Derivatives of First Solution in Reduction of Order

[beetle2]

Homework Statement

find the first and second derivative of first solution.

Homework Equations

$y(x)=m(x)y_1(x)$

$y'(x)=m'(x)y_1(x)+m(x)y_1'(x)$

The Attempt at a Solution

I have been given $y_1=\frac{1}{x^n}$

Which part is the m(x) and which is $y_1(x)$
I'm not sure how to do the substitution

 

[hunt_mat]

Can you give us a little more of the question.

 

[beetle2]

Sure the original question is Differential equation.

It asks find a pair of fundamentle soultions to the DE


$x^2y''-3xy'+y=0$


So I'm trying to find the two solutions I'm taking $y_1=\frac{1}{x^n}$
as one solution.

 

[hunt_mat]

This is known as Eulers equation, you are correct (partially) in looking for solutions of the form:
$$y=x^{n}$$
Insert this into your ODE and you will obtain a quadratic equation for $$n$$, this will have two solutions which correspond to the two solutions.

 

[beetle2]

So does

$x^2y''-3xy'+y=0$

become


$x^2(n^2-2)x^{n-2} -3xnx^{n-1} + x^n=0$

 

[beetle2]

I seem to be having trouble getting a quadratic for n

is this the right method?

I have

$y=x^n$
than
$y'=nx^{n-1}$
then
$y''=(n^2-n-1)x^{n-2}$


so do you substitute into

$x^2y''-3xy'+y=0$ to give

$x^2((n^2-n-1)x^{n-2})- 3x(nx^{n-1})+x^n=0$

I'm getting

$(n^2-4n+1)x^n$

I can see a characteristic equation there but not sure about the coefficient $x^n$?

 

[hunt_mat]

Almost:
$$y''=n(n-1)x^{n-2}$$
So
$$x^{2}(n(n-1)x^{n-2}-3xnx^{n-1}+x^{n}=0$$
which yields
$$(n^{2}-4n+1)x^{n}=0$$
So
$$n^{2}-4n+1=0$$
Can you calculate n from the above?

 

[beetle2]

Thanks,

I solve the quadratic $$n^{2}-4n+1=0$$
and get:

$$n_1=-\sqrt{3}-2$$ and $$n_2=\sqrt{3}+2$$


So are the two solutions...?

$$y_1=c_1x^{-\sqrt{3}-2}$$ and $$y_2=c_2x^{\sqrt{3}+2}$$

 

[vela]

You flipped a sign in n₁, but otherwise your solutions are correct.

 

[beetle2]

Sorry,

$$y_1=c_1x^{-\sqrt{3}+2}$$

thanks for your help guys