# Thread: Solving a 2nd order ODE with reduction of order.

1. ## Solving a 2nd order ODE with reduction of order.

I need help on solving this. Our professor requires that we must use this format:

W=u' and W'=u"

Picture of it is down below.

I do not know what to do once I have W'=0 my integrating factor would be weird

http://oi67.tinypic.com/2rdf2w6.jpg

2. ## Re: Solving a 2nd order ODE with reduction of order.

What's weird about it? It's about as straightforward as you are going to get.

-Dan

3. ## Re: Solving a 2nd order ODE with reduction of order.

Because W'=0. Therefore, that means my int. factor is e^(integral of 0) which = e^c.

so then we times the integrating factor by the whole equation....

then d/dx [e^c*W]=0
integrate.

W=c1*e^(-c)

W=u'... we need to find u because y=uy1

therefore...

integral of W...=?

I just feel like I am doing this wrong.

The solution to y_2= xe^2x

4. ## Re: Solving a 2nd order ODE with reduction of order.

Originally Posted by math951
Because W'=0. Therefore, that means my int. factor is e^(integral of 0) which = e^c.

so then we times the integrating factor by the whole equation....

then d/dx [e^c*W]=0
integrate.

W=c1*e^(-c)

W=u'... we need to find u because y=uy1

therefore...

integral of W...=?

I just feel like I am doing this wrong.

The solution to y_2= xe^2x
W' = 0 means that W = c = u' means that u = cx + C

Thus $\displaystyle y = (cx + C)e^{2x}$. Now plug that into your original equation.

-Dan

Thank you.