Second Order linear homog ODE

Find two linearly independent solutions of

x^{2}y''+x y'' + y = 0

Also find the solutions which satisfy

y_{1}(1) = 1, y'_{1}(1)=0, y_{2}(1) = 0, y'_{2}(1)=1,

So subbing in y(x)= x^{r }You get solutions y_{1}(x) = x and y_{2}(x)=x^{-1} which are linearly independent,

y_{1} satisfies the solutions provided by y_{2 }does not,

The y_{2} 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

Re: Second Order linear homog ODE

Quote:

Originally Posted by

**Linda200** Find two linearly independent solutions of

x^{2}y''+x y'' + y = 0

Also find the solutions which satisfy

y_{1}(1) = 1, y'_{1}(1)=0, y_{2}(1) = 0, y'_{2}(1)=1,

So subbing in y(x)= x^{r}

You get solutions y_{1}(x) = x and y_{2}(x)=x^{-1} which are linearly independent,

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^{r}r+ x^{r}= x^{r}(r^{2}+ 1)= 0.

In order that that be true for all x, you must have r^{2}+ 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:

y_{1} satisfies the solutions provided by y_{2 }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 y_{2} 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]

Re: Second Order linear homog ODE

x^{2}y''+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 x^{i}, 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 y_{2}=sin(ln(x)) which is consistent with the initial conditions

but with

x^{2}y''+x y'' - y = 0

i can't see a way to get x and x^{-1}into a form which satisfies the initial conditions

Re: Second Order linear homog ODE

Quote:

Originally Posted by

**Linda200** Find two linearly independent solutions of

x^{2}y''+x y'' + y = 0

Also find the solutions which satisfy

y_{1}(1) = 1, y'_{1}(1)=0, y_{2}(1) = 0, y'_{2}(1)=1,

So subbing in y(x)= x^{r }You get solutions y_{1}(x) = x and y_{2}(x)=x^{-1} which are linearly independent,

y_{1} satisfies the solutions provided by y_{2 }does not,

The y_{2} 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 $\displaystyle \displaystyle \begin{align*} x^2y'' + x\,y'' + y = 0 \end{align*}$ or $\displaystyle \displaystyle \begin{align*} x^2y'' + x\,y' + y = 0 \end{align*}$?

Re: Second Order linear homog ODE

The two odes your asked about are the same,

the original was x^{2}y''+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!