Can a Substitution Solve a Linear 2nd Order ODE with Non-Constant Coefficients?

In summary, the conversation is discussing different methods for solving a differential equation, specifically using a substitution to change it to a simpler form. One person suggests using u=ln(x) and another suggests using Cauchy Euler DE.
  • #1
kyin01
47
0

Homework Statement


[tex]y''x^{2}[/tex] + 4xy' -y = ln(x)

The Attempt at a Solution


-I considered the quadratic characteristic equation, but it won't work because of the x^2
-I also tried variation of parameters.
so i have v = y' and v'=y'' but i have no idea what i would sub when I get to y.

any ideas?
 
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  • #2
Hi kyin01! :smile:

Try changing x

look for a u(x) such that the equation becomes something simple in dy/du and d2y/du2. :wink:
 
  • #3
kyin01 said:

Homework Statement


[tex]y''x^{2}[/tex] + 4xy' -y = ln(x)

The Attempt at a Solution


-I considered the quadratic characteristic equation, but it won't work because of the x^2
-I also tried variation of parameters.
so i have v = y' and v'=y'' but i have no idea what i would sub when I get to y.

any ideas?

I think the standard way is to first find the solutions of the assoc. homogeneous DE [tex]y''x^{2}[/tex] + 4xy' -y =0 using Cauchy Euler DE by letting [tex]y=x^{m}[/tex].

Then use variation of parameter etc. to find the particular solution.
 
  • #4
Yes, although that can make the calculations pretty complicated. As tiny-time suggested, the substitution u= ln(x) will change any "Euler-type" or "Equipotential" equation to a very simple equation with constant coefficients.

I rather suspect that if kyin01 were to check his recent notes or textbook where this problem is given, he would find a discussion of "Euler-type" or "Equipotential" equations. (The name varies from textbook to textbook.)
 

1. What is a linear 2nd order ODE?

A linear 2nd order ODE (ordinary differential equation) is a mathematical equation that involves a function and its derivatives up to the second order. It can be written in the form of a*x'' + b*x' + c*x = f(x), where a, b, and c are constants and f(x) is a function of x. This type of ODE is "linear" because the function and its derivatives are only multiplied by constants, and there are no higher powers or products of the function and its derivatives.

2. What are some real-world applications of linear 2nd order ODEs?

Linear 2nd order ODEs can be used to model a variety of physical phenomena, such as the motion of a pendulum, the oscillations of a spring, and the growth or decay of populations. They are also commonly used in engineering for analyzing electrical circuits, vibrations in structures, and control systems.

3. How do you solve a linear 2nd order ODE?

The process of solving a linear 2nd order ODE involves finding a particular solution that satisfies the given equation, as well as the general solution which includes all possible solutions. This can be done using various methods, such as separation of variables, substitution, and using specific formulas for certain types of ODEs.

4. What are the initial conditions for a linear 2nd order ODE?

Initial conditions refer to the values of the function and its derivatives at a specific point in the domain. For a linear 2nd order ODE, the initial conditions typically include the value of the function and its first derivative at a given point. These conditions are necessary for finding a unique solution to the ODE.

5. What is the significance of the characteristic equation in a linear 2nd order ODE?

The characteristic equation is an auxiliary equation obtained by substituting the trial solution y = e^(rx) into the linear 2nd order ODE. The roots of this equation, also known as the characteristic roots, help determine the form of the general solution. They can also provide insight into the behavior of the solution, such as whether it will exhibit exponential growth or decay.

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