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)= x^r

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

Quote:

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

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No, you don't. If y= x^r then y'= rx^r-1 and y''= r(r-1)x^r-2. Putting that into your equation gives
r(r-1)x^rr+ x^r= x^r(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= e^t which changes the d.e. into a "constant coefficients" equation that has the same characteristic equation.

Quote:

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y₁ satisfies the solutions provided by y₂ does not,

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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]

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

Quote:

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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)= x^r

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

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Is your original DE or ?

The two odes your asked about are the same,

the original was x²y''+x y'' - y = 0

I solved it earlier today (Bigsmile) so it doesn't,

My mistyped equation was good for practice anyway he he

Thanks for your help guys!