Reduction of order differential eq'n

[rock.freak667]

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

$$y'=\frac{vxcosx-vsinx+v'xsinx}{x^3}$$

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

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

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.

 

[HallsofIvy]

Your first derivative is incorrect. The denominator should be x², not x³.

 

[rock.freak667]

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?

 

[Office_Shredder]

If you have a possible solution, just plug it into your differential equation and see if it works

 

[rock.freak667]

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.

 

[rock.freak667]

Ok here is my attempt (again)

y=vsinx/x

$$y'=(\frac{sinx}{x})v'+v(\frac{cosx}{x}-\frac{sinx}{x^2})$$

$$y''=v''(\frac{sinx}{x}+\frac{cosx}{x}-\frac{sinx}{x^2})+v'(\frac{cosx}{x}-\frac{sinx}{x^2})+v(-\frac{sinx}{x}-\frac{cosx}{x^2}-\frac{cosx}{x^3}+\frac{2sinx}{x^3}$$

I then put that into the equation and got

$$v''(\frac{sinx}{x}+\frac{cosx}{x}-\frac{sinx}{x^2})+v'(\frac{2cosx}{x})=0$$

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

thus getting
$$w'(\frac{sinx}{x}+\frac{cosx}{x}-\frac{sinx}{x^2})+w(\frac{2cosx}{x})=0$$

$$\times x^2$$

$$w'(xsinx+xcosx-sinx)+(2xcosx)w=0$$

Now

$$\int \frac{1}{w}dw= \int \frac{-2xcosx}{xsinx+xcosx-sinx}dx$$

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.