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:
<br /> y=x^{n}<br />
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&#039;&#039;-3xy&#039;+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&#039;=nx^{n-1}
then
y&#039;&#039;=(n^2-n-1)x^{n-2}


so do you substitute into

x^2y&#039;&#039;-3xy&#039;+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:
<br /> y&#039;&#039;=n(n-1)x^{n-2}<br />
So
<br /> x^{2}(n(n-1)x^{n-2}-3xnx^{n-1}+x^{n}=0<br />
which yields
<br /> (n^{2}-4n+1)x^{n}=0<br />
So
<br /> n^{2}-4n+1=0<br />
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,

<br /> y_1=c_1x^{-\sqrt{3}+2}<br />

thanks for your help guys