reduction of order

**poirot** · March 8th 2011, 03:59 PM

show that y=1/x is a solution to

[LaTeX ERROR: Convert failed]

Use reduction of order to find the second solution. Find the wronskian and hence show the solutions are linearly independant.

I have done the first part. I think this is a cauchy equation so i know the general solution. I obtained [LaTeX ERROR: Convert failed] but it didn't work when I subbed it in.

**topsquark** · March 8th 2011, 07:21 PM

Originally Posted by poirot

show that y=1/x is a solution to

[LaTeX ERROR: Convert failed]

Use reduction of order to find the second solution. Find the wronskian and hence show the solutions are linearly independant.

I have done the first part. I think this is a cauchy equation so i know the general solution. I obtained [LaTeX ERROR: Convert failed] but it didn't work when I subbed it in.

Well I come up with something that can't be integrated (by any method I know anyway.) I'll post what I have and hope that you (or someone else) can find my mistake.
[LaTeX ERROR: Convert failed]

We know that f(x) = 1/x is a solution, so I will assume a solution y = f(x)v(x) = v/x.

This gives me
$\displaystyle y = \frac{v}{x}$

$\displaystyle y' = \frac{v'}{x} - \frac{v}{x^2}$

$\displaystyle y'' = \frac{v''}{x} - \frac{2v'}{x^2} + \frac{2v}{x^3}$

Plugging this into the orginal equation:
$\displaystyle (4x - 1)v'' + \left ( \frac{-2(4x - 1) + 2(2x - 1)}{x} \right ) v + \left ( \frac{2(4x - 1) - 2(2x - 1) - 4x}{x^2} \right ) v = 0$

Upon simplifying:
$\displaystyle (4x - 1)v'' - 4xv' = 0$

Letting w = v' gives
$\displaystyle (4x - 1)w' - 4xw = 0$

Now, this is separable and leads to
$\displaystyle \int \frac{dw}{w} = \int \frac{4x~dx}{4x - 1}$

which leads to
$\displaystyle w = ke^{x}(4x - 1)^{1/4}$

However I cannot integrate this to find v(x)....

-Dan

**Danny** · March 9th 2011, 04:15 AM

Check the $-4x v'$ term - I think it should be $-4v'$ only.

**poirot** · March 9th 2011, 05:54 AM

Originally Posted by topsquark

Well I come up with something that can't be integrated (by any method I know anyway.) I'll post what I have and hope that you (or someone else) can find my mistake.
[LaTeX ERROR: Convert failed]

We know that f(x) = 1/x is a solution, so I will assume a solution y = f(x)v(x) = v/x.

This gives me

$\displaystyle y = \frac{v}{x}$

$\displaystyle y' = \frac{v'}{x} - \frac{v}{x^2}$

$\displaystyle y'' = \frac{v''}{x} - \frac{2v'}{x^2} + \frac{2v}{x^3}$

Plugging this into the orginal equation:
$\displaystyle (4x - 1)v'' + \left ( \frac{-2(4x - 1) + 2(2x - 1)}{x} \right ) v + \left ( \frac{2(4x - 1) - 2(2x - 1) - 4x}{x^2} \right ) v = 0$

Upon simplifying:
$\displaystyle (4x - 1)v'' - 4xv' = 0$

Letting w = v' gives
$\displaystyle (4x - 1)w' - 4xw = 0$

Now, this is separable and leads to
$\displaystyle \int \frac{dw}{w} = \int \frac{4x~dx}{4x - 1}$

which leads to
$\displaystyle w = ke^{x}(4x - 1)^{1/4}$

However I cannot integrate this to find v(x)....

-Dan

I think I worked it out. Like the person above said, you should have 4v' rather than 4xv'. Can anyone confirm that the answer is [LaTeX ERROR: Convert failed] . I wasn't not sure if I needed 2 constants since you integrate twice.

