# reduction of order

• Mar 8th 2011, 03:59 PM
poirot
reduction of order
show that y=1/x is a solution to

$\displaystyle x(4x-1)d^2y/dx^2+2(2x-1)dy/dx-4y=0$

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 $\displaystyle y=lnx/x$ but it didn't work when I subbed it in.
• Mar 8th 2011, 07:21 PM
topsquark
Quote:

Originally Posted by poirot
show that y=1/x is a solution to

$\displaystyle x(4x-1)d^2y/dx^2+2(2x-1)dy/dx-4y=0$

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 $\displaystyle y=lnx/x$ 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.
$\displaystyle \displaystyle x(4x-1)d^2y/dx^2+2(2x-1)dy/dx-4y=0$

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

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

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

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

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

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

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

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

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

-Dan
• Mar 9th 2011, 04:15 AM
Jester
Check the $\displaystyle -4x v'$ term - I think it should be $\displaystyle -4v'$ only.
• Mar 9th 2011, 05:54 AM
poirot
Quote:

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.
$\displaystyle \displaystyle x(4x-1)d^2y/dx^2+2(2x-1)dy/dx-4y=0$

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

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

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

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

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

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

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

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

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

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 $\displaystyle e^c(2x^2-x)$. I wasn't not sure if I needed 2 constants since you integrate twice.
• Mar 9th 2011, 06:04 AM
Ackbeet
You can re-label $\displaystyle e^{C}\mapsto C.$ Yes, you do need another constant. I get the following:

$\displaystyle v=C(2x^{2}-x)+B.$
• Mar 9th 2011, 06:19 AM
Chris L T521
Quote:

Originally Posted by poirot
show that y=1/x is a solution to

$\displaystyle x(4x-1)d^2y/dx^2+2(2x-1)dy/dx-4y=0$

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 $\displaystyle y=lnx/x$ 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 $\displaystyle y_1(x)$ is a solution, I state that the second solution is

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

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

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

I hope this helps (in addition to the other responses you have received) !!!
• Mar 9th 2011, 06:25 AM
poirot
Quote:

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

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

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

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

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

so everyone is wrong then lol?
• Mar 9th 2011, 06:31 AM
Chris L T521
Quote:

Originally Posted by poirot
so everyone is wrong then lol?

No... :P

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

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

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 $\displaystyle y_1=\frac{1}{x}$. Note that arbitrary constants would be necessary when you superimpose the two solutions to arrive at the general solution.
• Mar 9th 2011, 01:42 PM
poirot
Quote:

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

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

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

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

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

Can that formula be used for every type DE? Thanks
• Mar 9th 2011, 04:21 PM
Chris L T521
Quote:

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

I hope this helps!