# Thread: A Question about Reduction of Order

1. ## A Question about Reduction of Order

For an example, we were given the problem:

Consider the DE $(x-1) y''(x) - x y'(x) + y(x) = 0$ where x > 1 and y(2) = 1, y'(2) = 0

a. Verify that $y_1 = e^x$ is a solution of this DE for x > 1.

I've done this, my big question is in part b.

b. Construct a second solution by reduction of order: $y_2 = ve^x$

Now, me and my professor arrived at two different solutions. I couldn't ask him in lecture because it was a 55 minute class and he had a lot to cover, I didn't ask after class because I have a class 15 minutes afterwards, so I'm hoping you guys can help me.

My professor chose to work out the method entirely even though we had already done so generally, I chose to use the generalized result for $y_2$

My professor reduced the equation on the board and arrived at an integral that had no closed form, which I believe was:

$\int \frac{1}{e^{2x}(x-1)} ~dx$

It may be slightly different than that, but the point is it has no closed form. And while this may be correct and arrive at a proper solution, my method came up with something entirely different.

I used the generalized form since we had already proven it in class and my professor already says if it has been shown once it can be used for the rest of the semester without rederivation.

$y_2 = y_1 \int \frac{e^{\int P ~dx}}{y^2_1} ~dx$

And I got that $y_2 = -x$

Which is a linearly independent second solution to the differential equation.

And the only difference (besides the general solution) in our end result (IVP's) is in the first arbitrary constant:

Me: $c_1 = -e^{-2}$

My Professor: $c_1 = e^{-2}$

So I'm wondering if I'm doing it right or not?

2. Originally Posted by Aryth
For an example, we were given the problem:

Consider the DE $(x-1) y''(x) - x y'(x) + y(x) = 0$ where x > 1 and y(2) = 1, y'(2) = 0

a. Verify that $y_1 = e^x$ is a solution of this DE for x > 1.

I've done this, my big question is in part b.

b. Construct a second solution by reduction of order: $y_2 = ve^x$

Now, me and my professor arrived at two different solutions. I couldn't ask him in lecture because it was a 55 minute class and he had a lot to cover, I didn't ask after class because I have a class 15 minutes afterwards, so I'm hoping you guys can help me.

My professor chose to work out the method entirely even though we had already done so generally, I chose to use the generalized result for $y_2$

My professor reduced the equation on the board and arrived at an integral that had no closed form, which I believe was:

$\int \frac{1}{e^{2x}(x-1)} ~dx$

It may be slightly different than that, but the point is it has no closed form. And while this may be correct and arrive at a proper solution, my method came up with something entirely different.

I used the generalized form since we had already proven it in class and my professor already says if it has been shown once it can be used for the rest of the semester without rederivation.

$y_2 = y_1 \int \frac{e^{\int P ~dx}}{y^2_1} ~dx$

And I got that $y_2 = -x$

Which is a linearly independent second solution to the differential equation.

And the only difference (besides the general solution) in our end result (IVP's) is in the first arbitrary constant:

Me: $c_1 = -e^{-2}$

My Professor: $c_1 = e^{-2}$

So I'm wondering if I'm doing it right or not?
You are. If you let $y = v e^x$ then after substitution you get

$\frac{v''}{v'} = - \frac{x-2}{x-1}$ which after one integration gives

$v' = (x-1)e^{-x}$ and after a second $v = - x e^{-x}$ and so $y = ve^x = -x$.

On a side note. After the first integration one obtains

$\ln v' = - x + \ln (x-1)$. However, with a sign change $\ln v' = - x - \ln (x-1)$
you would then have to integrate $v = \int \frac{1}{e^x(x-1)}dx$ which is maybe what they did.

3. That's exactly what they did. I can see the difference now. Thanks for the help.