Solving second order linear homogeneous differential equation

In summary: If you substituted x=cos(\theta) correctly, then you would get:cos(θ)\frac{d^{2}y}{dθ^{2}}-cos(θ)\frac{dy}{dθ}-4sin(θ)y = 0
  • #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)y''+xy'-4y = 0[/itex]

Homework Equations



The Attempt at a Solution



OK, I'm not really sure how to go about solving this equation, I have only previously attempted problems where the functions in [itex]x[/itex] are constant.

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

doing this I get:

(1): [itex](cos^{2}(θ)-1)y''+cos(θ)y'-4y = 0[/itex]

which can be rearranged to give:

(2): [itex]sin^{2}(θ)y''-cos(θ)y'+4y = 0[/itex]
or
(3): [itex](\frac{cos(2θ)-1}{2})y''+cos(θ)y'-4y = 0[/itex]

No idea what to do next!
Any pointers would be great, thank you!
 
Physics news on Phys.org
  • #2
Looks like Heun's Differential Equation to me.
 
  • #3
I've never come across that before :/

How can I use that to solve it?

rearrange to give:

[itex]y''+\frac{x}{(x+1)(x-1)}y'-\frac{4}{(x+1)(x-1)} = 0[/itex] ?

Then I can't see why the change of variable hint has been given.
 
  • #4
the0 said:

Homework Statement



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

[itex](x^{2}-1)y''+xy'-4y = 0[/itex]

Homework Equations



The Attempt at a Solution



OK, I'm not really sure how to go about solving this equation, I have only previously attempted problems where the functions in [itex]x[/itex] are constant.

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

doing this I get:

(1): [itex](cos^{2}(θ)-1)y''+cos(θ)y'-4y = 0[/itex]
This is completely wrong. You have replaced x with [itex]cos(\theta)[/itex] but you have left y' and y'' as derivatives with respect to x.
If [itex]x= cos(\theta)[/itex] then [itex]\theta= cos^{-1}(x)[/itex] so that
[tex]\frac{d\theta}{dx}= \frac{1}{\sqrt{1- x^2}}[/tex]
and
[tex]\frac{dy}{dx}= \frac{dy}{d\theta}\frac{d\theta}{dx}= \frac{1}{\sqrt{1- x^2}}\frac{dy}{d\theta}[/tex]

which can be rearranged to give:

(2): [itex]sin^{2}(θ)y''-cos(θ)y'+4y = 0[/itex]
or
(3): [itex](\frac{cos(2θ)-1}{2})y''+cos(θ)y'-4y = 0[/itex]

No idea what to do next!
Any pointers would be great, thank you!
 
  • #5
the0 said:
I've never come across that before :/

How can I use that to solve it?

rearrange to give:

[itex]y''+\frac{x}{(x+1)(x-1)}y'-\frac{4}{(x+1)(x-1)} = 0[/itex] ?

Then I can't see why the change of variable hint has been given.
This "rearrangement" is wrong.

The correct arrangement is
[tex]y''+\frac{x}{(x+1)(x-1)}y'-\frac{4y}{(x+1)(x-1)} = 0[/tex]

Now, how does that help?

The substitution, [itex]x= cos(\theta)[/itex], if done correctly, simplifies this a great deal.
 
Last edited by a moderator:
  • #6
OK, thanks a lot!

Right, am I correct in substituting the following:

[itex]x = cos(θ)[/itex]

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

[itex]\frac{d^{2}y}{dx^{2}} = (\frac{-cos(θ)}{sin^{3}(θ)})\frac{d^{2}y}{dθ^{2}}[/itex]

?


If so, I get:

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

What next?
Or have I made some stupid error?
 

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

A second order linear homogeneous differential equation is an equation that describes the relationship between a function and its derivatives. It is called "linear" because the function and its derivatives appear in a linear form, and "homogeneous" because all terms in the equation have the same degree.

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

To solve a second order linear homogeneous differential equation, you need to find the general solution, which is a function that satisfies the equation for all possible values of the independent variable. This can be done by using techniques such as separation of variables, undetermined coefficients, or variation of parameters.

3. What are the initial conditions in solving a second order linear homogeneous differential equation?

The initial conditions refer to the known values of the function and its derivatives at a specific point. These conditions are necessary to find the particular solution, which is a specific function that satisfies both the equation and the initial conditions.

4. Can a second order linear homogeneous differential equation have complex solutions?

Yes, a second order linear homogeneous differential equation can have complex solutions. This is because the coefficients in the equation can be complex numbers, and the general solution can involve complex functions such as trigonometric functions and exponential functions.

5. What is the significance of second order linear homogeneous differential equations in science?

Second order linear homogeneous differential equations are widely used in science to model a variety of phenomena, such as oscillations, vibrations, and growth processes. They are also important in engineering, physics, and other fields where the relationships between quantities and their rates of change are studied.

Similar threads

  • Calculus and Beyond Homework Help
Replies
21
Views
736
  • Calculus and Beyond Homework Help
Replies
7
Views
624
  • Calculus and Beyond Homework Help
Replies
10
Views
388
  • Calculus and Beyond Homework Help
Replies
14
Views
160
  • Calculus and Beyond Homework Help
Replies
2
Views
133
  • Calculus and Beyond Homework Help
Replies
33
Views
3K
  • Calculus and Beyond Homework Help
Replies
2
Views
833
  • Calculus and Beyond Homework Help
Replies
1
Views
130
  • Calculus and Beyond Homework Help
Replies
5
Views
996
  • Calculus and Beyond Homework Help
Replies
6
Views
776
Back
Top