**Ackbeet** · March 9th 2011, 06:04 AM

You can re-label $e^{C}\mapsto C.$ Yes, you do need another constant. I get the following:

$v=C(2x^{2}-x)+B.$

**Chris L T521** · March 9th 2011, 06:19 AM

Originally Posted by poirot

show that y=1/x is a solution to

[LaTeX ERROR: Convert failed]

Use reduction of order to find the second solution. Find the wronskian and hence show the solutions are linearly independant.

I have done the first part. I think this is a cauchy equation so i know the general solution. I obtained [LaTeX ERROR: Convert failed] but it didn't work when I subbed it in.

I happened to go through reduction of order in my differential equations tutorial: http://www.mathhelpforum.com/math-he...tml#post145671

Given that $y_1(x)$ is a solution, I state that the second solution is

$\displaystyle y_2=\left[\int\left(\frac{e^{-\int P(x)\,dx}}{y_1^2(x)}\right)\,dx\right]y_1(x)$

where in our case, $P(x)=\dfrac{2(2x-1)}{x(4x-1)}=\dfrac{2}{x}-\dfrac{4}{4x-1}$ and $y_1(x)=\dfrac{1}{x}$ .

I leave it for you to verify that $y_2(x)=2x-1$ .

I hope this helps (in addition to the other responses you have received) !!!

**poirot** · March 9th 2011, 06:25 AM

Originally Posted by Chris L T521

I happened to go through reduction of order in my differential equations tutorial: http://www.mathhelpforum.com/math-he...tml#post145671

Given that $y_1(x)$ is a solution, I state that the second solution is

$\displaystyle y_2=\left[\int\left(\frac{e^{-\int P(x)\,dx}}{y_1^2(x)}\right)\,dx\right]y_1(x)$

where in our case, $P(x)=\dfrac{2(2x-1)}{x(4x-1)}=\dfrac{2}{x}-\dfrac{4}{4x-1}$ and $y_1(x)=\dfrac{1}{x}$ .

I leave it for you to verify that $y_2(x)=2x-1$ .

I hope this helps (in addition to the other responses you have received) !!!

I do agree with your answer except you need a constant
so everyone is wrong then lol?

**Chris L T521** · March 9th 2011, 06:31 AM

Originally Posted by poirot

so everyone is wrong then lol?

No... :P

What they found was $v=C(2x^2-x)+B$ . However, for the sake of finding another solution via reduction of order, I would ignore the constants. So $v(x)=C(2x^2-x)+B\sim 2x^2-x$ .

But from topsquark's post, we assumed the solution was of the form $y=f(x)v(x)=\frac{v(x)}{x}$ . So it now follows that your other solution is $\frac{2x^2-x}{x}=2x-1$ .

I hope this clarifies things!

EDIT: The reason why we're not keeping the arbitrary constants is due to the fact that we're not finding the general solution to the ODE in question, but just finding another possible (partial) solution like $y_1=\frac{1}{x}$ . Note that arbitrary constants would be necessary when you superimpose the two solutions to arrive at the general solution.

**poirot** · March 9th 2011, 01:42 PM

Originally Posted by Chris L T521

I happened to go through reduction of order in my differential equations tutorial: http://www.mathhelpforum.com/math-he...tml#post145671

Given that $y_1(x)$ is a solution, I state that the second solution is

$\displaystyle y_2=\left[\int\left(\frac{e^{-\int P(x)\,dx}}{y_1^2(x)}\right)\,dx\right]y_1(x)$

where in our case, $P(x)=\dfrac{2(2x-1)}{x(4x-1)}=\dfrac{2}{x}-\dfrac{4}{4x-1}$ and $y_1(x)=\dfrac{1}{x}$ .

I leave it for you to verify that $y_2(x)=2x-1$ .

I hope this helps (in addition to the other responses you have received) !!!

Can that formula be used for every type DE? Thanks

**Chris L T521** · March 9th 2011, 04:21 PM

Originally Posted by poirot

Can that formula be used for every type DE? Thanks

This formula only works for second order DEs of the type $y^{\prime\prime}+P(x)y^{\prime} + Q(x) y = 0$ . Furthermore, reduction of order only applies to second order ODEs; see: Reduction of order - Wikipedia, the free encyclopedia.

I hope this helps!

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