# A problem about reduction of the order of a linear ODE

1. Nov 4, 2011

1. The problem statement, all variables and given/known data
Show that if y1 is a solution to the ODE y'''+a2y''+a1y'+a0y=0 then the substitution y=uy1 reduces the order of the equation to a 2nd order linear ODE.

3. The attempt at a solution
well, I calculated first, second and third derivatives of y and plugged them in the equation and after cancellation and some tedious algebraic operations I obtained this new equation:
(3y1''+2a2y1'+ a1y1)u' + (3y1' + a2y1)u'' + u'''y1=0

Now I'm stuck and don't know what I should do next at this point. We still have the third derivative of u in the equation and I don't know how to cancel that.

Last edited: Nov 4, 2011
2. Nov 4, 2011

### LCKurtz

The first term should be y'''.

Presumably you mean y = uy1.
I didn't work it all out but, assuming you have checked your work, you can just let v = u' and you will have a second order equation in v.

3. Nov 4, 2011

Yes.

Yes.

That's a good idea. but do you think the professor wanted us to do that at this step? I mean does he accept this as the solution to his problem?

Last edited: Nov 4, 2011
4. Nov 4, 2011

### LCKurtz

Of course, you need an "= 0" on the right side to make it an equation. But yes, or course, you have reduced the problem of solving a 3rd order equation to one of solving a 2nd order equation.

5. Nov 4, 2011