# Thread: Second Order linear homog ODE

1. ## Second Order linear homog ODE

Find two linearly independent solutions of

x2y''+x y'' + y = 0

Also find the solutions which satisfy

y1(1) = 1, y'1(1)=0, y2(1) = 0, y'2(1)=1,

So subbing in y(x)= xr

You get solutions y1(x) = x and y2(x)=x-1 which are linearly independent,

y1 satisfies the solutions provided by y2 does not,

The y2 solutions look like they're for ln(x)

Does anyone know how I would get ln(x) as a solutions from x-1? (I know ln(x) is just the integrate of x-1 but I don't know why you would integrate)

Thanks

Linda

2. ## Re: Second Order linear homog ODE

Originally Posted by Linda200
Find two linearly independent solutions of

x2y''+x y'' + y = 0

Also find the solutions which satisfy

y1(1) = 1, y'1(1)=0, y2(1) = 0, y'2(1)=1,

So subbing in y(x)= xr
You get solutions y1(x) = x and y2(x)=x-1 which are linearly independent,
No, you don't. If y= xr then y'= rxr-1 and y''= r(r-1)xr-2. Putting that into your equation gives
r(r-1)xrr+ xr= xr(r2+ 1)= 0.

In order that that be true for all x, you must have r2+ 1= 0 which does NOT have 1 and -1 as roots.

A different way of solving an equation like this is to make the change of variable x= et which changes the d.e. into a "constant coefficients" equation that has the same characteristic equation.

y1 satisfies the solutions provided by y2 does not,
No, neither satisfies the equation. If y= x then y'= 1, y''= 0 so your equation becomes x+ x= 2x, not 0.

The y2 solutions look like they're for ln(x)

Does anyone know how I would get ln(x) as a solutions from x-1? (I know ln(x) is just the integrate of x-1 but I don't know why you would integrate)

Thanks

Linda[/QUOTE]

3. ## Re: Second Order linear homog ODE

x2y''+x y'' - y = 0

I miswrote the equation,

I am very sorry,

I had a look at that equation for practice though and I got solutions that worked,

since its xi, x-i i used eulers formula to turn it into exp[i ln(x)] which is cos + isin
so the solutions are y=cos(ln(x)) and y2=sin(ln(x)) which is consistent with the initial conditions

but with
x2y''+x y'' - y = 0
i can't see a way to get x and x-1into a form which satisfies the initial conditions

4. ## Re: Second Order linear homog ODE

Originally Posted by Linda200
Find two linearly independent solutions of

x2y''+x y'' + y = 0

Also find the solutions which satisfy

y1(1) = 1, y'1(1)=0, y2(1) = 0, y'2(1)=1,

So subbing in y(x)= xr

You get solutions y1(x) = x and y2(x)=x-1 which are linearly independent,

y1 satisfies the solutions provided by y2 does not,

The y2 solutions look like they're for ln(x)

Does anyone know how I would get ln(x) as a solutions from x-1? (I know ln(x) is just the integrate of x-1 but I don't know why you would integrate)

Thanks

Linda
Is your original DE \displaystyle \begin{align*} x^2y'' + x\,y'' + y = 0 \end{align*} or \displaystyle \begin{align*} x^2y'' + x\,y' + y = 0 \end{align*}?