Bit hazy on reduction of order

1. Mar 31, 2004

Matt Jacques

Although on the whole my Boyce & DiPrima book is fairly good in of itself and compared with others, but of-coarse some areas are lacking.

vt'' - 3v' = 0

I suppose I integrate twice, but then what?

2. Mar 31, 2004

matt grime

vt''-3v'? You might want to check that, v and t are both functions of some paramater?

3. Mar 31, 2004

Matt Jacques

More specifically, v(t)*t'' - 3v(t) = 0

I'd go insane if I kept that (t) in there while differentiating and combining. :P

4. Mar 31, 2004

matt grime

And that factorizes as:

v(t)*(t''-3)=0

not sure that's right either as now the second v is not primed.

5. Mar 31, 2004

Matt Jacques

v''(t)*t - 3v'(t) = 0

Sorry, this is it now.

6. Mar 31, 2004

matt grime

Well, one solution is v(t)=t^3, don't you have to use this to get the other solution?

Edit, of course it isn't. There's the constant solution, duh, and that gives the answer in the next post.

Last edited: Apr 2, 2004
7. Apr 1, 2004

Max0526

The general solution

Hi;
The general solution is:
$$v(t)=C_1t^4+C_2$$
Good luck,
Max.

8. Feb 12, 2008

bowzerman

add the 3v' to the other side getting:
vt''=3v'
divide v to the other side and then integrate with respect to t and v, based on the side of the equation, being sure to keep the integration constants in.

9. Feb 13, 2008

HallsofIvy

Staff Emeritus
No, you cannot treat second derivatives like that. If you really had a single equation in two dependent variables, you will need another equation to get an explicite solutions just as you cannot solve one numeric equation in two variables.

The problem is, after editing, tv"j- 3v'= 0.
Matt Jacques, since you explicitely refer to "reduction of order", I think you mean this: set u= v' so that u'= v" and you have the first order equation tu'- 3u= 0. Now, you can do exactly what bowzerman suggested, since this is now a first order equation.
t du/dt= 3u so du/u= 3t dt. ln|u|= (3/2)t2+ C and then
$$u= C_1 e^{(3/2)t^2[itex]. Since u= dv/dt, we have [tex]dv= C_1 e^{3/2}t^2 dt$$
and only have to integerate again. I'll leave that to you!

Last edited: Feb 15, 2008
10. Feb 13, 2008

HallsofIvy

Staff Emeritus
No, you cannot treat second derivatives like that. If you really had a single equation in two dependent variables, you will need another equation to get an explicite solutions just as you cannot solve one numeric equation in two variables.

The problem is, after editing, tv"- 3v'= 0.
Matt Jacques, since you explicitely refer to "reduction of order", I think you mean this: set u= v' so that u'= v" and you have the first order equation tu'- 3u= 0. Now, you can do exactly what bowzerman suggested, since this is now a first order equation.
t du/dt= 3u so du/u= 3t dt. ln|u|= (3/2)t2+ C and then
$$u= C_1 e^{(3/2)t^2[tex]. Since u= dv/dt, we have [tex]dv= C_1 e^{3/2}t^2 dt$$
and only have to integerate again. I'll leave that to you!

11. Feb 14, 2008

bowzerman

yeah, sorry, I was just wrote that one too quick. Forgot to clarify.

12. Feb 14, 2008

bowzerman

oh, and shouldn't that be:
t du/dt = 3u
1/u du = 3/t ?
ln(u)=3ln(t) +C
u= c*t^3

, then sub, integrate, and then plug in?

Last edited: Feb 14, 2008
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