# Solving a Homogeneous Linear ODE: (x^2-1)y'' + 4xy' + 2y = 6x

• manenbu
In summary, the given problem involves solving a system of equations with two solutions, y1 and y2, for the original equation. The solution to the homogenous system is yh=C1(1/x-1) + C2(1/x+1) and the solution to the original equation is y=C1(x)(1/x-1) + C2(x)(1/x+1). To solve, the variation method is used by setting yp=Ax+B and solving the system of equations C1'(x)(1/x-1) + C2'(x)(1/x+1)=0 and -C1'(x)(1/(x-1)^2) - C2'(x)(1/(x+
manenbu

## Homework Statement

Solve:
$(x^2-1)y'' + 4xy' + 2y = 6x$, given that $y_1=\frac{1}{x-1}$ and $y_2=\frac{1}{x+1}$.

## The Attempt at a Solution

Since both solutions are given, the solution to the homogenous system is:
$$y_h=C_1\frac{1}{x-1} + C_2\frac{1}{x+1}$$
And the solution to the original equation would be:
$$y=C_1(x)\frac{1}{x-1} + C_2(x)\frac{1}{x+1}$$

To solve I use this system of equations:

$$C_1'(x)\frac{1}{x-1} + C_2'(x)\frac{1}{x+1} = 0$$
$$-C_1'(x)\frac{1}{(x-1)^2} - C_2'(x)\frac{1}{(x+1)^2} = 6x$$

Somehow, all of this should end up being: $y=\frac{C_1}{x-1} + \frac{C_2}{x+1} + x$, according to the answers, but I just can't get there. Was there anything wrong in the systems of equations? Or is it me solving for the constants (I didn't write it here)?

It's not clear in your OP, but apparently y1 and y2 are solutions to the homogeneous equation. If that's the case, then you should try yp = Ax + B for your particular solution.

Yes, those are solutions to the homogenous equation. I'm trying to solve it using the variation method, using y = c1(x) y1 + c2(x) y2.

manenbu said:
Yes, those are solutions to the homogenous equation. I'm trying to solve it using the variation method, using y = c1(x) y1 + c2(x) y2.
You could do that, but Mark44's suggestion for "undetermined coefficients" is much simpler.

Ok, so let's assume I still want to solve it using the variation method, just to see that I'm doing it ok (because obviously I went wrong there).

Is my system ok? Any recommendations on a preferred way to solve it?

## What is a homogeneous linear ODE?

A homogeneous linear ODE is a type of ordinary differential equation (ODE) in which all terms involve only the dependent variable and its derivatives. This means that the equation can be written in the form: ay'' + by' + cy = 0, where a, b, and c are constants and y is the dependent variable.

## What does it mean for an ODE to be "solved"?

Solving an ODE means finding a function that satisfies the equation. In the case of a homogeneous linear ODE, this means finding a function y(x) that when plugged into the equation, makes the equation true. This function is called the solution to the ODE.

## Can all homogeneous linear ODEs be solved?

Yes, all homogeneous linear ODEs can be solved. This is because they follow a specific form and have known methods for finding their solutions. However, the solutions may not always be expressible in terms of elementary functions and may require the use of more advanced techniques.

## What is the process for solving a homogeneous linear ODE?

The process for solving a homogeneous linear ODE involves first rewriting the equation in standard form (ay'' + by' + cy = 0), then finding the roots of the characteristic equation (ar^2 + br + c = 0). The solutions to this equation will then be used to form the general solution to the ODE. This general solution can then be used to find the particular solution, which satisfies any initial conditions given.

## Why is the general solution to a homogeneous linear ODE expressed as a linear combination?

The general solution to a homogeneous linear ODE is expressed as a linear combination (sum) of multiple solutions because the equation itself is linear. This means that if y1(x) and y2(x) are both solutions to the equation, then ay1(x) + by2(x) will also be a solution. This allows for an infinite number of possible solutions to the ODE.

• Calculus and Beyond Homework Help
Replies
5
Views
379
• Calculus and Beyond Homework Help
Replies
8
Views
885
• Calculus and Beyond Homework Help
Replies
7
Views
814
• Calculus and Beyond Homework Help
Replies
2
Views
623
• Calculus and Beyond Homework Help
Replies
10
Views
567
• Calculus and Beyond Homework Help
Replies
1
Views
902
• Calculus and Beyond Homework Help
Replies
2
Views
342
• Calculus and Beyond Homework Help
Replies
3
Views
640
• Calculus and Beyond Homework Help
Replies
3
Views
648
• Calculus and Beyond Homework Help
Replies
6
Views
2K