Author Topic: Inquiry about finding third solution using reduction of order  (Read 1458 times)

Hyunmin Jung

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Inquiry about finding third solution using reduction of order
« on: October 29, 2014, 08:26:50 AM »
If you are given two solutions that satisfies the ODE.

Without wronskian given, using two solutions and by using reduction of order,

is there a way where you are able to find third solutions to the equation?

I could not really get a clear answer from office hour and the problem I am referring is textbook p.228 #28
« Last Edit: October 29, 2014, 11:40:50 AM by Hyunmin Jung »

Victor Ivrii

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Re: Inquiry about finding third solution using reduction of order
« Reply #1 on: October 29, 2014, 11:09:18 AM »
If you are given two solutions that satisfies the ODE.

Without wronskian given, using two solutions and by using reduction of order,

is there a way where you are able to find third solutions to the equation?

I could not really get a clear answer from office hour and the problem I am referring is textbook p.228 #18

There is no problem 18 on p 228 (at least in 10th or 9th editions). What order ODE are you talking about?

Hyunmin Jung

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Re: Inquiry about finding third solution using reduction of order
« Reply #2 on: October 29, 2014, 11:41:04 AM »
sorry, corrected.

Victor Ivrii

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Re: Inquiry about finding third solution using reduction of order
« Reply #3 on: October 29, 2014, 12:14:24 PM »
So, we have a third order linear homogeneous linear ODE and know two linearly independent solutions. And we need to find the third one using the method of the reduction of an order.

If we plug $y=zy_1$ we get for $z$ another third order linear homogeneous linear ODE which, however does not contain $z$ without derivatives, so denoting $z'=u$ we get a second order linear homogeneous linear ODE for $u$. But we know one of its linearly independent solutions, namely $u_1=(y_2/y_1)'$ and then plugging $u=vu_1$ we get another second order linear homogeneous linear ODE for $v$ which does not contain $v$ without derivatives so denoting $v'=w$ we get a first order linear homogeneous linear ODE for $u$.

However there is a little shortcut: since $u_1=1$ due to $y_1=t^2$ and $y_2=t^3$ the third order equation for $z$ contains neither $z$ nor $z'$ but only $z''$ and $z'''$ so denoting $z''=w$ we drop to the first order ODE just in one step instead of two.