Solving second order linear homogeneous differential equation

In summary, the set of functions from (-1,1)→ℝ which are solutions of the given differential equation can be found using the change of variable x=cos(θ) and solving for y'' + cot(θ)y' + 4y = 0. However, it is important to note that the substitution for (b) should include dy/dθ d/dx(-1/sinθ) in order to be correct.
  • #1
the0
14
0

Homework Statement



Find the set of functions from [itex](-1,1)→ℝ[/itex] which are solutions of:

[itex](x^{2}-1)\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}-4y = 0[/itex]

Homework Equations



The Attempt at a Solution



There is a hint which says to use the change of variable:
[itex]x=cos(θ)[/itex]

doing this I get:

[itex]\frac{dx}{dθ} = -sin(θ)[/itex]

[itex]\Rightarrow[/itex]

[itex]dx = -sin(θ)dθ[/itex]

[itex]\Rightarrow[/itex]

[itex](a): \frac{dy}{dx} = \frac{dy}{dθ}\frac{1}{-sin(θ)}[/itex]

[itex](b): \frac{d^{2}y}{dx^{2}} = \frac{d^{2}y}{dθ^{2}}\frac{1}{sin^{2}(θ)}[/itex]

Can I do this?!?

If so, substituting everything in gives:

[itex]-sin^{2}(θ)\frac{d^{2}y}{dθ^{2}}\frac{1}{sin^{2}(θ)} + cos(θ)\frac{dy}{dθ}\frac{1}{-sin(θ)} - 4y = 0[/itex]

[itex]\Rightarrow[/itex]

[itex]y'' + cot(θ)y' + 4y = 0[/itex]

Now... I am not sure.
Have I made some mistake?
Or should I be able to solve this?
Could someone please point me in the right direction?
Thanks a lot!
 
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  • #2
hi the0! :smile:
the0 said:
[itex](a): \frac{dy}{dx} = \frac{dy}{dθ}\frac{1}{-sin(θ)}[/itex]

[itex](b): \frac{d^{2}y}{dx^{2}} = \frac{d^{2}y}{dθ^{2}}\frac{1}{sin^{2}(θ)}[/itex]

no, your (b) doesn't include dy/dθ d/dx(-1/sinθ)
 

1. What is a second order linear homogeneous differential equation?

A second order linear homogeneous differential equation is a mathematical equation that describes the relationship between a function, its first derivative, and its second derivative. It is "linear" because the function and its derivatives are raised to the first power and there are no higher powers or products. It is "homogeneous" because all terms in the equation involve the function and its derivatives, rather than any constants or other variables.

2. How do you solve a second order linear homogeneous differential equation?

To solve a second order linear homogeneous differential equation, you must first find the general solution by using the characteristic equation to find the roots, or solutions, of the equation. These roots will determine the form of the general solution. Then, you can use initial conditions or boundary conditions to find the specific solution that satisfies the given conditions.

3. What is the characteristic equation of a second order linear homogeneous differential equation?

The characteristic equation of a second order linear homogeneous differential equation is a polynomial equation that is formed by replacing each derivative term with its corresponding variable, and setting the equation equal to zero. The roots of this equation will determine the form of the general solution.

4. What is the difference between a second order linear homogeneous differential equation and a non-homogeneous one?

A second order linear homogeneous differential equation has all terms involving the function and its derivatives, while a non-homogeneous one also has terms that do not involve the function or its derivatives. This can include constants, other variables, or functions. Solving a non-homogeneous equation requires finding both the general solution and a particular solution that satisfies the non-homogeneous terms.

5. What are some real-world applications of solving second order linear homogeneous differential equations?

Second order linear homogeneous differential equations are used in many fields of science and engineering, including physics, chemistry, economics, and engineering. They can be used to model systems such as damped oscillators, electrical circuits, population growth, and chemical reactions. They are also essential in understanding the behavior of mechanical and electrical systems, as well as predicting future trends in economic and environmental systems.

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