and analyze data in our research.
 

What is elasticity and why is it important in solving problems using differential equations?

Elasticity is a measure of how much a quantity changes in response to a change in another quantity. In the context of differential equations, it refers to the relationship between the rate of change of a variable and the variable itself. It is important because it allows us to model and understand complex systems and predict their behavior over time.

What is a differential equation and how does it relate to solving elasticity problems?

A differential equation is an equation that relates a function to its derivatives. In the context of elasticity problems, it is used to describe the relationship between the rate of change of a variable and the variable itself. By solving the differential equation, we can find the function that describes the behavior of the system over time.

What are the steps involved in solving an elasticity problem using differential equations?

The first step is to define the problem and identify the variables involved. Then, we need to formulate the differential equation that represents the relationship between the variables. Next, we use mathematical techniques such as separation of variables or substitution to solve the equation and find the function that describes the system. Finally, we can use this function to make predictions and analyze the behavior of the system.

What are some common applications of solving elasticity problems using differential equations?

Elasticity problems and differential equations have a wide range of applications in various fields such as physics, engineering, economics, and biology. Some common examples include modeling the motion of a spring, calculating the growth rate of a population, predicting the spread of disease, and analyzing the behavior of financial markets.

Are there any limitations to using differential equations to solve elasticity problems?

While differential equations are a powerful tool for modeling and predicting the behavior of complex systems, they have some limitations. For example, they may not be able to accurately capture the behavior of systems with discontinuities or sudden changes. Additionally, the solutions to differential equations may not always have a closed form, making it difficult to find exact solutions. In these cases, numerical methods may be used to approximate the solution.

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