Reduction of order differential eq'n

In summary, the student attempted to solve for y'' using an online differentiator but got extremely confused and instead solved for y'' using the first derivative and got w'=v'' which equals w.
  • #1
rock.freak667
Homework Helper
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Homework Statement



Solve : y''+(2/x)y'+y=0 given that y=sinx/x is a solution

Homework Equations





The Attempt at a Solution


y=vsinx/x is the other solution

I worked out

[tex]y'=\frac{vxcosx-vsinx+v'xsinx}{x^3}[/tex]

to work out y'' got extremely confusing for me,so I used an online differentiator to do it.
It gave y'' as

[tex]\frac{x[2(cosx-sinx)v'+v''xsinx]-v[2xcosx+(x^2-2)sinx]}{x^3}[/tex]

Now when I substitute it back into the equation, I keep getting the terms for v to not cancel and I get one complex differential equation.
 
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  • #2
Your first derivative is incorrect. The denominator should be x2, not x3.
 
  • #3
I will check it over, is there any site where I can input all my differential equations that I need to solve and I can check if I have the correct answer?
 
  • #4
If you have a possible solution, just plug it into your differential equation and see if it works
 
  • #5
Office_Shredder said:
If you have a possible solution, just plug it into your differential equation and see if it works

I can do that but I am afraid that I might make a mistake in differentiating the solution and it won't work with the equation and I'll think I did it wrong.
 
  • #6
Ok here is my attempt (again)

y=vsinx/x

[tex]y'=(\frac{sinx}{x})v'+v(\frac{cosx}{x}-\frac{sinx}{x^2})[/tex]

[tex]y''=v''(\frac{sinx}{x}+\frac{cosx}{x}-\frac{sinx}{x^2})+v'(\frac{cosx}{x}-\frac{sinx}{x^2})+v(-\frac{sinx}{x}-\frac{cosx}{x^2}-\frac{cosx}{x^3}+\frac{2sinx}{x^3}[/tex]

I then put that into the equation and got

[tex]v''(\frac{sinx}{x}+\frac{cosx}{x}-\frac{sinx}{x^2})+v'(\frac{2cosx}{x})=0[/tex]

Then let w=v' => w'=v''

thus getting
[tex]w'(\frac{sinx}{x}+\frac{cosx}{x}-\frac{sinx}{x^2})+w(\frac{2cosx}{x})=0[/tex]

[tex]\times x^2[/tex]

[tex]w'(xsinx+xcosx-sinx)+(2xcosx)w=0[/tex]

Now

[tex]\int \frac{1}{w}dw= \int \frac{-2xcosx}{xsinx+xcosx-sinx}dx[/tex]

and I have no idea how to integrate the term on the right side, I've tried substitutions but it didn't work out well.

EDIT: my differentiating over complicated the integral...I got it out now.
 
Last edited:

Related to Reduction of order differential eq'n

1. What is a reduction of order differential equation?

A reduction of order differential equation is a type of second-order ordinary differential equation that can be transformed into a first-order differential equation by substituting a new variable. This simplifies the equation and makes it easier to solve.

2. How do you solve a reduction of order differential equation?

To solve a reduction of order differential equation, you first need to substitute a new variable u = y'/y. This will transform the second-order equation into a first-order equation. Then, you can use standard methods such as separation of variables or integrating factors to solve for the original variable y.

3. When is it necessary to use reduction of order on a differential equation?

Reduction of order is typically used when the given differential equation is in standard form (y'' + p(x)y' + q(x)y = 0) and the coefficients p(x) and q(x) are not constants. This method is useful for simplifying the equation and finding a general solution.

4. Can a reduction of order differential equation be applied to higher-order equations?

Yes, a reduction of order can also be used for higher-order differential equations. However, it may require multiple substitutions to transform the equation into a solvable first-order equation. It is also important to note that this method may not always work for higher-order equations, and other techniques may be needed.

5. Are there any limitations to using reduction of order on differential equations?

While reduction of order is a useful method for simplifying and solving differential equations, it may not work for all types of equations. It is important to check if the given equation is in standard form and if the coefficients are not constants before attempting to use this method. Additionally, this method may not always provide a complete solution, and other techniques may be needed for a more accurate solution.

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